Class 12: Determinants – Miscellaneous Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1. }\left|\begin{matrix}x & y & z\\p & q & r\\a & b & c\end{matrix}\right|+\left|\begin{matrix}x & y & z\\s & t & u\\2a & 2b & 2c\end{matrix}\right|,\text{ which one of the following is equal to } \ \$
$\displaystyle \text{the above sum? } \ \$
$\displaystyle \text{(a) }\left|\begin{matrix}3x & 3y & 3z\\2p+s & 2q+t & 2r+u\\4a & 4b & 4c\end{matrix}\right| \ \$ $\displaystyle \text{(b) }\left|\begin{matrix}x & y & z\\p+s & q+t & r+u\\2a & 2b & 2c\end{matrix}\right| \ \$
$\displaystyle \text{(c) }\left|\begin{matrix}2x & 2y & 2z\\2p+s & 2q+t & 2r+u\\2a & 2b & 2c\end{matrix}\right| \ \$ $\displaystyle \text{(d) }\left|\begin{matrix}x & y & z\\p+s & q+t & r+u\\4a & 4b & 4c\end{matrix}\right| \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) We have,}$
$\displaystyle \begin{vmatrix}x&y&z\\2p&2q&2r\\a&b&c\end{vmatrix}+\begin{vmatrix}x&y&z\\s&t&u\\2a&2b&2c\end{vmatrix}$
$\displaystyle =\begin{vmatrix}x&y&z\\p&q&r\\2a&2b&2c\end{vmatrix}+\begin{vmatrix}x&y&z\\s&t&u\\2a&2b&2c\end{vmatrix}$
$\displaystyle =\begin{vmatrix}x&y&z\\p+s&q+t&r+u\\2a&2b&2c\end{vmatrix}$
$\\$

$\displaystyle \textbf{Question 2. }\text{If }\begin{pmatrix}3 & 5\\7 & 9\end{pmatrix}=X+Y,\text{ where }X\text{ is skew-symmetric matrix} \ \$
$\displaystyle \text{and }Y\text{ symmetric matrix. Find }|X|. \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } A=\begin{bmatrix}3&5\\7&9\end{bmatrix}=X+Y$
$\displaystyle \text{where, } X \text{ is skew-symmetric matrix, i.e. } X=\frac{1}{2}(A-A^{T})$
$\displaystyle \text{and } Y \text{ is symmetric matrix, i.e. } Y=\frac{1}{2}(A+A^{T})$
$\displaystyle \therefore A=\begin{bmatrix}3&5\\7&9\end{bmatrix} \text{ and } A^{T}=\begin{bmatrix}3&7\\5&9\end{bmatrix}$
$\displaystyle \Rightarrow X=\frac{1}{2}(A-A^{T})=\frac{1}{2}\left[\begin{bmatrix}3&5\\7&9\end{bmatrix}-\begin{bmatrix}3&7\\5&9\end{bmatrix}\right]$
$\displaystyle \Rightarrow X=\frac{1}{2}\begin{bmatrix}3-3&5-7\\7-5&9-9\end{bmatrix}=\frac{1}{2}\begin{bmatrix}0&-2\\2&0\end{bmatrix}$
$\displaystyle \Rightarrow X=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$
$\displaystyle \therefore |X|=0-(-1)=1$
$\\$

$\displaystyle \textbf{Question 3. }\text{The value of }\left|\begin{matrix}1 & \log_{a}b\\\log_{b}a & 1\end{matrix}\right|\text{ is } \ \$
$\displaystyle \text{(a) }1-\log(ab) \ \$
$\displaystyle \text{(b) }1-\frac{\log b}{\log a} \ \$
$\displaystyle \text{(c) }0 \ \$
$\displaystyle \text{(d) }\log(ab)-1 \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) Given, } \begin{vmatrix}1&\log_{a}b\\ \log_{b}a&1\end{vmatrix}$
$\displaystyle =1\times 1-(\log_{b}a)(\log_{a}b)$
$\displaystyle =1-\frac{\log a}{\log b}\times \frac{\log b}{\log a} \text{ [change of base formula]}$
$\displaystyle =1-1=0$
$\\$

