\displaystyle \textbf{Question 1:}\quad \text{Show that }\left|\begin{array}{ccc} b-c & c-a & a-b\\ c-a & a-b & b-c\\ a-b & b-c & c-a \end{array}\right|\qquad [\text{CBSE 2009}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} b-c & c-a & a-b\\ c-a & a-b & b-c\\ a-b & b-c & c-a \end{array}\right|. \\ \text{Applying }C_{1}\rightarrow C_{1}+C_{2}+C_{3},\ \text{we get}
\displaystyle \Delta=\left|\begin{array}{ccc} 0 & c-a & a-b\\ 0 & a-b & b-c\\ 0 & b-c & c-a \end{array}\right|
\displaystyle \Rightarrow\ \Delta=0\qquad [\because\ C_{1}\ \text{consists of all zeros}]

\displaystyle \textbf{Question 2:}\quad \text{Find the value of the determinant }\Delta= \left|\begin{array}{ccc} 2 & 3 & 4\\ 5 & 6 & 8\\ 6x & 9x & 12x \end{array}\right|\qquad [\text{CBSE 2009}]
\displaystyle \text{Answer:}
\displaystyle \text{Taking }3x\text{ common from }R_{3},\ \text{we get}
\displaystyle \Delta=3x\left|\begin{array}{ccc} 2 & 3 & 4\\ 5 & 6 & 8\\ 2 & 3 & 4 \end{array}\right|
\displaystyle \Rightarrow\ \Delta=3x\times 0=0\qquad [\because\ R_{1}\ \text{and }R_{3}\ \text{are identical}]

\displaystyle \textbf{Question 3:}\quad \text{Without expanding show that} \left|\begin{array}{cccc} b^{2} & c^{2} & bc & b+c\\ c^{2} & a^{2} & ca & c+a\\ a^{2} & b^{2} & ab & a+b \end{array}\right|=0\qquad [\text{CBSE 2001 C}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta= \left|\begin{array}{cccc} b^{2} & c^{2} & bc & b+c\\ c^{2} & a^{2} & ca & c+a\\ a^{2} & b^{2} & ab & a+b \end{array}\right|. \\ \text{Applying }R_{1}\rightarrow R_{1}(a),\ R_{2}\rightarrow R_{2}(b) ,\ R_{3}\rightarrow R_{3}(c),\ \text{we get}
\displaystyle \Delta=\frac{1}{abc}\left|\begin{array}{cccc} ab^{2} & ac^{2} & abc & ab+ac\\ bc^{2} & ba^{2} & abc & bc+ba\\ ca^{2} & cb^{2} & abc & ca+bc \end{array}\right|
\displaystyle [ \because \ R_{1},R_{2},R_{3}\ \text{are multiplied by } a,b,c \text{ respectively, therefore we divide by } abc ]
\displaystyle \Rightarrow\ \Delta=\frac{1}{abc}(abc)^{2} \left|\begin{array}{ccc} bc & 1 & ab+ac\\ ca & 1 & bc+ba\\ ab & 1 & ac+bc \end{array}\right|\qquad [\text{Taking out }abc\text{ common from }C_{1}\text{ and }C_{2}]
\displaystyle \Rightarrow\ \Delta=abc \left|\begin{array}{ccc} bc & 1 & ab+bc+ca\\ ca & 1 & ab+bc+ca\\ ab & 1 & ab+bc+ca \end{array}\right|\qquad [\text{Applying }C_{3}\rightarrow C_{3}+C_{1}]
\displaystyle \Rightarrow\ \Delta=abc(ab+bc+ca) \left|\begin{array}{ccc} bc & 1 & 1\\ ca & 1 & 1\\ ab & 1 & 1 \end{array}\right|\qquad [\text{Taking out }ab+bc+ca\text{ common from }C_{3}]
\displaystyle \Rightarrow\ \Delta=abc(ab+bc+ca)\times 0=0\qquad [\because\ C_{2}\ \text{and }C_{3}\ \text{are identical}]

\displaystyle \textbf{Question 4:}\quad \text{Prove that: } \left|\begin{array}{ccc} -a^{2} & ab & ac\\ ba & -b^{2} & bc\\ ac & bc & -c^{2} \end{array}\right|=4a^{2}b^{2}c^{2}\qquad [\text{CBSE 2011}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta= \left|\begin{array}{ccc} -a^{2} & ab & ac\\ ba & -b^{2} & bc\\ ac & bc & -c^{2} \end{array}\right|.\ \text{Then,}
\displaystyle \Delta=abc \left|\begin{array}{ccc} -a & b & c\\ a & -b & c\\ a & b & -c \end{array}\right|\qquad [\text{Taking }a,b\text{ and }c\text{ common from } R_{1},R_{2}\text{ and }R_{3}\text{ respectively}]
\displaystyle \Rightarrow\ \Delta=a^{2}b^{2}c^{2} \left|\begin{array}{ccc} -1 & 1 & 1\\ 1 & -1 & 1\\ 1 & 1 & -1 \end{array}\right|\qquad [\text{Taking }a,b\text{ and }c\text{ common from } C_{1},C_{2}\text{ and }C_{3}\text{ respectively}]
\displaystyle \Rightarrow\ \Delta=a^{2}b^{2}c^{2} \left|\begin{array}{ccc} -1 & 0 & 0\\ 1 & 0 & 2\\ 1 & 2 & 0 \end{array}\right|\qquad [\text{Applying }C_{2}\rightarrow C_{2}+C_{1}, \ C_{3}\rightarrow C_{3}+C_{1}]
\displaystyle \Rightarrow\ \Delta=a^{2}b^{2}c^{2}\times(-1) \left|\begin{array}{cc} 0 & 2\\ 2 & 0 \end{array}\right|\qquad [\text{Expanding along }R_{1}]
\displaystyle \Rightarrow\ \Delta=a^{2}b^{2}c^{2}(-1)(0-4) =4a^{2}b^{2}c^{2}

\displaystyle \textbf{Question 5: }\text{Show that: } \left|\begin{array}{ccc} x & y & z\\ x^{2} & y^{2} & z^{2}\\ x^{3} & y^{3} & z^{3} \end{array}\right|=xyz(x-y)(y-z)(z-x)\ [\text{CBSE 2000, 2010 C, 2011}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta= \left|\begin{array}{ccc} x & y & z\\ x^{2} & y^{2} & z^{2}\\ x^{3} & y^{3} & z^{3} \end{array}\right|.
\displaystyle \text{Taking }x,y\text{ and }z\text{ common from } C_{1},C_{2}\text{ and }C_{3}\text{ respectively, we get}
\displaystyle \Delta=xyz \left|\begin{array}{ccc} 1 & 1 & 1\\ x & y & z\\ x^{2} & y^{2} & z^{2} \end{array}\right|
\displaystyle \Rightarrow\ \Delta=xyz \left|\begin{array}{ccc} 1 & 0 & 0\\ x & y-x & z-x\\ x^{2} & y^{2}-x^{2} & z^{2}-x^{2} \end{array}\right|
\displaystyle [\text{Applying }C_{2}\rightarrow C_{2}-C_{1} \text{ and }C_{3}\rightarrow C_{3}-C_{1}]
\displaystyle \Rightarrow\ \Delta=xyz(y-x)(z-x) \left|\begin{array}{ccc} 1 & 0 & 0\\ x & 1 & 1\\ x^{2} & y+x & z+x \end{array}\right|
\displaystyle [\text{Taking }(y-x)\text{ and }(z-x)\text{ common from }C_{2}\text{ and }C_{3}\text{ respectively}]
\displaystyle \Rightarrow\ \Delta=xyz(y-x)(z-x)\times 1 \left|\begin{array}{cc} 1 & 1\\ y+x & z+x \end{array}\right|\qquad [\text{Expanding along }R_{1}]
\displaystyle \Rightarrow\ \Delta=xyz(y-x)(z-x)(z+x-y-x)
\displaystyle \Rightarrow\ \Delta=xyz(y-x)(z-x)(z-y)
\displaystyle \Rightarrow\ \Delta=xyz(x-y)(y-z)(z-x)

\displaystyle \textbf{Question 6: }\quad \text{Prove that: }\\ \left|\begin{array}{ccc} \alpha & \beta & \gamma\\ \alpha^{2} & \beta^{2} & \gamma^{2}\\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{array}\right|=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma).
\displaystyle \qquad [\text{CBSE 2007 C, 2008, 2010 C}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} \alpha & \beta & \gamma\\ \alpha^{2} & \beta^{2} & \gamma^{2}\\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{array}\right|.\ \text{Applying }R_{3}\rightarrow R_{1}+R_{3},\ \text{we get}
\displaystyle \Delta=\left|\begin{array}{ccc} \alpha & \beta & \gamma\\ \alpha^{2} & \beta^{2} & \gamma^{2}\\ \alpha+\beta+\gamma & \alpha+\beta+\gamma & \alpha+\beta+\gamma \end{array}\right|
\displaystyle \Rightarrow\ \Delta=(\alpha+\beta+\gamma)\left|\begin{array}{ccc} \alpha & \beta & \gamma\\ \alpha^{2} & \beta^{2} & \gamma^{2}\\ 1 & 1 & 1 \end{array}\right|\qquad [\text{Taking out }(\alpha+\beta+\gamma)\text{ common from }R_{3}]
\displaystyle \Rightarrow\ \Delta=(\alpha+\beta+\gamma)\left|\begin{array}{ccc} \alpha & \beta-\alpha & \gamma-\alpha\\ \alpha^{2} & \beta^{2}-\alpha^{2} & \gamma^{2}-\alpha^{2}\\ 1 & 0 & 0 \end{array}\right|
\displaystyle [\text{Applying }C_{2}\rightarrow C_{2}-C_{1} \text{ and }C_{3}\rightarrow C_{3}-C_{1}]
\displaystyle \Rightarrow\ \Delta=(\alpha+\beta+\gamma)(\beta-\alpha)(\gamma-\alpha)\left|\begin{array}{ccc} \alpha & 1 & 1\\ \alpha^{2} & \beta+\alpha & \gamma+\alpha\\ 1 & 0 & 0 \end{array}\right|
\displaystyle [\text{Taking }(\beta-\alpha)\text{ common from }C_{2} \text{ and }(\gamma-\alpha)\text{ from }C_{3}]
\displaystyle \Rightarrow\ \Delta=(\alpha+\beta+\gamma)(\beta-\alpha)(\gamma-\alpha)\times 1\times \left|\begin{array}{cc} 1 & 1\\ \beta+\alpha & \gamma+\alpha \end{array}\right|\qquad [\text{Expanding along }R_{3}]
\displaystyle \Rightarrow\ \Delta=(\alpha+\beta+\gamma)(\beta-\alpha)(\gamma-\alpha) \bigl(\gamma+\alpha-\beta-\alpha\bigr)
\displaystyle \Rightarrow\ \Delta=(\alpha+\beta+\gamma)(\beta-\alpha)(\gamma-\alpha)(\gamma-\beta) =(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma).

