Class 12: Determinants – Exercise 6.1

$\displaystyle \textbf{Question 1: } \text{Write the minor and cofactor of element of the first column of the} \\ \text{following matrix and hence evaluate the determinant:}$

$\displaystyle \text{(i) }A=\begin{bmatrix}5 & 20\\0 & -1\end{bmatrix}$ $\displaystyle \text{(ii) }A=\begin{bmatrix}-1 & 4\\2 & 3\end{bmatrix}$ $\displaystyle \text{(iii) }A=\begin{bmatrix}1 & -3 & 2\\4 & -1 & 2\\3 & 5 & 2\end{bmatrix}$

$\displaystyle \text{(iv) }A=\begin{bmatrix}1 & a & bc\\1 & b & ca\\1 & c & ab\end{bmatrix}$ $\displaystyle \text{(v) }A=\begin{bmatrix}0 & 2 & 6\\1 & 5 & 0\\3 & 7 & 1\end{bmatrix}$ $\displaystyle \text{(vi) }A=\begin{bmatrix}a & h & g\\h & b & f\\g & f & c\end{bmatrix}$

$\displaystyle \text{(vii) }A=\begin{bmatrix}2 & -1 & 0 & 1\\-3 & 0 & 1 & -2\\1 & 1 & -1 & 1\\2 & -1 & 5 & 0\end{bmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle M_{11}=-1$

$\displaystyle M_{21}=20$

$\displaystyle C_{ij}=(-1)^{i+j}M_{ij}$

$\displaystyle C_{11}=(-1)^{1+1}(-1)=-1$

$\displaystyle C_{21}=(-1)^{1+2}(20)=-20$

$\displaystyle D=(-1\times5)-(20\times0)=-5$

$\displaystyle \text{(ii) }$

$\displaystyle M_{11}=3$

$\displaystyle M_{21}=4$

$\displaystyle C_{ij}=(-1)^{i+j}M_{ij}$

$\displaystyle C_{11}=(-1)^{1+1}M_{11}=3$

$\displaystyle C_{21}=(-1)^{2+1}M_{21}=-(4)=-4$

$\displaystyle D=(3\times-1)-(4\times2)=-3-8=-11$

$\displaystyle \text{(iii) }$

$\displaystyle M_{11}=\begin{vmatrix}-1 & 2\\ 5 & 2\end{vmatrix}=-2-10=-12$

$\displaystyle M_{21}=\begin{vmatrix}-3 & 2\\ 5 & 2\end{vmatrix}=-6-10=-16$

$\displaystyle M_{31}=\begin{vmatrix}-3 & 2\\ -1 & 2\end{vmatrix}=-6+2=-4$

$\displaystyle C_{11}=(-1)^{1+1}M_{11}=-12$

$\displaystyle C_{21}=(-1)^{2+1}M_{21}=-(-16)=16$

$\displaystyle C_{31}=(-1)^{3+1}M_{31}=-4$

$\displaystyle D=1(-12)+3(8-6)+2(20+3)=-12+6+46=40$

$\displaystyle \text{(iv) }$

$\displaystyle M_{11}=\begin{vmatrix}b & ca\\ c & ab\end{vmatrix} =ab^{2}-c^{2}a=a(b^{2}-c^{2})$

$\displaystyle M_{21}=\begin{vmatrix}a & bc\\ c & ab\end{vmatrix} =a^{2}b-c^{2}b=b(a^{2}-c^{2})$

$\displaystyle M_{31}=\begin{vmatrix}a & bc\\ b & ca\end{vmatrix} =a^{2}c-b^{2}c=c(a^{2}-b^{2})$