$\displaystyle \textbf{Question 4. }\text{If }\Delta=\left|\begin{matrix}a & b & c\\x & y & z\\p & q & r\end{matrix}\right|,\text{ then }\left|\begin{matrix}ka & kb & kc\\kx & ky & kz\\kp & kq & kr\end{matrix}\right|\text{ is } \ \$
$\displaystyle \text{(a) }\Delta \ \$ $\displaystyle \text{(b) }k\Delta \ \$ $\displaystyle \text{(c) }3k\Delta \ \$ $\displaystyle \text{(d) }k^{3}\Delta \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) We know that if each element of a column is multiplied by the same constant,}$
$\displaystyle \text{then the determinant gets multiplied by the same constant.}$
$\displaystyle \text{So, } \begin{vmatrix}ka&kb&kc\\kx&ky&kz\\kp&kq&kr\end{vmatrix}=k^{3}\begin{vmatrix}a&b&c\\x&y&z\\p&q&r\end{vmatrix}=k^{3}\Delta$
$\\$

$\displaystyle \textbf{Question 5. }\text{What will be the value of }x\text{ for the determinant} \ \$
$\displaystyle \left|\begin{matrix}3-x & 6 & 3\\-6 & 3-x & 3\\3 & 3 & 3-x\end{matrix}\right|=0? \ \$
$\displaystyle \text{(a) }6 \ \$ $\displaystyle \text{(b) }3 \ \$ $\displaystyle \text{(c) }0 \ \$ $\displaystyle \text{(d) }-6 \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Let } x=3$
$\displaystyle \text{Now, } \Delta=\begin{vmatrix}3-x&6&3\\-6&3-x&3\\3&3&3-x\end{vmatrix}=\begin{vmatrix}0&6&3\\-6&0&3\\3&3&0\end{vmatrix}$
$\displaystyle \text{On expanding along } R_{1}, \text{ we get}$
$\displaystyle \Delta=0(0-9)-6(0-9)+3(-18-0)$
$\displaystyle \Delta=0+54-54=0$
$\displaystyle \therefore x=3$
$\\$

$\displaystyle \textbf{Question 6. }\text{Which of the following statement is correct? } \ \$
$\displaystyle \text{(a) Determinant is number associated to a matrix.} \ \$
$\displaystyle \text{(b) Determinant is a square matrix.} \ \$
$\displaystyle \text{(c) Determinant is a number associated to a square matrix.} \ \$
$\displaystyle \text{(d) None of the above} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) Determinant is a number associated to a square matrix.}$
$\\$

$\displaystyle \textbf{Question 7. }\text{If }A\text{ is square matrix of order }3,\text{ with }|A|=4,\text{ then write} \ \$
$\displaystyle \text{the value of }|-2A|. \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } A \text{ is a square matrix of order } 3 \text{ and also, } |A|=4$
$\displaystyle \text{Now, } |-2A|=(-2)^{3}|A|=-8|A|=-8\times 4=-32$
$\\$

$\displaystyle \textbf{Question 8. }\text{Using properties of determinants, prove that} \ \$
$\displaystyle \left|\begin{matrix}x & x^{2} & 1+px^{3}\\y & y^{2} & 1+py^{3}\\z & z^{2} & 1+pz^{3}\end{matrix}\right|=(1+pxyz)(x-y)(y-z)(z-x), \ \$
$\displaystyle \text{where }p\text{ is any scalar. } \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let LHS }=\begin{vmatrix}x&x^{2}&1+px^{3}\\y&y^{2}&1+py^{3}\\z&z^{2}&1+pz^{3}\end{vmatrix}$
$\displaystyle =\begin{vmatrix}x&x^{2}&1\\y&y^{2}&1\\z&z^{2}&1\end{vmatrix}+\begin{vmatrix}x&x^{2}&px^{3}\\y&y^{2}&py^{3}\\z&z^{2}&pz^{3}\end{vmatrix}$
$\displaystyle \text{On taking common } p \text{ from } C_{3},\ x \text{ from } R_{1},\ y \text{ from } R_{2} \text{ and } z \text{ from } R_{3} \text{ of second determinant, we get}$
$\displaystyle =\begin{vmatrix}x&x^{2}&1\\y&y^{2}&1\\z&z^{2}&1\end{vmatrix}+pxyz\begin{vmatrix}1&x&x^{2}\\1&y&y^{2}\\1&z&z^{2}\end{vmatrix}$
$\displaystyle =(1+pxyz)\begin{vmatrix}1&x&x^{2}\\1&y&y^{2}\\1&z&z^{2}\end{vmatrix}$
$\displaystyle \text{On expanding along } R_{1}, \text{ we get}$
$\displaystyle =(1+pxyz)\left[1(yz^{2}-zy^{2})-x(z^{2}-y^{2})+x^{2}(z-y)\right]$
$\displaystyle =(1+pxyz)\left[yz(z-y)-x(z-y)(z+y)+x^{2}(z-y)\right]$
$\displaystyle =(1+pxyz)(z-y)\left[yz-xz-xy+x^{2}\right]$
$\displaystyle =(1+pxyz)(z-y)\left[z(y-x)-x(y-x)\right]$
$\displaystyle =(1+pxyz)(z-y)(y-x)(z-x)$
$\displaystyle =(1+pxyz)(x-y)(y-z)(z-x)=\text{RHS}$
$\\$