\displaystyle \textbf{Question 7: }\quad \text{Prove that: }\left|\begin{array}{ccc} 1 & a & a^{3}\\ 1 & b & b^{3}\\ 1 & c & c^{3} \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c).
\displaystyle \qquad [\text{CBSE 2011, 2012, 2013}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} 1 & a & a^{3}\\ 1 & b & b^{3}\\ 1 & c & c^{3} \end{array}\right|.
\displaystyle \text{Applying }R_{2}\rightarrow R_{2}-R_{1} \text{ and }R_{3}\rightarrow R_{3}-R_{1},\ \text{we obtain}
\displaystyle \Delta=\left|\begin{array}{ccc} 1 & a & a^{3}\\ 0 & b-a & b^{3}-a^{3}\\ 0 & c-a & c^{3}-a^{3} \end{array}\right|
\displaystyle \Rightarrow\ \Delta=(b-a)(c-a)\left|\begin{array}{ccc} 1 & a & a^{3}\\ 0 & 1 & b^{2}+a^{2}+ab\\ 0 & 1 & c^{2}+a^{2}+ac \end{array}\right|
\displaystyle [\text{Taking out }(b-a)\text{ from }R_{2} \text{ and }(c-a)\text{ from }R_{3}]
\displaystyle \Rightarrow\ \Delta=(b-a)(c-a)\left|\begin{array}{ccc} 1 & a & a^{3}\\ 0 & 0 & (b^{2}-c^{2})+(ab-ac)\\ 0 & 1 & c^{2}+a^{2}+ac \end{array}\right|
\displaystyle [\text{Applying }R_{2}\rightarrow R_{2}-R_{3}]
\displaystyle \Rightarrow\ \Delta=(b-a)(c-a)\left|\begin{array}{ccc} 1 & a & a^{3}\\ 0 & 0 & (b-c)(b+c+a)\\ 0 & 1 & c^{2}+a^{2}+ac \end{array}\right|
\displaystyle \Rightarrow\ \Delta=(b-a)(c-a)(b-c)\left|\begin{array}{ccc} 1 & a & a^{3}\\ 0 & 0 & a+b+c\\ 0 & 1 & c^{2}+a^{2}+ac \end{array}\right|
\displaystyle [\text{Taking out }(b-c)\text{ common from }R_{2}]
\displaystyle \Rightarrow\ \Delta=(b-a)(c-a)(b-c)\times 1\times \left|\begin{array}{cc} 0 & a+b+c\\ 1 & c^{2}+a^{2}+ac \end{array}\right|\qquad [\text{Expanding along }C_{1}]
\displaystyle \Rightarrow\ \Delta=(b-a)(c-a)(b-c)\bigl(0-(a+b+c)\bigr) =(a-b)(b-c)(c-a)(a+b+c).

\displaystyle \textbf{Question 8: } \\ \text{Show that }\left|\begin{array}{ccc} a & b & c\\ a^{2} & b^{2} & c^{2}\\ bc & ca & ab \end{array}\right|=(a-b)(b-c)(c-a)(ab+bc+ca).
\displaystyle \qquad [\text{CBSE 2007, 2011, 2013, 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} a & b & c\\ a^{2} & b^{2} & c^{2}\\ bc & ca & ab \end{array}\right|.
\displaystyle \text{Multiplying }C_{1},C_{2}\text{ and }C_{3}\text{ by } a,b\text{ and }c\text{ respectively, we get}
\displaystyle \Delta=\frac{1}{abc}\left|\begin{array}{ccc} a^{2} & b^{2} & c^{2}\\ a^{3} & b^{3} & c^{3}\\ abc & abc & abc \end{array}\right|
\displaystyle \Rightarrow\ \Delta=\frac{abc}{abc}\left|\begin{array}{ccc} a^{2} & b^{2} & c^{2}\\ a^{3} & b^{3} & c^{3}\\ 1 & 1 & 1 \end{array}\right|\qquad [\text{Taking }abc\text{ common from }R_{3}]
\displaystyle \Rightarrow\ \Delta=-\left|\begin{array}{ccc} a^{2} & b^{2} & c^{2}\\ 1 & 1 & 1\\ a^{3} & b^{3} & c^{3} \end{array}\right|\qquad [\text{Applying }R_{2}\leftrightarrow R_{3}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} 1 & 1 & 1\\ a^{2} & b^{2} & c^{2}\\ a^{3} & b^{3} & c^{3} \end{array}\right|\qquad [\text{Applying }R_{1}\leftrightarrow R_{2}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} 1 & 0 & 0\\ a^{2} & b^{2}-a^{2} & c^{2}-a^{2}\\ a^{3} & b^{3}-a^{3} & c^{3}-a^{3} \end{array}\right|
\displaystyle [\text{Applying }C_{2}\rightarrow C_{2}-C_{1} \text{ and }C_{3}\rightarrow C_{3}-C_{1}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} 1 & 0 & 0\\ a^{2} & (b-a)(b+a) & (c-a)(c+a)\\ a^{3} & (b-a)(b^{2}+ba+a^{2}) & (c-a)(c^{2}+ca+a^{2}) \end{array}\right|
\displaystyle \Rightarrow\ \Delta=(b-a)(c-a)\left|\begin{array}{ccc} 1 & 0 & 0\\ a^{2} & b+a & c+a\\ a^{3} & b^{2}+a^{2}+ab & c^{2}+a^{2}+ac \end{array}\right|
\displaystyle [\text{Taking }(b-a)\text{ and }(c-a)\text{ common from }C_{2}\text{ and }C_{3}\text{ respectively}]
\displaystyle \Rightarrow\ \Delta=(b-a)(c-a)\times 1\times \left|\begin{array}{cc} b+a & c+a\\ b^{2}+a^{2}+ab & c^{2}+a^{2}+ac \end{array}\right|\qquad [\text{Expanding along }R_{1}]
\displaystyle \Rightarrow\ \Delta=(b-a)(c-a)\left|\begin{array}{cc} b+a & c+a\\ (b^{2}-c^{2})+a(b-c) & c^{2}+a^{2}+ac \end{array}\right|
\displaystyle [\text{Applying }C_{1}\rightarrow C_{1}-C_{2}]
\displaystyle \Rightarrow\ \Delta=(b-a)(c-a)\left|\begin{array}{cc} b-c & c+a\\ (b-c)(b+c+a) & c^{2}+a^{2}+ac \end{array}\right|
\displaystyle \Rightarrow\ \Delta=(b-a)(c-a)(b-c)\left|\begin{array}{cc} 1 & c+a\\ b+c+a & c^{2}+a^{2}+ac \end{array}\right|
\displaystyle [\text{Taking }(b-c)\text{ common from }C_{1}]
\displaystyle \Rightarrow\ \Delta=(b-a)(c-a)(b-c) \bigl(c^{2}+a^{2}+ac-(b+c+a)(c+a)\bigr)
\displaystyle \Rightarrow\ \Delta=(b-a)(c-a)(b-c)(-bc-ab-ac)
\displaystyle \Rightarrow\ \Delta=(a-b)(b-c)(c-a)(ab+bc+ca).

\displaystyle \textbf{Question 9: }\quad \text{If }x\neq y\neq z\text{ and } \left|\begin{array}{ccc}x&x^{2}&1+x^{3}\\y&y^{2}&1+y^{3}\\z&z^{2}&1+z^{3}\end{array}\right|=0, \\ \text{ then prove that }xyz=-1.\qquad [\text{CBSE 2011}]
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \Delta=\left|\begin{array}{ccc}x&x^{2}&1+x^{3}\\y&y^{2}&1+y^{3}\\z&z^{2}&1+z^{3}\end{array}\right|
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc}x&x^{2}&1\\y&y^{2}&1\\z&z^{2}&1\end{array}\right| +\left|\begin{array}{ccc}x&x^{2}&x^{3}\\y&y^{2}&y^{3}\\z&z^{2}&z^{3}\end{array}\right| [\text{Since each element of third column is sum of two elements}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc}x&x^{2}&1\\y&y^{2}&1\\z&z^{2}&1\end{array}\right| +xyz\left|\begin{array}{ccc}1&x&x^{2}\\1&y&y^{2}\\1&z&z^{2}\end{array}\right| [\text{Taking }x,y\text{ and }z\text{ common from }C_{1},C_{2},\text{ and }C_{3} \text{ in second determinant}]
\displaystyle \Rightarrow\ \Delta=-\left|\begin{array}{ccc}x&1&x^{2}\\y&1&y^{2}\\z&1&z^{2}\end{array}\right| +xyz\left|\begin{array}{ccc}1&x&x^{2}\\1&y&y^{2}\\1&z&z^{2}\end{array}\right| [\text{Interchanging }C_{2}\text{ and }C_{3}\text{ in first determinant}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc}1&x&x^{2}\\1&y&y^{2}\\1&z&z^{2}\end{array}\right| +xyz\left|\begin{array}{ccc}1&x&x^{2}\\1&y&y^{2}\\1&z&z^{2}\end{array}\right| [\text{Interchanging }C_{1}\text{ and }C_{2}\text{ in first determinant}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc}1&x&x^{2}\\1&y&y^{2}\\1&z&z^{2}\end{array}\right|(1+xyz)
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc}1&x&x^{2}\\0&y-x&y^{2}-x^{2}\\0&z-x&z^{2}-x^{2}\end{array}\right|(1+xyz)
\displaystyle [\text{Applying }R_{2}\rightarrow R_{2}-R_{1}\text{ and }R_{3}\rightarrow R_{3}-R_{1}]
\displaystyle \Rightarrow\ \Delta=(y-x)(z-x)\left|\begin{array}{ccc}1&x&x^{2}\\0&1&y+x\\0&1&z+x\end{array}\right|(1+xyz)
\displaystyle [\text{Taking }(y-x)\text{ and }(z-x)\text{ common }R_{2}\text{ and }R_{3}\text{ resp.}]
\displaystyle \Rightarrow\ \Delta=(y-x)(z-x)\times 1\times \left|\begin{array}{cc}1&y+x\\1&z+x\end{array}\right|(1+xyz) \qquad [\text{Expanding along }C_{1}]
\displaystyle \Rightarrow\ \Delta=(y-x)(z-x)(z+x-y-x)(1+xyz)
\displaystyle \Rightarrow\ \Delta=(y-x)(z-x)(z-y)(1+xyz)=(x-y)(y-z)(z-x)(1+xyz)
\displaystyle \Rightarrow\ \Delta=0
\displaystyle \Rightarrow\ (x-y)(y-z)(z-x)(1+xyz)=0
\displaystyle \Rightarrow\ 1+xyz=0\qquad [\because\ x\neq y\neq z\Rightarrow x-y\neq 0,\ y-z\neq 0\ \text{and }z-x\neq 0]
\displaystyle \Rightarrow\ xyz=-1.

