\displaystyle \textbf{Question 1: }\text{Write the minor and cofactor of each element of the first column of}
\displaystyle \text{the following matrices and hence evaluate the determinants:}
\displaystyle \text{(i) }A=\begin{bmatrix}5&20\\0&-1\end{bmatrix}\qquad \text{(ii) }A=\begin{bmatrix}-1&4\\2&3\end{bmatrix}
\displaystyle \text{(iii) }A=\begin{bmatrix}1&-3&2\\4&-1&2\\3&5&2\end{bmatrix}\qquad \text{(iv) }A=\begin{bmatrix}1&a&bc\\1&b&ca\\1&c&ab\end{bmatrix}
\displaystyle \text{(v) }A=\begin{bmatrix}0&2&6\\1&5&0\\3&7&1\end{bmatrix}\qquad \text{(vi) }A=\begin{bmatrix}a&h&g\\h&b&f\\g&f&c\end{bmatrix}
\displaystyle \text{(vii) }A=\begin{bmatrix}2&-1&0&1\\-3&0&1&-2\\1&1&-1&1\\2&-1&5&0\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{We use }C_{ij}=(-1)^{i+j}M_{ij}.
\displaystyle \text{(i)}
\displaystyle M_{11}=-1,\qquad M_{21}=20
\displaystyle C_{11}=(-1)^{1+1}M_{11}=-1
\displaystyle C_{21}=(-1)^{2+1}M_{21}=-20
\displaystyle |A|=a_{11}C_{11}+a_{21}C_{21}
\displaystyle =5(-1)+0(-20)=-5
\displaystyle \\

\displaystyle \text{(ii)}
\displaystyle M_{11}=3,\qquad M_{21}=4
\displaystyle C_{11}=(-1)^{1+1}M_{11}=3
\displaystyle C_{21}=(-1)^{2+1}M_{21}=-4
\displaystyle |A|=a_{11}C_{11}+a_{21}C_{21}
\displaystyle =(-1)(3)+2(-4)=-3-8=-11
\displaystyle \\

\displaystyle \text{(iii)}
\displaystyle M_{11}=\begin{vmatrix}-1&2\\5&2\end{vmatrix}=-2-10=-12
\displaystyle M_{21}=\begin{vmatrix}-3&2\\5&2\end{vmatrix}=-6-10=-16
\displaystyle M_{31}=\begin{vmatrix}-3&2\\-1&2\end{vmatrix}=-6+2=-4
\displaystyle C_{11}=(-1)^{1+1}M_{11}=-12
\displaystyle C_{21}=(-1)^{2+1}M_{21}=16
\displaystyle C_{31}=(-1)^{3+1}M_{31}=-4
\displaystyle |A|=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}
\displaystyle =1(-12)+4(16)+3(-4)
\displaystyle =-12+64-12=40
\displaystyle \\

\displaystyle \text{(iv)}
\displaystyle M_{11}=\begin{vmatrix}b&ca\\c&ab\end{vmatrix} =ab^2-ac^2=a(b^2-c^2)
\displaystyle M_{21}=\begin{vmatrix}a&bc\\c&ab\end{vmatrix} =a^2b-bc^2=b(a^2-c^2)
\displaystyle M_{31}=\begin{vmatrix}a&bc\\b&ca\end{vmatrix} =a^2c-b^2c=c(a^2-b^2)
\displaystyle C_{11}=a(b^2-c^2)
\displaystyle C_{21}=-b(a^2-c^2)
\displaystyle C_{31}=c(a^2-b^2)
\displaystyle |A|=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}
\displaystyle =a(b^2-c^2)-b(a^2-c^2)+c(a^2-b^2)
\displaystyle =ab^2-ac^2-a^2b+bc^2+a^2c-b^2c
\displaystyle =a^2(c-b)+b^2(a-c)+c^2(b-a)
\displaystyle \\

\displaystyle \text{(v)}
\displaystyle M_{11}=\begin{vmatrix}5&0\\7&1\end{vmatrix}=5
\displaystyle M_{21}=\begin{vmatrix}2&6\\7&1\end{vmatrix}=2-42=-40
\displaystyle M_{31}=\begin{vmatrix}2&6\\5&0\end{vmatrix}=0-30=-30
\displaystyle C_{11}=(-1)^{1+1}M_{11}=5
\displaystyle C_{21}=(-1)^{2+1}M_{21}=40
\displaystyle C_{31}=(-1)^{3+1}M_{31}=-30
\displaystyle |A|=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}
\displaystyle =0(5)+1(40)+3(-30)
\displaystyle =40-90=-50
\displaystyle \\