$\displaystyle \textbf{Question 9. }\text{Find the area of the triangle, whose vertices are }(3,8), \ \$
$\displaystyle (-4,2)\text{ and }(5,1). \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given vertices of a triangle are } (3,8),\ (-4,2) \text{ and } (5,1).$
$\displaystyle \text{Let } (x_{1},y_{1})=(3,8),\ (x_{2},y_{2})=(-4,2) \text{ and } (x_{3},y_{3})=(5,1)$
$\displaystyle \text{Then, area of triangle }=\frac{1}{2}\begin{vmatrix}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{vmatrix}$
$\displaystyle =\frac{1}{2}\begin{vmatrix}3&8&1\\-4&2&1\\5&1&1\end{vmatrix}$
$\displaystyle =\frac{1}{2}\left[3\begin{vmatrix}2&1\\1&1\end{vmatrix}-8\begin{vmatrix}-4&1\\5&1\end{vmatrix}+1\begin{vmatrix}-4&2\\5&1\end{vmatrix}\right]$
$\displaystyle =\frac{1}{2}\left[3(2-1)-8(-4-5)+1(-4-10)\right]$
$\displaystyle =\frac{1}{2}[3+72-14]=\frac{61}{2} \text{ sq units}$
$\\$

$\displaystyle \textbf{Question 10. }\text{If area of the triangle with vertices }(-3,0),\ (3,0)\text{ and} \ \$
$\displaystyle (0,k)\text{ is }9\text{ sq units, then find the value of }k. \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, area of a triangle with vertices } (-3,0),\ (3,0) \text{ and } (0,k) \text{ is } 9 \text{ sq units.}$
$\displaystyle \therefore \frac{1}{2}\begin{vmatrix}-3&0&1\\3&0&1\\0&k&1\end{vmatrix}=\pm 9 \Rightarrow \begin{vmatrix}-3&0&1\\3&0&1\\0&k&1\end{vmatrix}=\pm 18$
$\displaystyle \Rightarrow -3(0-k)-0+1(3k-0)=\pm 18$
$\displaystyle \Rightarrow 3k+3k=\pm 18$
$\displaystyle \Rightarrow 6k=\pm 18 \Rightarrow k=\pm 3$
$\\$

$\displaystyle \textbf{Question 11. }\text{If the points }(2,-3),\ (\lambda,-1)\text{ and }(0,4)\text{ are collinear,} \ \$
$\displaystyle \text{then find the value of }\lambda. \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given points } (2,-3),\ (\lambda,-1) \text{ and } (0,4) \text{ are collinear.}$
$\displaystyle \text{So, area of triangle formed by these three points will be zero.}$
$\displaystyle \therefore \frac{1}{2}\begin{vmatrix}2&-3&1\\ \lambda&-1&1\\0&4&1\end{vmatrix}=0$
$\displaystyle \Rightarrow \begin{vmatrix}2&-3&1\\ \lambda&-1&1\\0&4&1\end{vmatrix}=0$
$\displaystyle \text{Now, expanding along } R_{1}, \text{ we get}$
$\displaystyle 2(-1-4)+3(\lambda-0)+1(4\lambda-0)=0$
$\displaystyle -10+3\lambda+4\lambda=0$
$\displaystyle 7\lambda=10$
$\displaystyle \Rightarrow \lambda=\frac{10}{7}$
$\displaystyle \text{Hence, the required value of } \lambda \text{ is } \frac{10}{7}.$