\displaystyle \textbf{Question 10: }\quad \text{For any scalar }p\text{ prove that } \\ \Delta= \left|\begin{array}{ccc}x&x^{2}&1+px^{3}\\y&y^{2}&1+py^{3}\\z&z^{2}&1+pz^{3}\end{array}\right| =(1+pxyz)(x-y)(y-z)(z-x).\qquad [\text{CBSE 2010}]
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \Delta=\left|\begin{array}{ccc}x&x^{2}&1+px^{3}\\y&y^{2}&1+py^{3}\\z&z^{2}&1+pz^{3}\end{array}\right|
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc}x&x^{2}&1\\y&y^{2}&1\\z&z^{2}&1\end{array}\right| +\left|\begin{array}{ccc}x&x^{2}&px^{3}\\y&y^{2}&py^{3}\\z&z^{2}&pz^{3}\end{array}\right| \qquad [\because\ \text{Each element in III column is sum of two elements}]
\displaystyle \Rightarrow\ \Delta=-\left|\begin{array}{ccc}1&x^{2}&x\\1&y^{2}&y\\1&z^{2}&z\end{array}\right| +pxyz\left|\begin{array}{ccc}1&x&x^{2}\\1&y&y^{2}\\1&z&z^{2}\end{array}\right|
\displaystyle [\text{Interchanging }C_{1}\text{ and }C_{3}\text{ in first det.} \text{ Taking }x,y,z\text{ common from }R_{1},R_{2},R_{3} \\ \text{respectively and }p \text{ from }C_{3}\text{ in 2nd det.}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc}1&x&x^{2}\\1&y&y^{2}\\1&z&z^{2}\end{array}\right| +pxyz\left|\begin{array}{ccc}1&x&x^{2}\\1&y&y^{2}\\1&z&z^{2}\end{array}\right| \qquad [\text{Interchanging }C_{2}\text{ and }C_{3}\text{ in first determinant}]
\displaystyle \Rightarrow\ \Delta=(1+pxyz)\left|\begin{array}{ccc}1&x&x^{2}\\1&y&y^{2}\\1&z&z^{2}\end{array}\right|
\displaystyle \Rightarrow\ \Delta=(1+pxyz)\left|\begin{array}{ccc}1&x&x^{2}\\0&y-x&y^{2}-x^{2}\\0&z-x&z^{2}-x^{2}\end{array}\right|
\displaystyle [\text{Applying }R_{2}\rightarrow R_{2}-R_{1}\text{ and }R_{3}\rightarrow R_{3}-R_{1}]
\displaystyle \Rightarrow\ \Delta=(1+pxyz)(y-x)(z-x)\left|\begin{array}{ccc}1&x&x^{2}\\0&1&y+x\\0&1&z+x\end{array}\right|
\displaystyle [\text{Taking }(y-x)\text{ and }(z-x)\text{ common from }R_{2}\text{ and }R_{3} \text{ respectively}]
\displaystyle \Rightarrow\ \Delta=(1+pxyz)(y-x)(z-x)\left|\begin{array}{cc}1&y+x\\1&z+x\end{array}\right| \qquad [\text{Expanding along }C_{1}]
\displaystyle \Rightarrow\ \Delta=(1+pxyz)(y-x)(z-x)(z+x-y-x)=(1+pxyz)(x-y)(y-z)(z-x).

\displaystyle \textbf{Question 11: }\quad \text{Using properties of determinants, show that } \\ \left|\begin{array}{ccc} 1 & a & a^{2}-bc\\ 1 & b & b^{2}-ca\\ 1 & c & c^{2}-ab \end{array}\right|=0.\qquad [\text{CBSE 2002}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} 1 & a & a^{2}-bc\\ 1 & b & b^{2}-ca\\ 1 & c & c^{2}-ab \end{array}\right|.\ \text{Then,}
\displaystyle \Delta=\left|\begin{array}{ccc} 1 & a & a^{2}\\ 1 & b & b^{2}\\ 1 & c & c^{2} \end{array}\right|+\left|\begin{array}{ccc} 1 & a & -bc\\ 1 & b & -ca\\ 1 & c & -ab \end{array}\right|\qquad [\text{Each element of third column is sum of two elements}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} 1 & a & a^{2}\\ 1 & b & b^{2}\\ 1 & c & c^{2} \end{array}\right|-\left|\begin{array}{ccc} 1 & a & bc\\ 1 & b & ca\\ 1 & c & ab \end{array}\right|\qquad [\text{Taking }(-1)\text{ common from }C_{3}\text{ of second determinant}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} 1 & a & a^{2}\\ 1 & b & b^{2}\\ 1 & c & c^{2} \end{array}\right|-\frac{1}{abc}\left|\begin{array}{ccc} a & a^{2} & abc\\ b & b^{2} & abc\\ c & c^{2} & abc \end{array}\right|
\displaystyle [\text{Multiplying }R_{1},R_{2}\text{ and }R_{3}\text{ of second} \text{ determinant by }a,b\text{ and }c\text{ respectively}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} 1 & a & a^{2}\\ 1 & b & b^{2}\\ 1 & c & c^{2} \end{array}\right|-\frac{abc}{abc}\left|\begin{array}{ccc} a & a^{2} & 1\\ b & b^{2} & 1\\ c & c^{2} & 1 \end{array}\right|\qquad [\text{Taking }abc\text{ common from }C_{3}\text{ of second determinant}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} 1 & a & a^{2}\\ 1 & b & b^{2}\\ 1 & c & c^{2} \end{array}\right|+\left|\begin{array}{ccc} a & 1 & a^{2}\\ b & 1 & b^{2}\\ c & 1 & c^{2} \end{array}\right|\qquad [\text{Applying }C_{2}\leftrightarrow C_{3}\text{ in second determinant}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} 1 & a & a^{2}\\ 1 & b & b^{2}\\ 1 & c & c^{2} \end{array}\right|-\left|\begin{array}{ccc} 1 & a & a^{2}\\ 1 & b & b^{2}\\ 1 & c & c^{2} \end{array}\right|\qquad [\text{Applying }C_{1}\leftrightarrow C_{2}\text{ in second determinant}]
\displaystyle \Rightarrow\ \Delta=0.

\displaystyle \textbf{Question 12: }\quad \text{If }f(x)=\left|\begin{array}{ccc} a & -1 & 0\\ ax & a & -1\\ ax^{2} & ax & a \end{array}\right|, \\ \text{using properties of determinants, find the value of }f(2x)-f(x). \qquad [\text{CBSE 2015}]
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle f(x)=\left|\begin{array}{ccc} a & -1 & 0\\ ax & a & -1\\ ax^{2} & ax & a \end{array}\right|
\displaystyle \Rightarrow\ f(x)=\left|\begin{array}{ccc} a & -1 & 0\\ 0 & a+x & -1\\ 0 & 0 & a+x \end{array}\right|
\displaystyle [\text{Applying }R_{3}\rightarrow R_{3}-xR_{2},\ R_{2}\rightarrow R_{2}-xR_{1}]
\displaystyle \Rightarrow\ f(x)=a\left|\begin{array}{cc} a+x & -1\\ 0 & a+x \end{array}\right|\qquad [\text{Expanding along }C_{1}]
\displaystyle \Rightarrow\ f(x)=a(a+x)^{2}
\displaystyle \Rightarrow\ f(2x)=a(a+2x)^{2}\qquad [\text{Replacing }x\text{ by }2x]
\displaystyle \therefore\ f(2x)-f(x)=a(a+2x)^{2}-a(a+x)^{2} =a\{(a+2x+a+x)(a+2x-a-x)\}=ax(2a+3x)

\displaystyle \textbf{Question 13: }\quad \text{Prove that: } \left|\begin{array}{ccc} x+y & x & x\\ 5x+4y & 4x & 2x\\ 10x+8y & 8x & 3x \end{array}\right|=x^{3}.\qquad [\text{CBSE 2002 C, 2009, 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} x+y & x & x\\ 5x+4y & 4x & 2x\\ 10x+8y & 8x & 3x \end{array}\right|.
\displaystyle \text{Since each element in the first column of }\Delta\text{ is the sum of two elements, therefore }\\ \Delta\text{ can be expressed as the sum of two determinants given by}
\displaystyle \Delta=\left|\begin{array}{ccc} x & x & x\\ 5x & 4x & 2x\\ 10x & 8x & 3x \end{array}\right| +\left|\begin{array}{ccc} y & x & x\\ 4y & 4x & 2x\\ 8y & 8x & 3x \end{array}\right|
\displaystyle \Rightarrow\ \Delta=x^{3}\left|\begin{array}{ccc} 1 & 1 & 1\\ 5 & 4 & 2\\ 10 & 8 & 3 \end{array}\right|+yx^{2}\left|\begin{array}{ccc} 1 & 1 & 1\\ 4 & 4 & 2\\ 8 & 8 & 3 \end{array}\right|
\displaystyle \Rightarrow\ \Delta=x^{3}\left|\begin{array}{ccc} 1 & 1 & 1\\ 5 & 4 & 2\\ 10 & 8 & 3 \end{array}\right|+yx^{2}\times 0\qquad [\because\ C_{1}\text{ and }C_{2}\text{ are identical in the second determinant}]
\displaystyle \Rightarrow\ \Delta=x^{3}\left|\begin{array}{ccc} 0 & 0 & 1\\ 3 & 2 & 2\\ 7 & 5 & 3 \end{array}\right|\qquad [\text{Applying }C_{1}\rightarrow C_{1}-C_{3},\ C_{2}\rightarrow C_{2}-C_{3}]
\displaystyle \Rightarrow\ \Delta=x^{3}\times 1\times \left|\begin{array}{cc} 3 & 2\\ 7 & 5 \end{array}\right|\qquad [\text{Expanding along }R_{1}]
\displaystyle \Rightarrow\ \Delta=x^{3}(15-14)=x^{3}.