$\displaystyle C_{11}=(-1)^{1+1}M_{11}=a(b^{2}-c^{2})$

$\displaystyle C_{21}=(-1)^{2+1}M_{21}=-b(a^{2}-c^{2})$

$\displaystyle C_{31}=(-1)^{3+1}M_{31}=c(a^{2}-b^{2})$

$\displaystyle D=1\cdot a(b^{2}-c^{2})-a(ab-ca)+b\cdot c(c-b)$

$\displaystyle =ab^{2}-ac^{2}-a^{2}b+a^{2}c+c^{2}b-b^{2}c$

$\displaystyle =a^{2}(c-b)+b^{2}(a-c)+c^{2}(b-a)$

$\displaystyle \text{(v) }$

$\displaystyle M_{11}=\begin{vmatrix}5 & 0\\ 7 & 1\end{vmatrix}=5-0=5$

$\displaystyle M_{21}=\begin{vmatrix}2 & 6\\ 7 & 1\end{vmatrix}=2-42=-40$

$\displaystyle M_{31}=\begin{vmatrix}2 & 6\\ 5 & 0\end{vmatrix}=0-30=-30$

$\displaystyle C_{11}=(-1)^{1+1}M_{11}=5$

$\displaystyle C_{21}=(-1)^{2+1}M_{21}=-(-40)$

$\displaystyle C_{31}=(-1)^{3+1}M_{31}=-30$

$\displaystyle D=0(5-0)-2(1-0)+6(7-15)=-2-48=-50$

$\displaystyle \text{(vi) }$

$\displaystyle M_{11}=\begin{vmatrix}b & f\\ f & c\end{vmatrix}=bc-f^{2}$

$\displaystyle M_{21}=\begin{vmatrix}h & g\\ f & c\end{vmatrix}=hc-fg$

$\displaystyle M_{31}=\begin{vmatrix}h & g\\ b & f\end{vmatrix}=hf-gb$

$\displaystyle C_{11}=(-1)^{1+1}M_{11}=bc-f^{2}$

$\displaystyle C_{21}=(-1)^{2+1}M_{21}=-(hc-fg)=fg-hc$

$\displaystyle C_{31}=(-1)^{3+1}M_{31}=hf-gb$

$\displaystyle D=a(bc-f^{2})-h(hc-fg)+g(fh-bg)$

$\displaystyle =abc-af^{2}-h^{2}c+fgh+fgh-bg^{2}$

$\displaystyle =abc+2hfg-af^{2}-bg^{2}-ch^{2}$

$\displaystyle \text{(vii) }$

$\displaystyle M_{11}=0(0-5)-1(0+1)-2(5-1)=-1-8=-9$

$\displaystyle M_{21}=-1(0-5)+1(5-1)=5+4=9$

$\displaystyle M_{31}=-1(0+10)+1(0+1)=-10+1=-9$

$\displaystyle M_{41}=-1(1-2)+1(0-1)=1-1=0$

$\displaystyle C_{11}=(-1)^{1+1}M_{11}=-9$

$\displaystyle C_{21}=(-1)^{2+1}M_{21}=(-1)\times9$

$\displaystyle C_{31}=(-1)^{3+1}M_{31}=-9$

$\displaystyle C_{41}=(-1)^{4+1}M_{41}=0$

$\displaystyle D= 2\begin{vmatrix}0 & 1 & -2\\ 1 & -1 & 1\\ -1 & 5 & 0\end{vmatrix} + 1\begin{vmatrix}-3 & 1 & -2\\ 1 & -1 & 1\\ 2 & 5 & 0\end{vmatrix} - 1\begin{vmatrix}-3 & 0 & 1\\ 1 & 1 & -1\\ 2 & -1 & 5\end{vmatrix}$

$\displaystyle =-18-27+15=30$

$\\$

$\displaystyle \textbf{Question 2: } \text{Evaluate the following determinant: }$

$\displaystyle \text{(i) }\begin{vmatrix}x & -7\\x & 5x+1\end{vmatrix}$ $\displaystyle \text{(ii) }\begin{vmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{vmatrix}$

$\displaystyle \text{(iii) }\begin{vmatrix}\cos 15^\circ & \sin 15^\circ\\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}$ $\displaystyle \text{(iv) }\begin{vmatrix}a+ib & c+id\\ -c+id & a-ib\end{vmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \Delta=x(5x+1)+7x=5x^{2}+x+7x=5x^{2}+8x$

$\displaystyle \text{(ii) }$

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

$\displaystyle \text{(iii) }$

$\displaystyle \Delta=\cos15^\circ\cos75^\circ-\sin15^\circ\sin75^\circ$

$\displaystyle =\cos15^\circ\cos75^\circ-\sin(90^\circ-75^\circ)\sin(90^\circ-15^\circ)\;[\because\ \sin(90^\circ-\theta)=\cos\theta]$