$\\$
$\displaystyle \textbf{Question 12. }\text{Find the equation of line joining }(2,3)\text{ and }(-1,2). \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Equation of line is given by } \begin{vmatrix}x&y&1\\x_{1}&y_{1}&1\\x_{2}&y_{2}&1\end{vmatrix}=0$
$\displaystyle \text{Consider, } (x_{1},y_{1})=(2,3) \text{ and } (x_{2},y_{2})=(-1,2)$
$\displaystyle \therefore \begin{vmatrix}x&y&1\\2&3&1\\-1&2&1\end{vmatrix}=0$
$\displaystyle \Rightarrow x(3-2)-y(2+1)+1(4+3)=0$
$\displaystyle \Rightarrow x-3y+7=0$
$\displaystyle \text{which is the required equation.}$
$\\$

$\displaystyle \textbf{Question 13. }\text{If }A\text{ is a }3\times 3\text{ matrix such that }A(\mathrm{adj}\ A)=\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}, \ \$
$\displaystyle \text{then the value of }|A^{T}|\text{ will be } \ \$
$\displaystyle \text{(a) }27 \ \$
$\displaystyle \text{(b) }9 \ \$
$\displaystyle \text{(c) }3 \ \$
$\displaystyle \text{(d) }-3 \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) Given, } A \text{ is } 3\times 3 \text{ matrix and } A(\mathrm{adj}\ A)=\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}$
$\displaystyle \text{We know that } A(\mathrm{adj}\ A)=|A|I_{3}$
$\displaystyle \therefore \begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}=|A|\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\displaystyle \Rightarrow 3I_{3}=|A|I_{3} \Rightarrow |A|=3$
$\displaystyle \therefore |A^{T}|=|A|=3$
$\\$

$\displaystyle \textbf{Question 4. }\text{What is the multiplicative inverse of matrix }A\text{? } \ \$
$\displaystyle \text{(a) }A \ \$
$\displaystyle \text{(b) }A^{2} \ \$
$\displaystyle \text{(c) }|A| \ \$
$\displaystyle \text{(d) }\frac{\mathrm{adj}\ A}{|A|} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) The multiplicative inverse of matrix } A \text{ is } \frac{\mathrm{adj}\ A}{|A|}$
$\\$

$\displaystyle \textbf{Question 15. }\text{For what value of }k\text{ inverse does not exist for the matrix }\begin{bmatrix}1&2\\k&6\end{bmatrix}\text{? } \ \$
$\displaystyle \text{(a) }2 \ \$ $\displaystyle \text{(b) }3 \ \$ $\displaystyle \text{(c) }6 \ \$ $\displaystyle \text{(d) }5 \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) For } A=\begin{bmatrix}1&2\\k&6\end{bmatrix}$
$\displaystyle \text{If } |A|=0, \text{ then inverse does not exist}$
$\displaystyle \therefore \begin{vmatrix}1&2\\k&6\end{vmatrix}=0$
$\displaystyle \Rightarrow 6-2k=0 \Rightarrow 2k=6 \Rightarrow k=3$
$\\$