\displaystyle \textbf{Question 14: }\quad \text{Show that: } \left|\begin{array}{ccc} 1 & 1+p & 1+p+q\\ 2 & 3+2p & 1+3p+2q\\ 3 & 6+3p & 1+6p+3q \end{array}\right|=1.\qquad [\text{CBSE 2009}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} 1 & 1+p & 1+p+q\\ 2 & 3+2p & 1+3p+2q\\ 3 & 6+3p & 1+6p+3q \end{array}\right|.
\displaystyle \text{Applying }C_{2}\rightarrow C_{2}-pC_{1}\text{ and }C_{3}\rightarrow C_{3}-qC_{1},\ \text{we get}
\displaystyle \Delta=\left|\begin{array}{ccc} 1 & 1 & 1+p\\ 2 & 3 & 1+3p\\ 3 & 6 & 1+6p \end{array}\right|
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} 1 & 1 & 1\\ 2 & 3 & 1\\ 3 & 6 & 1 \end{array}\right|\qquad [\text{Applying }C_{3}\rightarrow C_{3}-pC_{2}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} 0 & 0 & 1\\ 1 & 2 & 1\\ 2 & 5 & 1 \end{array}\right|\qquad [\text{Applying }C_{1}\rightarrow C_{1}-C_{3},\ C_{2}\rightarrow C_{2}-C_{3}]
\displaystyle \Rightarrow\ \Delta=1\times\left|\begin{array}{cc} 1 & 2\\ 2 & 5 \end{array}\right|\qquad [\text{Expanding along }R_{1}]
\displaystyle \Rightarrow\ \Delta=5-4=1.

\displaystyle \textbf{Question 15: }\quad \text{Show that: }\left|\begin{array}{ccc} a & a+b & a+b+c\\ 2a & 3a+2b & 4a+3b+2c\\ 3a & 6a+3b & 10a+6b+3c \end{array}\right|=a^{3}.\qquad [\text{CBSE 2012}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} a & a+b & a+b+c\\ 2a & 3a+2b & 4a+3b+2c\\ 3a & 6a+3b & 10a+6b+3c \end{array}\right|.
\displaystyle \text{Since each element of the second column is sum of two elements. Therefore, }\\ \Delta\text{ can be written as the sum of two determinants as follows:}
\displaystyle \Delta=\left|\begin{array}{ccc} a & a & a+b+c\\ 2a & 3a & 4a+3b+2c\\ 3a & 6a & 10a+6b+3c \end{array}\right|+\left|\begin{array}{ccc} a & b & a+b+c\\ 2a & 2b & 4a+3b+2c\\ 3a & 3b & 10a+6b+3c \end{array}\right|
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} a & a & a+b+c\\ 2a & 3a & 4a+3b+2c\\ 3a & 6a & 10a+6b+3c \end{array}\right|+ab\left|\begin{array}{ccc} 1 & 1 & a+b+c\\ 2 & 2 & 4a+3b+2c\\ 3 & 3 & 10a+6b+3c \end{array}\right|
\displaystyle [\text{Taking }a\text{ and }b\text{ common from }C_{1}\text{ and }C_{2}\text{ of second determinant}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} a & a & a+b+c\\ 2a & 3a & 4a+3b+2c\\ 3a & 6a & 10a+6b+3c \end{array}\right|+ab\times 0\qquad [\because\ C_{2}\text{ and }C_{3}\text{ are identical in second determinant}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} a & a & a\\ 2a & 3a & 4a\\ 3a & 6a & 10a \end{array}\right|+\left|\begin{array}{ccc} a & a & b\\ 2a & 3a & 3b\\ 3a & 6a & 6b \end{array}\right|+\left|\begin{array}{ccc} a & a & c\\ 2a & 3a & 2c\\ 3a & 6a & 3c \end{array}\right|
\displaystyle \Rightarrow\ \Delta=a^{3}\left|\begin{array}{ccc} 1 & 1 & 1\\ 2 & 3 & 4\\ 3 & 6 & 10 \end{array}\right|+a^{2}b\left|\begin{array}{ccc} 1 & 1 & 1\\ 2 & 3 & 3\\ 3 & 6 & 6 \end{array}\right|+a^{2}c\left|\begin{array}{ccc} 1 & 1 & 1\\ 2 & 3 & 2\\ 3 & 6 & 3 \end{array}\right|
\displaystyle \Rightarrow\ \Delta=a^{3}\left|\begin{array}{ccc} 1 & 1 & 1\\ 2 & 3 & 4\\ 3 & 6 & 10 \end{array}\right|+a^{2}b\times 0+a^{2}c\times 0
\displaystyle [\because\ C_{2}\text{ and }C_{3}\text{ are identical in second det. and }C_{1}\text{ and }C_{3}\text{ are identical in third det.}]
\displaystyle \Rightarrow\ \Delta=a^{3}\left|\begin{array}{ccc} 1 & 0 & 0\\ 2 & 1 & 2\\ 3 & 3 & 7 \end{array}\right|\qquad [\text{Applying }C_{2}\rightarrow C_{2}-C_{1},\ C_{3}\rightarrow C_{3}-C_{1}]
\displaystyle \Rightarrow\ \Delta=a^{3}\times 1\times \left|\begin{array}{cc} 1 & 2\\ 3 & 7 \end{array}\right|\qquad [\text{Expanding along }R_{1}]
\displaystyle \Rightarrow\ \Delta=a^{3}(7-6)=a^{3}.

\displaystyle \textbf{Question 16: }\quad \text{Show that: }\left|\begin{array}{ccc} b+c & c+a & a+b\\ q+r & r+p & p+q\\ y+z & z+x & x+y \end{array}\right|=2\left|\begin{array}{ccc} a & b & c\\ p & q & r\\ x & y & z \end{array}\right|.
\displaystyle [\text{CBSE 2004, 2006, 2010, 2012, 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} b+c & c+a & a+b\\ q+r & r+p & p+q\\ y+z & z+x & x+y \end{array}\right|.\ \text{Applying }C_{1}\rightarrow C_{1}+C_{2}+C_{3},\ \text{we get}
\displaystyle \Delta=\left|\begin{array}{ccc} 2(a+b+c) & c+a & a+b\\ 2(p+q+r) & r+p & p+q\\ 2(x+y+z) & z+x & x+y \end{array}\right|
\displaystyle \Rightarrow\ \Delta=2\left|\begin{array}{ccc} a+b+c & c+a & a+b\\ p+q+r & r+p & p+q\\ x+y+z & z+x & x+y \end{array}\right|\qquad [\text{Taking }2\text{ common from }C_{1}]
\displaystyle \Rightarrow\ \Delta=2\left|\begin{array}{ccc} a+b+c & -b & -c\\ p+q+r & -q & -r\\ x+y+z & -y & -z \end{array}\right|
\displaystyle [\text{Applying }C_{2}\rightarrow C_{2}-C_{1},\ C_{3}\rightarrow C_{3}-C_{1}]
\displaystyle \Rightarrow\ \Delta=2\left|\begin{array}{ccc} a & -b & -c\\ p & -q & -r\\ x & -y & -z \end{array}\right|\qquad [\text{Applying }C_{1}\rightarrow C_{1}+C_{2}+C_{3}]
\displaystyle \Rightarrow\ \Delta=2(-1)^{2}\left|\begin{array}{ccc} a & b & c\\ p & q & r\\ x & y & z \end{array}\right|=2\left|\begin{array}{ccc} a & b & c\\ p & q & r\\ x & y & z \end{array}\right|\qquad [\text{Taking }(-1)\text{ common from both }C_{2}\text{ and }C_{3}]

\displaystyle \textbf{Question 17: }\quad \text{Prove that } \\ \left|\begin{array}{ccc} 1+a & 1 & 1\\ 1 & 1+b & 1\\ 1 & 1 & 1+c \end{array}\right|=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=abc+bc+ca+ab.
\displaystyle \qquad [\text{CBSE 2004, 2009, 2012, 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} 1+a & 1 & 1\\ 1 & 1+b & 1\\ 1 & 1 & 1+c \end{array}\right|.
\displaystyle \text{Taking }a,b\text{ and }c\text{ common from }C_{1},C_{2}\text{ and }C_{3}\text{ respectively, we obtain}
\displaystyle \Delta=abc\left|\begin{array}{ccc} \frac{1}{a}+1 & \frac{1}{b} & \frac{1}{c}\\ \frac{1}{a} & 1+\frac{1}{b} & \frac{1}{c}\\ \frac{1}{a} & \frac{1}{b} & 1+\frac{1}{c} \end{array}\right|
\displaystyle \Rightarrow\ \Delta=abc\left|\begin{array}{ccc} 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & \frac{1}{b} & \frac{1}{c}\\ 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{b} & \frac{1}{c}\\ 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & \frac{1}{b} & 1+\frac{1}{c} \end{array}\right|\qquad [\text{Applying }C_{1}\rightarrow C_{1}+C_{2}+C_{3}]
\displaystyle \Rightarrow\ \Delta=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left|\begin{array}{ccc} 1 & \frac{1}{b} & \frac{1}{c}\\ 1 & 1+\frac{1}{b} & \frac{1}{c}\\ 1 & \frac{1}{b} & 1+\frac{1}{c} \end{array}\right|\qquad \left[\text{Taking }\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\text{ common from }C_{1}\right]
\displaystyle \Rightarrow\ \Delta=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left|\begin{array}{ccc} 1 & \frac{1}{b} & \frac{1}{c}\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right|\qquad [\text{Applying }R_{2}\rightarrow R_{2}-R_{1},\ R_{3}\rightarrow R_{3}-R_{1}]
\displaystyle \Rightarrow\ \Delta=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\times 1\times \left|\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right|\qquad [\text{Expanding along }C_{1}]
\displaystyle \Rightarrow\ \Delta=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=abc+bc+ca+ab.