$\displaystyle =\cos15^\circ\cos75^\circ-\cos75^\circ\cos15^\circ$

$\displaystyle =\cos15^\circ\cos75^\circ-\cos15^\circ\cos75^\circ=0$

$\displaystyle \text{(iv) }$

$\displaystyle \Delta=a^{2}-i^{2}b^{2}-(i^{2}d^{2}-c^{2})$

$\displaystyle =a^{2}-i^{2}b^{2}-i^{2}d^{2}+c^{2}$

$\displaystyle =a^{2}+c^{2}-i^{2}(b^{2}+d^{2})\;[\because\ i^{2}=-1]$

$\displaystyle =a^{2}+c^{2}+b^{2}+d^{2}$

$\\$

$\displaystyle \textbf{Question 3:} \text{ Evaluate } \left|\begin{matrix}2 & 3 & 7\\13 & 17 & 5\\15 & 20 & 12\end{matrix}\right|^{2}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let }A=\begin{vmatrix}2 & 3 & 7\\ 13 & 17 & 5\\ 15 & 20 & 12\end{vmatrix} =2(204-100)-3(156-75)+7(260-255)$

$\displaystyle \Rightarrow A=2(104)-3(81)+7(5)$

$\displaystyle \Rightarrow A=208-243+35$

$\displaystyle \Rightarrow A=243-243=0$

$\displaystyle \therefore \begin{vmatrix}2 & 3 & 7\\ 13 & 17 & 5\\ 15 & 20 & 12\end{vmatrix}=0$

$\displaystyle \Rightarrow \left|\begin{matrix}2 & 3 & 7\\ 13 & 17 & 5\\ 15 & 20 & 12\end{matrix}\right|^{2} =0^{2}=0\;[\because\ \det A^{2}=(\det A)^{2}]$

$\\$

$\displaystyle \textbf{Question 4: } \text{ Show that } \left|\begin{matrix}\sin 10^\circ & -\cos 10^\circ\\ \sin 80^\circ & \cos 80^\circ\end{matrix}\right| = 1$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let }\Delta=\begin{vmatrix}\sin10^\circ & -\cos10^\circ\\ \sin80^\circ & \cos80^\circ\end{vmatrix}$

$\displaystyle \Rightarrow \Delta=\sin10^\circ\cos80^\circ+\cos10^\circ\sin80^\circ$

$\displaystyle =\sin10^\circ\cos(90^\circ-10^\circ)+\cos10^\circ\sin(90^\circ-10^\circ)\;[\because\ \cos\theta=\sin(90^\circ-\theta)]$

$\displaystyle \Rightarrow \Delta=\sin10^\circ\sin10^\circ+\cos10^\circ\cos10^\circ$

$\displaystyle =\sin^{2}10^\circ+\cos^{2}10^\circ\;[\because\ \sin^{2}\theta+\cos^{2}\theta=1]$

$\displaystyle \Rightarrow \Delta=1$

$\\$

$\displaystyle \textbf{Question 5:} \text{ Evaluate } \begin{vmatrix}2 & 3 & -5\\ 7 & 1 & -2\\ -3 & 4 & 1\end{vmatrix} \text{ by two methods.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let}$

$\displaystyle \Delta=\begin{vmatrix}2 & 3 & -5\\ 7 & 1 & -2\\ -3 & 4 & 1\end{vmatrix}$

$\displaystyle \text{First method}$

$\displaystyle \Delta=(-1)^{1+1}2(1+8)+(-1)^{1+2}3(7-6)+(-1)^{1+3}(-5)(28+3)$

$\displaystyle =2(1+8)-3(7-6)-5(28+3)$

$\displaystyle =18-3-155$

$\displaystyle =-140$

$\displaystyle \text{Second method is the Sarrus Method, where we adjoin the first two columns to the right to get}$

$\displaystyle \begin{vmatrix}2 & 3 & -5 & 2 & 3\\ 7 & 1 & -2 & 7 & 1\\ -3 & 4 & 1 & -3 & 4\end{vmatrix}$

$\displaystyle =(2\times1\times1+3\times(-2)\times(-3)+(-5)\times7\times4) -( (-5)\times1\times(-3)+2\times(-2)\times4+3\times7\times1 )$

$\displaystyle =(2+18-140)-(15-16+21)$

$\displaystyle =-120-20$

$\displaystyle =-140$

$\\$

$\displaystyle \textbf{Question 6:} \text{ Evaluate } \Delta = \begin{vmatrix}0 & \sin\alpha & -\cos\alpha\\ -\sin\alpha & 0 & \sin\beta\\ \cos\alpha & -\sin\beta & 0\end{vmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let}$

$\displaystyle \Delta=\begin{vmatrix} 0 & \sin\alpha & -\cos\alpha\\ -\sin\alpha & 0 & \sin\beta\\ \cos\alpha & -\sin\beta & 0 \end{vmatrix}$