$\displaystyle \textbf{Question 16. }\text{Garima owns a cafe that serves burgers, samosas and juice bottles.} \ \$
$\displaystyle \text{On a particular day, }100\text{ items in total were sold. The number of burgers sold was } \ \$
$\displaystyle \text{thrice the number of samosas sold. Also, the number of samosas sold was 10 more than} \ \$
$\displaystyle \text{the number of juice bottles sold.} \ \$
$\displaystyle \text{(i) Form a set of simultaneous equations for the } \text{above information.} \ \$
$\displaystyle \text{(ii) Solve the set of equations formed in (i) by matrix } \text{method.} \ \$
$\displaystyle \text{(iii) Hence, find the number of items sold in each } \text{category.} \ \$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let } x,\ y \text{ and } z \text{ be number of burgers, samosas and juice bottles}$
$\displaystyle \text{(i) } x+y+z=100,\ x-3y=0,\ y-z=10$
$\displaystyle \text{(ii) Writing in matrix form } AX=B$
$\displaystyle \begin{bmatrix}1&1&1\\1&-3&0\\0&1&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}100\\0\\10\end{bmatrix}$
$\displaystyle \text{Now, } |A|=1(3+0)-1(-1-0)+1(1-0)=3+1+1=5\neq 0$
$\displaystyle \text{So, } A^{-1} \text{ exists}$
$\displaystyle \text{Cofactors are } A_{11}=3,\ A_{12}=1,\ A_{13}=-1,$
$\displaystyle A_{21}=-1,\ A_{22}=-1,\ A_{23}=-1,\ A_{31}=3,\ A_{32}=1,\ A_{33}=-4$
$\displaystyle \therefore \mathrm{adj}(A)=\begin{bmatrix}3&-1&3\\1&-1&1\\-1&-1&-4\end{bmatrix}$
$\displaystyle \therefore A^{-1}=\frac{1}{5}\begin{bmatrix}3&-1&3\\1&-1&1\\-1&-1&-4\end{bmatrix}$
$\displaystyle \text{Now, } X=A^{-1}B$
$\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{5}\begin{bmatrix}3&-1&3\\1&-1&1\\-1&-1&-4\end{bmatrix}\begin{bmatrix}100\\0\\10\end{bmatrix}$
$\displaystyle =\frac{1}{5}\begin{bmatrix}300+0+30\\100+0+10\\-100+0-40\end{bmatrix}=\frac{1}{5}\begin{bmatrix}330\\110\\-140\end{bmatrix}$
$\displaystyle \Rightarrow x=66,\ y=22,\ z=-28$
$\displaystyle \text{(iii) Hence, burgers }=66,\ \text{samosas }=22,\ \text{juice bottles }=-28$
$\\$

$\displaystyle \textbf{Question 17. }\text{Find }A^{-1}\text{, if }A=\begin{bmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{bmatrix}.\text{ Using }A^{-1}\text{, solve the} \ \$
$\displaystyle \text{system of linear equations } x-2y=10,\ 2x-y-z=8,\ -2y+z=7 \ \$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Here, } A=\begin{bmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{bmatrix}$
$\displaystyle |A|=1(-1-2)-2(-2+0)+0=-3+4=1$
$\displaystyle \text{Let } C_{ij} \text{ be the cofactors of } a_{ij} \text{ in } A$
$\displaystyle C_{11}=-3,\ C_{12}=2,\ C_{13}=2,\ C_{21}=-2,\ C_{22}=-1,\ C_{23}=1,$
$\displaystyle C_{31}=-4,\ C_{32}=2,\ C_{33}=3$
$\displaystyle \therefore \mathrm{adj}(A)=\begin{bmatrix}-3&-2&-4\\2&-1&2\\2&1&3\end{bmatrix}$
$\displaystyle \therefore A^{-1}=\mathrm{adj}(A)=\begin{bmatrix}-3&-2&-4\\2&-1&2\\2&1&3\end{bmatrix}$
$\displaystyle \text{We know that } (A^{T})^{-1}=(A^{-1})^{T}$
$\displaystyle \text{Hence, required relation holds.}$
$\displaystyle C=\begin{bmatrix}1&-2&0\\2&-1&-1\\0&-2&1\end{bmatrix}$
$\displaystyle \therefore C^{-1}=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix} \text{ or } CX=B$
$\displaystyle \text{where, } C=\begin{bmatrix}1&-2&0\\2&-1&-1\\0&-2&1\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix} \text{ and } B=\begin{bmatrix}10\\8\\7\end{bmatrix}$
$\displaystyle \text{Now, } X=C^{-1}B$
$\displaystyle \Rightarrow X=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix}$
$\displaystyle \Rightarrow X=\begin{bmatrix}-30+16+14\\-20+8+7\\-40+16+21\end{bmatrix}$
$\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\-5\\-3\end{bmatrix}$
$\displaystyle \therefore x=0,\ y=-5 \text{ and } z=-3$
$\\$