\displaystyle \textbf{Question 18: }\quad \text{Prove that: } \\  \\ \left|\begin{array}{ccc} (b+c)^{2} & a^{2} & a^{2}\\ b^{2} & (c+a)^{2} & b^{2}\\ c^{2} & c^{2} & (a+b)^{2} \end{array}\right|=2abc(a+b+c)^{3}.\qquad [\text{CBSE 2006 C, 2010}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} (b+c)^{2} & a^{2} & a^{2}\\ b^{2} & (c+a)^{2} & b^{2}\\ c^{2} & c^{2} & (a+b)^{2} \end{array}\right|.
\displaystyle \text{Applying }C_{1}\rightarrow C_{1}-C_{3}\ \text{and }C_{2}\rightarrow C_{2}-C_{3},
\displaystyle \text{we get}
\displaystyle \Delta=\left|\begin{array}{ccc} (b+c)^{2}-a^{2} & 0 & a^{2}\\ 0 & (c+a)^{2}-b^{2} & b^{2}\\ c^{2}-(a+b)^{2} & c^{2}-(a+b)^{2} & (a+b)^{2} \end{array}\right|
\displaystyle \Rightarrow\ \Delta=(a+b+c)^{2}\left|\begin{array}{ccc} b+c-a & 0 & a^{2}\\ 0 & c+a-b & b^{2}\\ c-a-b & c-a-b & (a+b)^{2} \end{array}\right|
\displaystyle [\text{Taking }(a+b+c)\text{ common from }C_{1}\ \&\ C_{2}]
\displaystyle \Rightarrow\ \Delta=(a+b+c)^{2}\left|\begin{array}{ccc} b+c-a & 0 & a^{2}\\ 0 & c+a-b & b^{2}\\ -2b & -2a & 2ab \end{array}\right|
\displaystyle [\text{Applying }R_{3}\rightarrow R_{3}-(R_{1}+R_{2})]
\displaystyle \Rightarrow\ \Delta=\frac{(a+b+c)^{2}}{ab}\left|\begin{array}{ccc} ab+ac-a^{2} & 0 & a^{2}\\ 0 & bc+ba-b^{2} & b^{2}\\ -2ab & -2ab & 2ab \end{array}\right|
\displaystyle [\text{Applying }C_{1}\rightarrow C_{1}(a),\ C_{2}\rightarrow C_{2}(b)]
\displaystyle \Rightarrow\ \Delta=\frac{(a+b+c)^{2}}{ab}\left|\begin{array}{ccc} ab+ac & a^{2} & a^{2}\\ b^{2} & bc+ba & b^{2}\\ 0 & 0 & 2ab \end{array}\right|
\displaystyle [\text{Applying }C_{1}\rightarrow C_{1}+C_{3},\ C_{2}\rightarrow C_{2}+C_{3}]
\displaystyle \Rightarrow\ \Delta=\frac{(a+b+c)^{2}}{ab}\times ab\times 2ab\left|\begin{array}{ccc} b+c & a & a\\ b & c+a & b\\ 0 & 0 & 1 \end{array}\right|
\displaystyle [\text{Taking }a,b\text{ and }2ab\text{ common from }R_{1},R_{2}\text{ and }R_{3}\text{ respectively}]
\displaystyle \Rightarrow\ \Delta=2ab(a+b+c)^{2}\times 1\times\left|\begin{array}{cc} b+c & a\\ b & c+a \end{array}\right|\qquad [\text{Expanding along }R_{3}]
\displaystyle \Rightarrow\ \Delta=2ab(a+b+c)^{2}\bigl((b+c)(c+a)-ab\bigr)=2abc(a+b+c)^{3}

\displaystyle \textbf{Question 19: } \\ \\ \text{Show that: }\left|\begin{array}{ccc} (b+c)^{2} & ba & ca\\ ab & (c+a)^{2} & cb\\ ac & bc & (a+b)^{2} \end{array}\right|=2abc(a+b+c)^{3}.\qquad [\text{CBSE 2006, 10}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} (b+c)^{2} & ba & ca\\ ab & (c+a)^{2} & cb\\ ac & bc & (a+b)^{2} \end{array}\right|.
\displaystyle \text{Multiplying }R_{1},R_{2}\text{ and }R_{3}\text{ by }a,b\text{ and }c\text{ respectively, we get}
\displaystyle \Delta=\frac{1}{abc}\left|\begin{array}{ccc} a(b+c)^{2} & ba^{2} & ca^{2}\\ ab^{2} & b(c+a)^{2} & cb^{2}\\ ac^{2} & bc^{2} & c(a+b)^{2} \end{array}\right|
\displaystyle \Rightarrow\ \Delta=\frac{1}{abc}\left|\begin{array}{ccc} a(b+c)^{2} & a^{2}b & a^{2}c\\ ab^{2} & b(c+a)^{2} & b^{2}c\\ ac^{2} & bc^{2} & c(a+b)^{2} \end{array}\right|
\displaystyle \Rightarrow\ \Delta=\frac{1}{abc}\cdot abc\left|\begin{array}{ccc} (b+c)^{2} & a^{2} & a^{2}\\ b^{2} & (c+a)^{2} & b^{2}\\ c^{2} & c^{2} & (a+b)^{2} \end{array}\right|
\displaystyle [\text{Taking }a,b\text{ and }c\text{ common from }C_{1},C_{2}\text{ and }C_{3}\text{ respectively}]
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} (b+c)^{2} & a^{2} & a^{2}\\ b^{2} & (c+a)^{2} & b^{2}\\ c^{2} & c^{2} & (a+b)^{2} \end{array}\right|
\displaystyle \Rightarrow\ \Delta=2abc(a+b+c)^{3}

\displaystyle \textbf{Question 20: } \\ \text{Show that }\left|\begin{array}{ccc} 1+a^{2}-b^{2} & 2ab & -2b\\ 2ab & 1-a^{2}+b^{2} & 2a\\ 2b & -2a & 1-a^{2}-b^{2} \end{array}\right|=(1+a^{2}+b^{2})^{3}.\qquad [\text{CBSE 2009, 2010 C}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} 1+a^{2}-b^{2} & 2ab & -2b\\ 2ab & 1-a^{2}+b^{2} & 2a\\ 2b & -2a & 1-a^{2}-b^{2} \end{array}\right|.
\displaystyle \text{We shall try to introduce zeros at as many places as possible keeping in mind} \\ \text{that we have to introduce the factor }1+a^{2}+b^{2}.
\displaystyle \text{Applying }C_{1}\rightarrow C_{1}-bC_{3}\text{ and }C_{2}\rightarrow C_{2}+aC_{3},\ \text{we get}
\displaystyle \Delta=\left|\begin{array}{ccc} 1+a^{2}+b^{2} & 0 & -2b\\ 0 & 1+a^{2}+b^{2} & 2a\\ b(1+a^{2}+b^{2}) & -a(1+a^{2}+b^{2}) & 1-a^{2}-b^{2} \end{array}\right|
\displaystyle \Rightarrow\ \Delta=(1+a^{2}+b^{2})^{2}\left|\begin{array}{ccc} 1 & 0 & -2b\\ 0 & 1 & 2a\\ b & -a & 1-a^{2}-b^{2} \end{array}\right|
\displaystyle [\text{Taking }(1+a^{2}+b^{2})\text{ common from both }C_{1}\text{ and }C_{2}]
\displaystyle \Rightarrow\ \Delta=(1+a^{2}+b^{2})^{2}\left|\begin{array}{ccc} 1 & 0 & -2b\\ 0 & 1 & 2a\\ 0 & 0 & 1+a^{2}+b^{2} \end{array}\right|
\displaystyle [\text{Applying }R_{3}\rightarrow R_{3}-bR_{1}+aR_{2}]
\displaystyle \Rightarrow\ \Delta=(1+a^{2}+b^{2})^{2}\times 1\times \left|\begin{array}{cc} 1 & 2a\\ 0 & 1+a^{2}+b^{2} \end{array}\right|\qquad [\text{Expanding along }C_{1}]
\displaystyle \Rightarrow\ \Delta=(1+a^{2}+b^{2})^{3}.