\displaystyle \text{(vi)}
\displaystyle M_{11}=\begin{vmatrix}b&f\\f&c\end{vmatrix}=bc-f^2
\displaystyle M_{21}=\begin{vmatrix}h&g\\f&c\end{vmatrix}=hc-fg
\displaystyle M_{31}=\begin{vmatrix}h&g\\b&f\end{vmatrix}=hf-bg
\displaystyle C_{11}=bc-f^2
\displaystyle C_{21}=-(hc-fg)=fg-hc
\displaystyle C_{31}=hf-bg
\displaystyle |A|=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}
\displaystyle =a(bc-f^2)+h(fg-hc)+g(hf-bg)
\displaystyle =abc-af^2+hfg-h^2c+ghf-bg^2
\displaystyle =abc+2fgh-af^2-bg^2-ch^2
\displaystyle \\

\displaystyle \text{(vii)}
\displaystyle M_{11}=\begin{vmatrix}0&1&-2\\1&-1&1\\-1&5&0\end{vmatrix}
\displaystyle =0-1(0+1)-2(5-1)=-1-8=-9
\displaystyle M_{21}=\begin{vmatrix}-1&0&1\\1&-1&1\\-1&5&0\end{vmatrix}
\displaystyle =(-1)(0-5)+1(5-1)=5+4=9
\displaystyle M_{31}=\begin{vmatrix}-1&0&1\\0&1&-2\\-1&5&0\end{vmatrix}
\displaystyle =(-1)(0+10)+1(0+1)=-10+1=-9
\displaystyle M_{41}=\begin{vmatrix}-1&0&1\\0&1&-2\\1&-1&1\end{vmatrix}
\displaystyle =(-1)(1-2)+1(0-1)=1-1=0
\displaystyle C_{11}=(-1)^{1+1}M_{11}=-9
\displaystyle C_{21}=(-1)^{2+1}M_{21}=-9
\displaystyle C_{31}=(-1)^{3+1}M_{31}=-9
\displaystyle C_{41}=(-1)^{4+1}M_{41}=0
\displaystyle |A|=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}+a_{41}C_{41}
\displaystyle =2(-9)+(-3)(-9)+1(-9)+2(0)
\displaystyle =-18+27-9=0
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Evaluate the following determinants:}
\displaystyle \text{(i) }\begin{vmatrix}x&-7\\x&5x+1\end{vmatrix}\qquad \text{(ii) }\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}
\displaystyle \text{(iii) }\begin{vmatrix}\cos15^\circ&\sin15^\circ\\ \sin75^\circ&\cos75^\circ\end{vmatrix}\qquad \text{(iv) }\begin{vmatrix}a+ib&c+id\\-c+id&a-ib\end{vmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \Delta=x(5x+1)-(-7)(x)
\displaystyle =5x^2+x+7x
\displaystyle =5x^2+8x
\displaystyle \\

\displaystyle \text{(ii)}
\displaystyle \Delta=\cos\theta\cdot\cos\theta-(-\sin\theta)(\sin\theta)
\displaystyle =\cos^2\theta+\sin^2\theta
\displaystyle =1
\displaystyle \\

\displaystyle \text{(iii)}
\displaystyle \Delta=\cos15^\circ\cos75^\circ-\sin15^\circ\sin75^\circ
\displaystyle \text{Since }\sin15^\circ=\cos75^\circ\text{ and }\sin75^\circ=\cos15^\circ,
\displaystyle \Delta=\cos15^\circ\cos75^\circ-\cos75^\circ\cos15^\circ
\displaystyle =0
\displaystyle \\