$\displaystyle \textbf{Question 18. }\text{Ramu purchased }5\text{ pens, }3\text{ bags and }1\text{ instrument box } \text{and paid Rs }16.$
$\displaystyle \text{From the same shop Venkat purchased } 2\text{ pens, }1\text{ bag and 3 instrument boxes and paid}$
$\displaystyle \text{Rs }19 \text{while Gopi purchased }1\text{ pen, }2\text{ bags and }4\text{ instrument } \text{boxes and paid Rs }25.$
$\displaystyle \text{Using the concepts of matrices } \text{and determinants to answer the following questions by} \ \$
$\displaystyle \text{choosing the correct option:} \ \$
$\displaystyle \text{(i) If }x,\ y\text{ and }z\text{ respectively denotes the cost of pen, } \text{bag and instrument box, then } \ \$
$\displaystyle \text{which of the following is true?} \ \$
$\displaystyle \text{(a) }5x+3y+z=16 \ \$
$\displaystyle \text{(b) }2x+y+3z=19 \ \$
$\displaystyle \text{(c) }x+2y+4z=25 \ \$
$\displaystyle \text{(d) All of these} \ \$
$\displaystyle \text{(ii) If }A=\begin{bmatrix}5&3&1\\2&1&3\\1&2&4\end{bmatrix},\ |A|\text{ is} \ \$
$\displaystyle \text{(a) }-22 \ \$ $\displaystyle \text{(b) }22 \ \$ $\displaystyle \text{(c) }0 \ \$ $\displaystyle \text{(d) }20 \ \$
$\displaystyle \text{(iii) If }A=\begin{bmatrix}5&3&1\\2&1&3\\1&2&4\end{bmatrix}\text{ and }\mathrm{adj}\ A=\begin{bmatrix}-2&x&8\\-5&19&-13\\3&-7&y\end{bmatrix},\text{ then} \ \$
$\displaystyle \text{missing value of }x\text{ and }y\text{ are} \ \$
$\displaystyle \text{(a) }x=-10\text{ and }y=-1 \ \$
$\displaystyle \text{(b) }x=10\text{ and }y=-1 \ \$
$\displaystyle \text{(c) }x=-10\text{ and }y=1 \ \$
$\displaystyle \text{(d) }x=10\text{ and }y=1 \ \$
$\displaystyle \text{(iv) The cost of one pen is } \ \$
$\displaystyle \text{(a) Rs }2 \ \$ $\displaystyle \text{(b) Rs }5 \ \$ $\displaystyle \text{(c) Rs }1 \ \$ $\displaystyle \text{(d) Rs }3 \ \$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{(i) (a) Ramu paid Rs } 16 \text{ for 5 pens, 3 bags and 1 instrument box.}$
$\displaystyle \text{If } x,\ y \text{ and } z \text{ represent the cost of a pen, a bag and an instrument box, respectively,}$
$\displaystyle \therefore 5x+3y+z=16$
$\displaystyle \text{(ii) (a) Given, } A=\begin{bmatrix}5&3&1\\2&1&3\\1&2&4\end{bmatrix}$
$\displaystyle \text{So, } |A|=5(4-6)-3(8-3)+1(4-1)$
$\displaystyle =5(-2)-3(5)+1(3)$
$\displaystyle =-10-15+3=-22$
$\displaystyle \text{(iii) (a) Given, } A=\begin{bmatrix}5&3&1\\2&1&3\\1&2&4\end{bmatrix}$
$\displaystyle \text{and } \mathrm{adj}\ A=\begin{bmatrix}-2&x&8\\-5&19&-13\\3&-7&y\end{bmatrix}$
$\displaystyle \text{We know that } A_{12}=C_{21} \text{ and } A_{33}=C_{33}$
$\displaystyle \therefore x=(-1)^{2+1}\begin{vmatrix}3&1\\2&4\end{vmatrix} \text{ and } y=(-1)^{3+3}\begin{vmatrix}5&3\\2&1\end{vmatrix}$
$\displaystyle \Rightarrow x=-1(12-2) \text{ and } y=1(5-6)$
$\displaystyle \Rightarrow x=-10 \text{ and } y=-1$
$\displaystyle \text{(iv) (c) Let } A=\begin{bmatrix}5&3&1\\2&1&3\\1&2&4\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix} \text{ and } B=\begin{bmatrix}16\\19\\25\end{bmatrix}$
$\displaystyle \text{Now, } AX=B \Rightarrow X=A^{-1}B=\frac{1}{|A|}(\mathrm{adj}\ A)B$
$\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{-22}\begin{bmatrix}-2&-10&8\\-5&19&-13\\3&-7&-1\end{bmatrix}\begin{bmatrix}16\\19\\25\end{bmatrix}$
$\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{-22}\begin{bmatrix}-32-190+200\\-80+361-325\\48-133-25\end{bmatrix}$
$\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{-22}\begin{bmatrix}-22\\-44\\-110\end{bmatrix}=\begin{bmatrix}1\\2\\5\end{bmatrix}$
$\displaystyle \therefore x=1$
$\displaystyle \text{Hence, the cost of one pen is Rs } 1.$
$\\$