\displaystyle \textbf{Question 21: }\quad \text{If }a,b,c\text{ are positive and unequal, show that the value of the determinant } \\ \left|\begin{array}{ccc} a & b & c\\ b & c & a\\ c & a & b \end{array}\right|\text{ is always negative.}\qquad [\text{CBSE 2010}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} a & b & c\\ b & c & a\\ c & a & b \end{array}\right|.\ \text{Applying }C_{1}\rightarrow C_{1}+C_{2}+C_{3},\ \text{we get}
\displaystyle \Delta=\left|\begin{array}{ccc} a+b+c & b & c\\ b+c+a & c & a\\ c+a+b & a & b \end{array}\right|
\displaystyle \Rightarrow\ \Delta=(a+b+c)\left|\begin{array}{ccc} 1 & b & c\\ 1 & c & a\\ 1 & a & b \end{array}\right|\qquad [\text{Taking }(a+b+c)\text{ common from }C_{1}]
\displaystyle \Rightarrow\ \Delta=(a+b+c)\left|\begin{array}{ccc} 1 & b & c\\ 0 & c-b & a-c\\ 0 & a-b & b-c \end{array}\right|
\displaystyle [\text{Applying }R_{2}\rightarrow R_{2}-R_{1}\ \text{and }R_{3}\rightarrow R_{3}-R_{1}]
\displaystyle \Rightarrow\ \Delta=(a+b+c)\left|\begin{array}{cc} c-b & a-c\\ a-b & b-c \end{array}\right|\qquad [\text{Expanding along }C_{1}]
\displaystyle \Rightarrow\ \Delta=(a+b+c)\bigl(-(c-b)^{2}-(a-b)(a-c)\bigr) =(a+b+c)(-a^{2}-b^{2}-c^{2}+ab+bc+ca)
\displaystyle \Rightarrow\ \Delta=-\frac{1}{2}(a+b+c)(2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca)
\displaystyle \Rightarrow\ \Delta=-\frac{1}{2}(a+b+c)\bigl\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\bigr\}
\displaystyle \Rightarrow\ \Delta<0\qquad [\because\ a+b+c>0,\ (a-b)^{2}>0,\ (b-c)^{2}>0,\ (c-a)^{2}>0]

\displaystyle \textbf{Question 22: }\quad \text{Show that: }\left|\begin{array}{ccc} 3a & -a+b & -a+c\\ -b+a & 3b & -b+c\\ -c+a & -c+b & 3c \end{array}\right|=3(a+b+c)(ab+bc+ca).\qquad [\text{CBSE 2006 C, 2013}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} 3a & -a+b & -a+c\\ -b+a & 3b & -b+c\\ -c+a & -c+b & 3c \end{array}\right|.\ \text{Applying }C_{1}\rightarrow C_{1}+C_{2}+C_{3},\ \text{we get}
\displaystyle \Delta=\left|\begin{array}{ccc} a+b+c & -a+b & -a+c\\ a+b+c & 3b & -b+c\\ a+b+c & -c+b & 3c \end{array}\right|
\displaystyle \Rightarrow\ \Delta=(a+b+c)\left|\begin{array}{ccc} 1 & -a+b & -a+c\\ 1 & 3b & -b+c\\ 1 & -c+b & 3c \end{array}\right|\qquad [\text{Applying }C_{1}\rightarrow C_{1}+C_{2}+C_{3}]
\displaystyle \Rightarrow\ \Delta=(a+b+c)\left|\begin{array}{ccc} 1 & -a+b & -a+c\\ 0 & 2b+a & -b+a\\ 0 & -c+a & 2c+a \end{array}\right|\qquad [\text{Applying }R_{2}\rightarrow R_{2}-R_{1},\ R_{3}\rightarrow R_{3}-R_{1}]
\displaystyle \Rightarrow\ \Delta=(a+b+c)\left|\begin{array}{cc} 2b+a & -b+a\\ -c+a & 2c+a \end{array}\right|\qquad [\text{Expanding along }C_{1}]
\displaystyle \Rightarrow\ \Delta=(a+b+c)\bigl\{(2b+a)(2c+a)-(-b+a)(-c+a)\bigr\}
\displaystyle \Rightarrow\ \Delta=(a+b+c)\bigl\{(4bc+2ab+2ca+a^{2})-(bc-ab-ac+a^{2})\bigr\}
\displaystyle \Rightarrow\ \Delta=(a+b+c)(3bc+3ab+3ca)
\displaystyle \Rightarrow\ \Delta=3(a+b+c)(ab+bc+ca).

\displaystyle \textbf{Question 23: }\quad \text{Solve: }\left|\begin{array}{ccc} a+x & a-x & a-x\\ a-x & a+x & a-x\\ a-x & a-x & a+x \end{array}\right|=0.\qquad [\text{CBSE 2004, 2005, 2011}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} a+x & a-x & a-x\\ a-x & a+x & a-x\\ a-x & a-x & a+x \end{array}\right|.\ \text{Applying }C_{1}\rightarrow C_{1}+C_{2}+C_{3},\ \text{we get}
\displaystyle \Delta=\left|\begin{array}{ccc} 3a-x & a-x & a-x\\ 3a-x & a+x & a-x\\ 3a-x & a-x & a+x \end{array}\right|
\displaystyle \Rightarrow\ \Delta=(3a-x)\left|\begin{array}{ccc} 1 & a-x & a-x\\ 1 & a+x & a-x\\ 1 & a-x & a+x \end{array}\right|\qquad [\text{Taking }(3a-x)\text{ common from }C_{1}]
\displaystyle \Rightarrow\ \Delta=(3a-x)\left|\begin{array}{ccc} 1 & a-x & a-x\\ 0 & 2x & 0\\ 0 & 0 & 2x \end{array}\right|

\displaystyle [\text{Applying }R_{2}\rightarrow R_{2}-R_{1},\ R_{3}\rightarrow R_{3}-R_{1}]
\displaystyle \Rightarrow\ \Delta=(3a-x)\times 1\times \left|\begin{array}{cc} 2x & 0\\ 0 & 2x \end{array}\right|\qquad [\text{Expanding along }C_{1}]
\displaystyle \Rightarrow\ \Delta=(3a-x)4x^{2}
\displaystyle \therefore\ \Delta=0\Rightarrow (3a-x)4x^{2}=0\Rightarrow x=0,\ 3a.

\displaystyle \textbf{Question 24: }\quad \text{Solve: }\left|\begin{array}{ccc} x-2 & 2x-3 & 3x-4\\ x-4 & 2x-9 & 3x-16\\ x-8 & 2x-27 & 3x-64 \end{array}\right|=0.\qquad [\text{CBSE 2011}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\left|\begin{array}{ccc} x-2 & 2x-3 & 3x-4\\ x-4 & 2x-9 & 3x-16\\ x-8 & 2x-27 & 3x-64 \end{array}\right|.
\displaystyle \text{Applying }C_{2}\rightarrow C_{2}-2C_{1}\text{ and }C_{3}\rightarrow C_{3}-3C_{1},\ \text{we get}
\displaystyle \Delta=\left|\begin{array}{ccc} x-2 & 1 & 2\\ x-4 & -1 & -4\\ x-8 & -11 & -40 \end{array}\right|
\displaystyle \Rightarrow\ \Delta=\left|\begin{array}{ccc} x-2 & 1 & 2\\ -2 & -2 & -6\\ -6 & -12 & -42 \end{array}\right|\qquad [\text{Applying }R_{2}\rightarrow R_{2}-R_{1},\ R_{3}\rightarrow R_{3}-R_{1}]
\displaystyle \Rightarrow\ \Delta=(-2)(-6)\left|\begin{array}{ccc} x-2 & 1 & 2\\ 1 & 1 & 3\\ 1 & 2 & 7 \end{array}\right|
\displaystyle [\text{Taking }(-2)\text{ and }(-6)\text{ common from }R_{2}\text{ and }R_{3}\text{ respectively}]
\displaystyle \Rightarrow\ \Delta=12\left|\begin{array}{ccc} x-2 & 1 & 2\\ 1 & 1 & 3\\ 0 & 1 & 4 \end{array}\right|\qquad [\text{Applying }R_{3}\rightarrow R_{3}-R_{2}]
\displaystyle \Rightarrow\ \Delta=12\left\{(x-2)\left|\begin{array}{cc} 1 & 3\\ 1 & 4 \end{array}\right|-1\left|\begin{array}{cc} 1 & 2\\ 1 & 4 \end{array}\right|\right\}\qquad [\text{Expanding along }C_{1}]
\displaystyle \Rightarrow\ \Delta=12\{(x-2)(4-3)-(4-2)\}=12(x-4)
\displaystyle \therefore\ \Delta=0\Rightarrow 12(x-4)=0\Rightarrow x=4.

\displaystyle \textbf{Question 25: }\quad \text{Prove that: }\left|\begin{array}{ccc} bc-a^{2} & ca-b^{2} & ab-c^{2}\\ ca-b^{2} & ab-c^{2} & bc-a^{2}\\ ab-c^{2} & bc-a^{2} & ca-b^{2} \end{array}\right|\ \text{is divisible by }a+b+c\ \text{and find the quotient.}\qquad [\text{CBSE 2016}]
\displaystyle \text{Answer:}
\displaystyle \left|\begin{array}{ccc} bc-a^{2} & ca-b^{2} & ab-c^{2}\\ ca-b^{2} & ab-c^{2} & bc-a^{2}\\ ab-c^{2} & bc-a^{2} & ca-b^{2} \end{array}\right|=\left|\begin{array}{ccc} a & b & c\\ b & c & a\\ c & a & b \end{array}\right|^{2}
\displaystyle \left|\begin{array}{ccc} a & b & c\\ b & c & a\\ c & a & b \end{array}\right|=-(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)
\displaystyle \left|\begin{array}{ccc} bc-a^{2} & ca-b^{2} & ab-c^{2}\\ ca-b^{2} & ab-c^{2} & bc-a^{2}\\ ab-c^{2} & bc-a^{2} & ca-b^{2} \end{array}\right|=(a+b+c)^{2}(a^{2}+b^{2}+c^{2}-ab-bc-ca)^{2}
\displaystyle \text{Clearly, RHS is divisible by }(a+b+c)\ \text{and the quotient is } (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)^{2}.
\displaystyle \text{Hence, LHS is divisible by }a+b+c\ \text{and }(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)^{2} \text{ is the quotient.}