\displaystyle \text{(iv)}
\displaystyle \Delta=(a+ib)(a-ib)-(c+id)(-c+id)
\displaystyle =\left(a^2-i^2b^2\right)-\left(-c^2+i^2d^2\right)
\displaystyle =a^2+b^2-(-c^2-d^2)\qquad[\because\ i^2=-1]
\displaystyle =a^2+b^2+c^2+d^2
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Evaluate }\left|\begin{matrix}2&3&7\\13&17&5\\15&20&12\end{matrix}\right|^2.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\begin{vmatrix}2&3&7\\13&17&5\\15&20&12\end{vmatrix}.
\displaystyle \text{Here, }R_3=R_1+R_2,
\displaystyle \text{since }(15,20,12)=(2,3,7)+(13,17,5).
\displaystyle \therefore \text{The rows are linearly dependent, and hence }\Delta=0.
\displaystyle \therefore \left|\begin{matrix}2&3&7\\13&17&5\\15&20&12\end{matrix}\right|^2
\displaystyle =\Delta^2=0^2=0.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Show that }\left|\begin{matrix}\sin10^\circ&-\cos10^\circ\\\sin80^\circ&\cos80^\circ\end{matrix}\right|=1.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\begin{vmatrix}\sin10^\circ&-\cos10^\circ\\\sin80^\circ&\cos80^\circ\end{vmatrix}.
\displaystyle \Delta=\sin10^\circ\cos80^\circ-(-\cos10^\circ)\sin80^\circ
\displaystyle =\sin10^\circ\cos80^\circ+\cos10^\circ\sin80^\circ
\displaystyle =\sin10^\circ\sin10^\circ+\cos10^\circ\cos10^\circ
\displaystyle \qquad[\because\ \cos80^\circ=\sin10^\circ,\ \sin80^\circ=\cos10^\circ]
\displaystyle =\sin^210^\circ+\cos^210^\circ=1.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Evaluate }\begin{vmatrix}2&3&-5\\7&1&-2\\-3&4&1\end{vmatrix}\text{ by two methods.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\Delta=\begin{vmatrix}2&3&-5\\7&1&-2\\-3&4&1\end{vmatrix}.
\displaystyle \text{First method (Expansion along the first row)}
\displaystyle \Delta=2\begin{vmatrix}1&-2\\4&1\end{vmatrix}-3\begin{vmatrix}7&-2\\-3&1\end{vmatrix}+(-5)\begin{vmatrix}7&1\\-3&4\end{vmatrix}
\displaystyle =2(1+8)-3(7-6)-5(28+3)
\displaystyle =18-3-155
\displaystyle =-140
\displaystyle \text{Second method (Sarrus' Rule)}
\displaystyle \text{Adjoin the first two columns to the right:}
\displaystyle \begin{vmatrix}2&3&-5&2&3\\7&1&-2&7&1\\-3&4&1&-3&4\end{vmatrix}
\displaystyle \Delta=(2\times1\times1+3\times(-2)\times(-3)+(-5)\times7\times4)
\displaystyle \qquad-\left[(-5)\times1\times(-3)+2\times(-2)\times4+3\times7\times1\right]
\displaystyle =(2+18-140)-(15-16+21)
\displaystyle =-120-20
\displaystyle =-140
\displaystyle \therefore \begin{vmatrix}2&3&-5\\7&1&-2\\-3&4&1\end{vmatrix}=-140.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Evaluate }\Delta=\begin{vmatrix}0&\sin\alpha&-\cos\alpha\\-\sin\alpha&0&\sin\beta\\\cos\alpha&-\sin\beta&0\end{vmatrix}.
\displaystyle \text{Answer:}
\displaystyle \Delta=\begin{vmatrix}0&\sin\alpha&-\cos\alpha\\-\sin\alpha&0&\sin\beta\\\cos\alpha&-\sin\beta&0\end{vmatrix}
\displaystyle \text{Expanding along }R_1,
\displaystyle \Delta=0\begin{vmatrix}0&\sin\beta\\-\sin\beta&0\end{vmatrix} -\sin\alpha\begin{vmatrix}-\sin\alpha&\sin\beta\\\cos\alpha&0\end{vmatrix}
\displaystyle \qquad-\cos\alpha\begin{vmatrix}-\sin\alpha&0\\\cos\alpha&-\sin\beta\end{vmatrix}
\displaystyle =0-\sin\alpha\left(0-\sin\beta\cos\alpha\right) -\cos\alpha\left(\sin\alpha\sin\beta-0\right)