$\displaystyle \textbf{Question 19. }\text{Using matrices, solve the following system of equations} \ \$
$\displaystyle 2x-3y+5z=11,\ 3x+2y-4z=-5 \text{ and }x+y-2z=-3. \ \$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{The given system of equation is}$
$\displaystyle 2x-3y+5z=11,$
$\displaystyle 3x+2y-4z=-5$
$\displaystyle \text{and } x+y-2z=-3$
$\displaystyle \text{In matrix form, } AX=B$
$\displaystyle \text{where, } A=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix} \text{ and } B=\begin{bmatrix}11\\-5\\-3\end{bmatrix}$
$\displaystyle \text{Now, } |A|=2(-4+4)+3(-6+4)+5(3-2)$
$\displaystyle =2(0)+3(-2)+5(1)$
$\displaystyle =-1\neq 0$
$\displaystyle \text{Thus, } A \text{ is a non-singular matrix.}$
$\displaystyle \text{So, } A^{-1} \text{ exists. Cofactors of elements of } |A| \text{ are}$
$\displaystyle C_{11}=(-1)^{1+1}\begin{vmatrix}2&-4\\1&-2\end{vmatrix}=(-4+4)=0$
$\displaystyle C_{12}=(-1)^{1+2}\begin{vmatrix}3&-4\\1&-2\end{vmatrix}=-( -6+4)=2$
$\displaystyle C_{13}=(-1)^{1+3}\begin{vmatrix}3&2\\1&1\end{vmatrix}=3-2=1$
$\displaystyle C_{21}=(-1)^{2+1}\begin{vmatrix}3&5\\1&-2\end{vmatrix}=-( -6-5)=1$
$\displaystyle C_{22}=(-1)^{2+2}\begin{vmatrix}2&5\\1&-2\end{vmatrix}=(-4-5)=-9$
$\displaystyle C_{23}=(-1)^{2+3}\begin{vmatrix}2&3\\1&1\end{vmatrix}=-(2-3)=1$
$\displaystyle C_{31}=(-1)^{3+1}\begin{vmatrix}-3&5\\2&-4\end{vmatrix}=(12-10)=2$
$\displaystyle C_{32}=(-1)^{3+2}\begin{vmatrix}-3&5\\3&-4\end{vmatrix}=-(12-15)=3$
$\displaystyle C_{33}=(-1)^{3+3}\begin{vmatrix}-3&2\\3&2\end{vmatrix}=(-6-6)=-12$
$\displaystyle \text{Then, } \mathrm{adj}(A)=\begin{bmatrix}C_{11}&C_{21}&C_{31}\\C_{12}&C_{22}&C_{32}\\C_{13}&C_{23}&C_{33}\end{bmatrix}^{T}$
$\displaystyle =\begin{bmatrix}0&-1&2\\-1&-9&3\\2&3&-12\end{bmatrix}$
$\displaystyle \text{Now, } A^{-1}=\frac{1}{|A|}\mathrm{adj}(A)$
$\displaystyle \text{Now, } X=A^{-1}B$
$\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}\begin{bmatrix}11\\-5\\-3\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0-5+6\\-22-45+69\\-11-25+39\end{bmatrix}$
$\displaystyle =\begin{bmatrix}1\\2\\3\end{bmatrix}$
$\displaystyle \therefore x=1,\ y=2,\ z=3$