\displaystyle \textbf{Question 26: }\quad \text{Find the equation of the line joining }A(1,3) \text{ and }B(0,0)\text{ using} \\ \text{determinants and find }k\text{ if }D(k,0)\text{ is a point such that area of } \triangle ABD\text{ is }3\ \text{sq. units.}
\displaystyle [\text{CBSE 2013}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y)\text{ be any point on line }AB.\ \text{Then,}
\displaystyle \text{Area of }\triangle ABP=0
\displaystyle \Rightarrow\ \frac{1}{2}\left|\begin{array}{ccc} 1 & 3 & 1\\ 0 & 0 & 1\\ x & y & 1 \end{array}\right|=0
\displaystyle \Rightarrow\ \frac{1}{2}\{1(0-y)-3(0-x)+1(0-0)\}=0
\displaystyle \Rightarrow\ 3x-y=0,\ \text{which is the required equation of }AB.
\displaystyle \text{Now,}\quad \text{Area of }\triangle ABD=3\ \text{sq. units}
\displaystyle \Rightarrow\ \frac{1}{2}\left|\begin{array}{ccc} 1 & 3 & 1\\ 0 & 0 & 1\\ k & 0 & 1 \end{array}\right|=\pm 3

\displaystyle \textbf{Question 27. }\text{The value of the determinant }\begin{vmatrix}2&7&1\\1&1&1\\10&8&1\end{vmatrix}\text{ is} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }47\qquad \text{(b) }-79\qquad \text{(c) }49\qquad \text{(d) }-51
\displaystyle \text{Answer:}
\displaystyle (a)\ \text{Let }\Delta=\begin{vmatrix}2&7&1\\1&1&1\\10&8&1\end{vmatrix}
\displaystyle \text{Expanding the determinant along }R_1,\text{ we get}
\displaystyle \Delta=2(1-8)-7(1-10)+1(8-10)
\displaystyle =2(-7)-7(-9)+1(-2)
\displaystyle =-14+63-2=47
\\

\displaystyle \textbf{Question 28. }\text{If }\begin{vmatrix}\alpha&3&4\\1&2&1\\1&4&1\end{vmatrix}=0,\text{ then the value of }\alpha\text{ is} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }1\qquad \text{(b) }2\qquad \text{(c) }3\qquad \text{(d) }4
\displaystyle \text{Answer:}
\displaystyle (d)\ \text{Given, }\begin{vmatrix}\alpha&3&4\\1&2&1\\1&4&1\end{vmatrix}=0
\displaystyle \text{On expanding along }R_1,\text{ we get}
\displaystyle \alpha(2-4)-3(1-1)+4(4-2)=0
\displaystyle \Rightarrow \alpha(-2)-3(0)+4(2)=0
\displaystyle \Rightarrow -2\alpha+8=0
\displaystyle \Rightarrow -2\alpha=-8
\displaystyle \Rightarrow \alpha=4
\\

\displaystyle \textbf{Question 29. }\text{Let }A\text{ be a skew-symmetric matrix of order }3.\text{ If } |A|=x,\text{ then }(2023)^x \\ \text{ is equal to} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }2023\qquad \text{(b) }\frac{1}{2023}\qquad \text{(c) }(2023)^2\qquad \text{(d) }1
\displaystyle \text{Answer:}
\displaystyle (d)\ \text{We know that, the determinant of an odd order skew-symmetric matrix is zero.}
\displaystyle \therefore |A|=0=x
\displaystyle \therefore (2023)^x=(2023)^0=1
\\

\displaystyle \textbf{Question 30. }\text{If }\begin{bmatrix}1&2&1\\2&3&1\\3&a&1\end{bmatrix}\text{ is non-singular matrix and }a\in A, \text{ then the set }A\text{ is} \\  \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }R\qquad \text{(b) }\{0\}\qquad \text{(c) }\{4\}\qquad \text{(d) }R-\{4\}
\displaystyle \text{Answer:}
\displaystyle (d)\ \text{Given, }\begin{bmatrix}1&2&1\\2&3&1\\3&a&1\end{bmatrix}\text{ is non-singular matrix.}
\displaystyle \therefore \begin{vmatrix}1&2&1\\2&3&1\\3&a&1\end{vmatrix}\ne0
\displaystyle \Rightarrow 1(3-a)-2(2-3)+1(2a-9)\ne0
\displaystyle \Rightarrow 3-a+2+2a-9\ne0
\displaystyle \Rightarrow -4+a\ne0
\displaystyle \Rightarrow a\ne4
\\

\displaystyle \textbf{Question 31. }\text{Let }A\text{ be the area of a triangle with vertices }(x_1,y_1),(x_2,y_2)
\displaystyle \text{ and }(x_3,y_3). \text{ Which of the following is correct?} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\pm A\qquad \text{(b) }\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\pm2A
\displaystyle \text{(c) }\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\pm\frac{A}{2}\qquad \text{(d) }\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}^{2}=A^2
\displaystyle \text{Answer:}
\displaystyle (b)\ \text{The area of a triangle whose vertices are }(x_1,y_1),(x_2,y_2)\text{ and }(x_3,y_3)\text{ is}
\displaystyle \Delta=\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}
\displaystyle \text{According to the question,}
\displaystyle \frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\pm A
\displaystyle \Rightarrow \begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\pm2A
\\

\displaystyle \textbf{Question 32. }\text{If }\begin{vmatrix}2&3&2\\x&x&x\\4&9&1\end{vmatrix}+3=0,\text{ then the value of }x\text{ is} \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }3\qquad \text{(b) }0\qquad \text{(c) }-1\qquad \text{(d) }1
\displaystyle \text{Answer:}
\displaystyle (c)\ \text{We have, }\begin{vmatrix}2&3&2\\x&x&x\\4&9&1\end{vmatrix}+3=0
\displaystyle \Rightarrow 2(x-9x)-3(x-4x)+2(9x-4x)+3=0
\displaystyle \Rightarrow -16x+9x+10x+3=0
\displaystyle \Rightarrow 3x+3=0
\displaystyle \Rightarrow 3x=-3
\displaystyle \Rightarrow x=-1
\\

\displaystyle \textbf{Question 33. }\text{Let }A=\begin{bmatrix}200&50\\10&2\end{bmatrix}\text{ and }B=\begin{bmatrix}50&40\\2&3\end{bmatrix},\text{ then }|AB|\text{ is equal to} \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }460\qquad \text{(b) }2000\qquad \text{(c) }3000\qquad \text{(d) }-7000
\displaystyle \text{Answer:}
\displaystyle (d)\ \text{We have, }A=\begin{bmatrix}200&50\\10&2\end{bmatrix}
\displaystyle \Rightarrow |A|=400-500=-100
\displaystyle \text{and }B=\begin{bmatrix}50&40\\2&3\end{bmatrix}
\displaystyle \Rightarrow |B|=150-80=70
\displaystyle \therefore |AB|=|A|\cdot|B|
\displaystyle =-100\times70=-7000
\\

\displaystyle \textbf{Question 34. }\text{Find }|AB|,\text{ if }A=\begin{bmatrix}0&-1\\0&2\end{bmatrix}\text{ and }B=\begin{bmatrix}3&5\\0&0\end{bmatrix}. \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}0&-1\\0&2\end{bmatrix}\text{ and }B=\begin{bmatrix}3&5\\0&0\end{bmatrix}
\displaystyle \text{Now, }AB=\begin{bmatrix}0&-1\\0&2\end{bmatrix}\begin{bmatrix}3&5\\0&0\end{bmatrix}
\displaystyle =\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \therefore |AB|=0
\\

\displaystyle \textbf{Question 35. }\text{Find the maximum value of }\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta\end{vmatrix}. \hspace{1.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta\end{vmatrix}
\displaystyle \text{On expanding along first row i.e. }R_1,\text{ we get}
\displaystyle \Delta=1\begin{vmatrix}1+\sin\theta&1\\1&1+\cos\theta\end{vmatrix}-1\begin{vmatrix}1&1\\1&1+\cos\theta\end{vmatrix}+1\begin{vmatrix}1&1+\sin\theta\\1&1\end{vmatrix}
\displaystyle =[(1+\sin\theta)(1+\cos\theta)-1]-[1+\cos\theta-1]+[1-1-\sin\theta]
\displaystyle =1+\cos\theta+\sin\theta+\sin\theta\cos\theta-1-\cos\theta-\sin\theta
\displaystyle =\sin\theta\cos\theta
\displaystyle =\frac{1}{2}(2\sin\theta\cos\theta)=\frac{1}{2}\sin2\theta
\displaystyle \text{We know that, maximum value of }\sin2\theta\text{ is }1.
\displaystyle \therefore \Delta_{\max}=\frac{1}{2}\times1=\frac{1}{2}
\\

\displaystyle \textbf{Question 36. }\text{If }\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}=8,\text{ write the value of }x. \hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}=8
\displaystyle \text{On expanding along }R_1,\text{ we get}
\displaystyle x(-x^2-1)-\sin\theta(-x\sin\theta-\cos\theta)+\cos\theta(-\sin\theta+x\cos\theta)=8
\displaystyle \Rightarrow -x^3-x+x\sin^2\theta+\sin\theta\cos\theta-\sin\theta\cos\theta+x\cos^2\theta=8
\displaystyle \Rightarrow -x^3-x+x(\sin^2\theta+\cos^2\theta)=8
\displaystyle \Rightarrow -x^3-x+x=8\qquad [\because \sin^2\theta+\cos^2\theta=1]
\displaystyle \Rightarrow -x^3=8\Rightarrow x^3+8=0\Rightarrow x^3+2^3=0
\displaystyle \Rightarrow (x+2)(x^2+4-2x)=0\Rightarrow x=-2
\displaystyle \left[\because x^2-2x+4=0,\text{ gives imaginary values}\right]
\\

\displaystyle \textbf{Question 37. }\text{If }A=\begin{bmatrix}5&6&-3\\-4&3&2\\-4&-7&3\end{bmatrix},\text{ then write the cofactor of the element }a_{21}
\displaystyle \text{of its 2nd row.} \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}5&6&-3\\-4&3&2\\-4&-7&3\end{bmatrix}
\displaystyle \text{Now, cofactor of }a_{21}=(-1)^{2+1}\begin{vmatrix}6&-3\\-7&3\end{vmatrix}
\displaystyle =-\,(18-21)=3
\\