$\displaystyle \Delta=(-1)^{1+1}0(0+\sin^{2}\beta)+(-1)^{1+2}\sin\alpha(0-\sin\beta\cos\alpha)+(-1)^{1+3}(-\cos\alpha)(\sin\alpha\sin\beta-0)\;[\text{Expanding along }R_{1}]$

$\displaystyle =0(0+\sin^{2}\beta)-\sin\alpha(0-\sin\beta\cos\alpha)-\cos\alpha(\sin\alpha\sin\beta-0)$

$\displaystyle =\sin\alpha\sin\beta\cos\alpha-\sin\alpha\sin\beta\cos\alpha$

$\displaystyle =0$

$\\$

$\displaystyle \textbf{Question 7: } \text{Evaluate } \Delta=\begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha\\ -\sin\beta & \cos\beta & 0\\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle \Delta=\begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha\\ -\sin\beta & \cos\beta & 0\\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix}$

$\displaystyle \Rightarrow \Delta=(-1)^{1+1}\cos\alpha\cos\beta(\cos\alpha\cos\beta-0) +(-1)^{1+2}\cos\alpha\sin\beta(-\sin\beta\cos\alpha-0) +(-1)^{1+3}(-\sin\alpha)(-\sin^{2}\beta\sin\alpha-\sin\alpha\cos^{2}\beta) \;[\text{Expanding along }R_{1}]$

$\displaystyle =\cos\alpha\cos\beta(\cos\alpha\cos\beta-0) -\cos\alpha\sin\beta(-\sin\beta\cos\alpha-0) -\sin\alpha(-\sin^{2}\beta\sin\alpha-\sin\alpha\cos^{2}\beta)$

$\displaystyle =\cos^{2}\alpha\cos^{2}\beta+\cos^{2}\alpha\sin^{2}\beta +\sin^{2}\alpha\sin^{2}\beta+\sin^{2}\alpha\cos^{2}\beta$

$\displaystyle =\cos^{2}\alpha(\cos^{2}\beta+\sin^{2}\beta) +\sin^{2}\alpha(\sin^{2}\beta+\cos^{2}\beta)$

$\displaystyle \Rightarrow \Delta=\cos^{2}\alpha+\sin^{2}\alpha \;[\because\ \sin^{2}\theta+\cos^{2}\theta=1]$

$\displaystyle \Rightarrow \Delta=1 \;[\because\ \sin^{2}\theta+\cos^{2}\theta=1]$

$\\$

$\displaystyle \textbf{Question 8: } \text{ If } A=\begin{bmatrix}2 & 5\\2 & 1\end{bmatrix} \text{ and} B=\begin{bmatrix}4 & -3\\2 & 5\end{bmatrix} \text{ verify that } |AB|=|A|\;|B|.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Consider LHS}$

$\displaystyle AB=\begin{bmatrix}2 & 5\\ 2 & 1\end{bmatrix} \begin{bmatrix}4 & -3\\ 2 & 5\end{bmatrix}$

$\displaystyle =\begin{bmatrix} 8+10 & -6+25\\ 8+2 & -6+5 \end{bmatrix} =\begin{bmatrix} 18 & 19\\ 10 & -1 \end{bmatrix}$

$\displaystyle |AB|=-18-190=-208$

$\displaystyle \text{Consider RHS}$

$\displaystyle |A|=2-10=-8$

$\displaystyle |B|=20-(-6)=26$

$\displaystyle |A||B|=-8\times26=-208$

$\displaystyle \therefore\ \text{LHS}=\text{RHS}$

$\\$

$\displaystyle \textbf{Question 9:} \text{ If } A=\begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}, \text{ then show that } |3A|=27\,|A|.$

$\displaystyle \text{Answer:}$

$\displaystyle A=\begin{bmatrix}1 & 0 & 1\\ 0 & 1 & 2\\ 0 & 0 & 4\end{bmatrix}$

$\displaystyle \Rightarrow 3A=\begin{bmatrix}3 & 0 & 3\\ 0 & 3 & 6\\ 0 & 0 & 12\end{bmatrix} \;[\text{Multiplying each element of }A\text{ by }3]$

$\displaystyle \Rightarrow |3A| =(-1)^{1+1}3(36-0)+(-1)^{1+2}0(0-0)+(-1)^{1+3}3(0-0)$
$\displaystyle =3(36-0)-0(0-0)+3(0-0) \;[\text{Expanding along }R_{1}]$