\displaystyle =\sin\alpha\sin\beta\cos\alpha-\sin\alpha\sin\beta\cos\alpha
\displaystyle =0
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Evaluate}
\displaystyle \Delta=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha\end{vmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Expanding along }R_1,
\displaystyle \Delta=\cos\alpha\cos\beta\begin{vmatrix}\cos\beta&0\\\sin\alpha\sin\beta&\cos\alpha\end{vmatrix}
\displaystyle \qquad-\cos\alpha\sin\beta\begin{vmatrix}-\sin\beta&0\\\sin\alpha\cos\beta&\cos\alpha\end{vmatrix}
\displaystyle \qquad-\sin\alpha\begin{vmatrix}-\sin\beta&\cos\beta\\\sin\alpha\cos\beta&\sin\alpha\sin\beta\end{vmatrix}
\displaystyle =\cos\alpha\cos\beta(\cos\alpha\cos\beta)
\displaystyle \qquad-\cos\alpha\sin\beta(-\cos\alpha\sin\beta)
\displaystyle \qquad-\sin\alpha(-\sin\alpha\sin^2\beta-\sin\alpha\cos^2\beta)
\displaystyle =\cos^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta
\displaystyle \qquad+\sin^2\alpha\sin^2\beta+\sin^2\alpha\cos^2\beta
\displaystyle =\cos^2\alpha(\cos^2\beta+\sin^2\beta)
\displaystyle \qquad+\sin^2\alpha(\sin^2\beta+\cos^2\beta)
\displaystyle =\cos^2\alpha+\sin^2\alpha
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }A=\begin{bmatrix}2&5\\2&1\end{bmatrix}\text{ and }B=\begin{bmatrix}4&-3\\2&5\end{bmatrix},\text{ verify that }|AB|=|A||B|.
\displaystyle \text{Answer:}
\displaystyle \text{Consider LHS}
\displaystyle AB=\begin{bmatrix}2&5\\2&1\end{bmatrix}\begin{bmatrix}4&-3\\2&5\end{bmatrix}
\displaystyle =\begin{bmatrix}8+10&-6+25\\8+2&-6+5\end{bmatrix}=\begin{bmatrix}18&19\\10&-1\end{bmatrix}
\displaystyle |AB|=(18)(-1)-(19)(10)
\displaystyle =-18-190=-208
\displaystyle \text{Consider RHS}
\displaystyle |A|=(2)(1)-(5)(2)=2-10=-8
\displaystyle |B|=(4)(5)-(-3)(2)=20+6=26
\displaystyle |A||B|=(-8)(26)=-208
\displaystyle \therefore |AB|=|A||B|.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }A=\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix},\text{ then show that }|3A|=27|A|.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix}
\displaystyle \therefore |A|=1\times1\times4=4
\displaystyle 3A=\begin{bmatrix}3&0&3\\0&3&6\\0&0&12\end{bmatrix}
\displaystyle \therefore |3A|=3\times3\times12=108
\displaystyle 27|A|=27\times4=108
\displaystyle \therefore |3A|=27|A|.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the value of }x,\text{ if}
\displaystyle \text{(i) }\begin{vmatrix}2&4\\5&1\end{vmatrix}=\begin{vmatrix}2x&4\\6&x\end{vmatrix}\qquad \text{(ii) }\begin{vmatrix}2&3\\4&5\end{vmatrix}=\begin{vmatrix}x&3\\2x&5\end{vmatrix}
\displaystyle \text{(iii) }\begin{vmatrix}3&x\\x&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}\qquad \text{(iv) }\begin{vmatrix}3x&7\\2&4\end{vmatrix}=10
\displaystyle \text{(v) }\begin{vmatrix}x+1&x-1\\x-3&x+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix}\qquad \text{(vi) }\begin{vmatrix}2x&5\\8&x\end{vmatrix}=\begin{vmatrix}6&5\\8&3\end{vmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle 2(1)-4(5)=2x(x)-4(6)
\displaystyle \Rightarrow 2-20=2x^2-24
\displaystyle \Rightarrow -18=2x^2-24
\displaystyle \Rightarrow 2x^2=6
\displaystyle \Rightarrow x^2=3
\displaystyle \therefore x=\pm\sqrt{3}
\displaystyle \\