\displaystyle \textbf{Question 38. }\text{If }A=\begin{bmatrix}1&2\\3&-1\end{bmatrix}\text{ and }B=\begin{bmatrix}1&3\\-1&1\end{bmatrix},\text{ write the value of }|AB|. \hspace{0.2cm}\text{[CBSE 2015C]}
\displaystyle \text{Answer:}
\displaystyle \text{Clearly, }|A|=\begin{vmatrix}1&2\\3&-1\end{vmatrix}=-1-6=-7
\displaystyle \text{and }|B|=\begin{vmatrix}1&3\\-1&1\end{vmatrix}=1+3=4
\displaystyle \therefore |AB|=|A|\cdot|B|=(-7)(4)=-28
\\

\displaystyle \textbf{Question 39. }\text{In the interval }\pi/2<x<\pi,\text{ find the value of }x\text{ for which the matrix}
\displaystyle \begin{bmatrix}2\sin x&3\\1&2\sin x\end{bmatrix}\text{ is singular.} \hspace{2.2cm}\text{[CBSE 2015C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\begin{bmatrix}2\sin x&3\\1&2\sin x\end{bmatrix}
\displaystyle \therefore A\text{ is a singular matrix.}
\displaystyle \therefore |A|=0\Rightarrow \begin{vmatrix}2\sin x&3\\1&2\sin x\end{vmatrix}=0
\displaystyle \Rightarrow 4\sin^2x-3=0\Rightarrow \sin^2x=\frac{3}{4}
\displaystyle \Rightarrow \sin x=\frac{\sqrt{3}}{2}
\displaystyle \therefore x=\frac{2\pi}{3}
\\

\displaystyle \textbf{Question 40. }\text{If }\begin{vmatrix}2x&5\\8&x\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix},\text{ then write the value of }x. \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{vmatrix}2x&5\\8&x\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix}
\displaystyle \Rightarrow 2x^2-40=18-(-14)
\displaystyle \Rightarrow 2x^2-40=32\Rightarrow 2x^2=72\Rightarrow x^2=36
\displaystyle \therefore x=\pm6
\\

\displaystyle \textbf{Question 41. }\text{If }\begin{vmatrix}3x&7\\-2&4\end{vmatrix}=\begin{vmatrix}8&7\\6&4\end{vmatrix}, \text{then find the value of }x.\hspace{0.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \Rightarrow (3x)(4)-(7)(-2)=(8)(4)-(7)(6)
\displaystyle \Rightarrow 12x+14=32-42
\displaystyle \Rightarrow 12x+14=-10
\displaystyle \Rightarrow 12x=-24
\displaystyle \Rightarrow x=-2
\\

\displaystyle \textbf{Question 42. }\text{Write the value of the determinant }\begin{vmatrix}p&p+1\\p-1&p\end{vmatrix}. \hspace{2.2cm}\text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\begin{vmatrix}p&p+1\\p-1&p\end{vmatrix}
\displaystyle \text{On expanding, we get}
\displaystyle \Delta=p^2-(p-1)(p+1)
\displaystyle \Rightarrow \Delta=p^2-(p^2-1)\qquad [\because a^2-b^2=(a+b)(a-b)]
\displaystyle \Rightarrow \Delta=p^2-p^2+1
\displaystyle \therefore \Delta=1
\\

\displaystyle \textbf{Question 43. }\text{If }\begin{vmatrix}2x&x+3\\2(x+1)&x+1\end{vmatrix}=\begin{vmatrix}1&5\\3&3\end{vmatrix},\text{ then find the value of }x. \hspace{1.2cm}\text{[CBSE 2013C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{vmatrix}2x&x+3\\2(x+1)&x+1\end{vmatrix}=\begin{vmatrix}1&5\\3&3\end{vmatrix}
\displaystyle \Rightarrow 2x(x+1)-(x+3)(2x+2)=3-15
\displaystyle \Rightarrow 2x^2+2x-(2x^2+8x+6)=-12
\displaystyle \Rightarrow -6x-6=-12
\displaystyle \Rightarrow 6x=6
\displaystyle \therefore x=1
\\

\displaystyle \textbf{Question 44. }\text{If }\begin{vmatrix}x+1&x-1\\x-3&x+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix}, \text{then write the value of }x.\hspace{1.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \Rightarrow (x+1)(x+2)-(x-1)(x-3)=(4)(3)-(-1)(1)
\displaystyle \Rightarrow x^{2}+3x+2-(x^{2}-4x+3)=12+1
\displaystyle \Rightarrow x^{2}+3x+2-x^{2}+4x-3=13
\displaystyle \Rightarrow 7x-1=13
\displaystyle \Rightarrow 7x=14
\displaystyle \Rightarrow x=2
\\

\displaystyle \textbf{Question 45. }\text{If }A_{ij}\text{ is the cofactor of the element }a_{ij}\text{ of the determinant }\begin{vmatrix}2&-3&5\\6&0&4\\1&5&-7\end{vmatrix},
\displaystyle \text{then write the value of }a_{32}A_{32}. \hspace{2.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\begin{vmatrix}2&-3&5\\6&0&4\\1&5&-7\end{vmatrix}
\displaystyle \text{Here, }a_{32}=5
\displaystyle \text{Given, }A_{ij}\text{ is the cofactor of the element }a_{ij}\text{ of }\Delta.
\displaystyle \therefore A_{32}=(-1)^{3+2}\begin{vmatrix}2&5\\6&4\end{vmatrix}=-(8-30)=22
\displaystyle \Rightarrow a_{32}\cdot A_{32}=5\times22=110
\\

\displaystyle \textbf{Question 46. }\text{If }\Delta=\begin{vmatrix}5&3&8\\2&0&1\\1&2&3\end{vmatrix},\text{ write the cofactor of element }a_{32}. \hspace{2.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Cofactor of element }a_{32}
\displaystyle =(-1)^{3+2}\begin{vmatrix}5&8\\2&1\end{vmatrix}=(-1)(5-16)=11
\\

\displaystyle \textbf{Question 47. }\text{If }A=\begin{bmatrix}1&2&3\\2&0&1\\5&3&8\end{bmatrix},\text{ write the minor of element }a_{22}. \hspace{2.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Minor of element }a_{22}=\begin{vmatrix}1&3\\5&8\end{vmatrix}
\displaystyle =8-15=-7
\\

\displaystyle \textbf{Question 48. }\text{If }\Delta=\begin{vmatrix}5&3&8\\2&0&1\\1&2&3\end{vmatrix},\text{ then write the minor of the element }a_{23}. \hspace{1.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Minor of the element }a_{23}=\begin{vmatrix}5&3\\1&2\end{vmatrix}
\displaystyle =10-3=7
\\

\displaystyle \textbf{Question 49. }\text{For what value of }x,\ A=\begin{bmatrix}2(x+1)&2x\\x&x-2\end{bmatrix}\text{ is a singular matrix?} \hspace{0.2cm}\text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{We know that, a matrix }A\text{ is said to be singular, if }|A|=0.
\displaystyle \therefore \begin{vmatrix}2x+2&2x\\x&x-2\end{vmatrix}=0
\displaystyle \Rightarrow (2x+2)(x-2)-2x^2=0
\displaystyle \Rightarrow 2x^2-2x-4-2x^2=0
\displaystyle \Rightarrow -2x-4=0
\displaystyle \therefore x=-2
\\

\displaystyle \textbf{Question 50. }\text{For what value of }x,\text{ the matrix }\begin{bmatrix}2x+4&4\\x+5&3\end{bmatrix}\text{ is a singular matrix?} \hspace{0.2cm}\text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\begin{bmatrix}2x+4&4\\x+5&3\end{bmatrix}
\displaystyle \text{If matrix }A\text{ is singular, then}
\displaystyle |A|=0
\displaystyle \Rightarrow \begin{vmatrix}2x+4&4\\x+5&3\end{vmatrix}=0
\displaystyle \Rightarrow (2x+4)\times3-(x+5)\times4=0
\displaystyle \Rightarrow 6x+12-4x-20=0
\displaystyle \Rightarrow 2x=8
\displaystyle \therefore x=4
\\

\displaystyle \textbf{Question 51. }\text{For what value of }x,\text{ the matrix }\begin{bmatrix}2x&4\\x+2&3\end{bmatrix} \text{ is a singular matrix? }\hspace{0.2cm}\text{[CBSE 2011 C]}
\displaystyle \text{Answer:}
\displaystyle \text{For a singular matrix, determinant }=0
\displaystyle \therefore \begin{vmatrix}2x&4\\x+2&3\end{vmatrix}=0
\displaystyle \Rightarrow (2x)(3)-4(x+2)=0
\displaystyle \Rightarrow 6x-4x-8=0
\displaystyle \Rightarrow 2x=8
\displaystyle \Rightarrow x=4
\\

\displaystyle \textbf{Question 52. }\text{If }A=\begin{bmatrix}p&2\\2&p\end{bmatrix}\text{ and }|A^3|=125,\text{ then find the value of }p. \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}p&2\\2&p\end{bmatrix}
\displaystyle \therefore |A|=\begin{vmatrix}p&2\\2&p\end{vmatrix}=p^2-4
\displaystyle \text{and }|A^3|=125
\displaystyle \Rightarrow |A|^3=125\qquad [\because |A^3|=|A|^3]
\displaystyle \Rightarrow (p^2-4)^3=125
\displaystyle \Rightarrow p^2-4=5\Rightarrow p^2=9
\displaystyle \therefore p=\pm3
\\

\displaystyle \textbf{Question 53. }\text{Show that the determinant }\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}\text{ is independent of }\theta.
\displaystyle \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}
\displaystyle \text{On expanding along }R_1,\text{ we get}
\displaystyle \Delta=x(-x^2-1)-\sin\theta(-x\sin\theta-\cos\theta)+\cos\theta(-\sin\theta+x\cos\theta)
\displaystyle =-x^3-x+x(\sin^2\theta+\cos^2\theta)+\sin\theta\cos\theta-\cos\theta\sin\theta
\displaystyle =-x^3-x+x
\displaystyle =-x^3,\text{ which is independent of }\theta.\ \text{Hence proved.}
\\

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