\displaystyle \text{(ii)}
\displaystyle 2(5)-3(4)=x(5)-3(2x)
\displaystyle \Rightarrow 10-12=5x-6x
\displaystyle \Rightarrow -2=-x
\displaystyle \therefore x=2
\displaystyle \\

\displaystyle \text{(iii)}
\displaystyle 3(1)-x(x)=3(1)-2(4)
\displaystyle \Rightarrow 3-x^2=3-8
\displaystyle \Rightarrow -x^2=-8
\displaystyle \Rightarrow x^2=8
\displaystyle \therefore x=\pm2\sqrt{2}
\displaystyle \\

\displaystyle \text{(iv)}
\displaystyle (3x)(4)-7(2)=10
\displaystyle \Rightarrow 12x-14=10
\displaystyle \Rightarrow 12x=24
\displaystyle \therefore x=2
\displaystyle \\

\displaystyle \text{(v)}
\displaystyle (x+1)(x+2)-(x-3)(x-1)=4(3)-(-1)(1)
\displaystyle \Rightarrow x^2+3x+2-(x^2-4x+3)=13
\displaystyle \Rightarrow x^2+3x+2-x^2+4x-3=13
\displaystyle \Rightarrow 7x-1=13
\displaystyle \Rightarrow 7x=14
\displaystyle \therefore x=2
\displaystyle \\

\displaystyle \text{(vi)}
\displaystyle (2x)(x)-5(8)=6(3)-5(8)
\displaystyle \Rightarrow 2x^2-40=18-40
\displaystyle \Rightarrow 2x^2=18
\displaystyle \Rightarrow x^2=9
\displaystyle \therefore x=\pm3
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the integral value of }x\text{ if } \begin{vmatrix}x^2&x&1\\0&2&1\\3&1&4\end{vmatrix}=28.
\displaystyle \text{Answer:}
\displaystyle \begin{vmatrix}x^2&x&1\\0&2&1\\3&1&4\end{vmatrix}=28
\displaystyle \text{Expanding along }R_1,
\displaystyle x^2(8-1)-x(0-3)+1(0-6)=28
\displaystyle \Rightarrow 7x^2+3x-6=28
\displaystyle \Rightarrow 7x^2+3x-34=0
\displaystyle \Rightarrow (7x+17)(x-2)=0
\displaystyle \Rightarrow x=-\frac{17}{7}\text{ or }x=2
\displaystyle \therefore \text{The integral value of }x\text{ is }2.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{For what values of }x\text{ is }A\text{ singular?}
\displaystyle \text{(i) }A=\begin{bmatrix}1+x&7\\3-x&8\end{bmatrix}\qquad \text{(ii) }A=\begin{bmatrix}x-1&1&1\\1&x-1&1\\1&1&x-1\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{Matrix }A\text{ is singular if }|A|=0.
\displaystyle \begin{vmatrix}1+x&7\\3-x&8\end{vmatrix}=0
\displaystyle \Rightarrow 8(1+x)-7(3-x)=0
\displaystyle \Rightarrow 8+8x-21+7x=0
\displaystyle \Rightarrow 15x-13=0
\displaystyle \therefore x=\frac{13}{15}
\displaystyle \\

\displaystyle \text{(ii)}
\displaystyle \text{Matrix }A\text{ is singular if }|A|=0.
\displaystyle \begin{vmatrix}x-1&1&1\\1&x-1&1\\1&1&x-1\end{vmatrix}=0
\displaystyle \text{Expanding along }R_1,
\displaystyle (x-1)\begin{vmatrix}x-1&1\\1&x-1\end{vmatrix} -\begin{vmatrix}1&1\\1&x-1\end{vmatrix} +\begin{vmatrix}1&x-1\\1&1\end{vmatrix}=0
\displaystyle \Rightarrow (x-1)\left[(x-1)^2-1\right]-(x-2)+(2-x)=0
\displaystyle \Rightarrow (x-1)(x^2-2x)-2(x-2)=0
\displaystyle \Rightarrow x^3-3x^2+4=0
\displaystyle \Rightarrow (x-2)^2(x+1)=0
\displaystyle \therefore x=2\text{ or }x=-1
\displaystyle \\

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