$\displaystyle =3\times36=108\ldots(1)$

$\displaystyle \Rightarrow |A| =(-1)^{1+1}1(4-0)+(-1)^{1+2}0(0-0)+(-1)^{1+3}1(0-0)$
$\displaystyle =1(4-0)-0(0-0)+1(0-0)=4 \;[\text{Expanding along }R_{1}]$

$\displaystyle \Rightarrow 27|A|=27\times4=108\ldots(2)$

$\displaystyle \therefore |3A|=27|A| \;[\text{From eqs. }(1)\text{ and }(2)]$

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$\displaystyle \textbf{Question 10: } \text{ Find the value of } x, \text{ if }$

$\displaystyle \text{(i) } \begin{vmatrix}2 & 4\\ 5 & 1\end{vmatrix} = \begin{vmatrix}2x & 4\\ 6 & x\end{vmatrix}$ $\displaystyle \text{(ii) } \begin{vmatrix}2 & 3\\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3\\ 2x & 5\end{vmatrix}$

$\displaystyle \text{(iii) } \begin{vmatrix}3 & x\\ x & 1\end{vmatrix} = \begin{vmatrix}3 & 2\\ 4 & 1\end{vmatrix}$ $\displaystyle \text{(iv) } \begin{vmatrix}3x & 7\\ 2 & 4\end{vmatrix} = 10$

$\displaystyle \text{(v) } \begin{vmatrix}x+1 & x-1\\ x-3 & x+2\end{vmatrix} = \begin{vmatrix}4 & -1\\ 1 & 3\end{vmatrix}$ $\displaystyle \text{(vi) } \begin{vmatrix}2x & 5\\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & 5\\ 8 & 3\end{vmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \text{Given: }\begin{vmatrix}2 & 4\\ 5 & 1\end{vmatrix} =\begin{vmatrix}2x & 4\\ 6 & x\end{vmatrix}$

$\displaystyle \Rightarrow 2-20=2x^{2}-24$

$\displaystyle \Rightarrow -18=2x^{2}-24$

$\displaystyle \Rightarrow 2x^{2}=6$

$\displaystyle \Rightarrow x^{2}=3$

$\displaystyle \Rightarrow x=\pm\sqrt{3}$

$\displaystyle \text{(ii) }$

$\displaystyle \text{Given: }\begin{vmatrix}2 & 3\\ 4 & 5\end{vmatrix} =\begin{vmatrix}x & 3\\ 2x & 5\end{vmatrix}$

$\displaystyle \Rightarrow 10-12=5x-6x$

$\displaystyle \Rightarrow -2=-x$

$\displaystyle \Rightarrow x=2$

$\displaystyle \text{(iii) }$

$\displaystyle \text{Given: }\begin{vmatrix}3 & x\\ x & 1\end{vmatrix} =\begin{vmatrix}3 & 2\\ 4 & 1\end{vmatrix}$

$\displaystyle \Rightarrow 3-x^{2}=3-8$

$\displaystyle \Rightarrow -x^{2}=-8$

$\displaystyle \Rightarrow x^{2}=8$

$\displaystyle \Rightarrow x=\pm2\sqrt{2}$

$\displaystyle \text{(iv) }$

$\displaystyle \text{Given: }\begin{vmatrix}3x & 7\\ 2 & 4\end{vmatrix}=10$

$\displaystyle \Rightarrow 12x-14=10$

$\displaystyle \Rightarrow 12x=24$

$\displaystyle \Rightarrow x=2$

$\displaystyle \text{(v) }$

$\displaystyle \text{Given: }\begin{vmatrix}x+1 & x-1\\ x-3 & x+2\end{vmatrix} =\begin{vmatrix}4 & -1\\ 1 & 3\end{vmatrix}$

$\displaystyle \Rightarrow (x+1)(x+2)-(x-3)(x-1)=12+1$

$\displaystyle \Rightarrow x^{2}+3x+2-x^{2}+4x-3=13$

$\displaystyle \Rightarrow 7x-1=13$

$\displaystyle \Rightarrow 7x=14$

$\displaystyle \Rightarrow x=2$

$\displaystyle \text{(vi) }$

$\displaystyle \text{Given: }\begin{vmatrix}2x & 5\\ 8 & x\end{vmatrix} =\begin{vmatrix}6 & 5\\ 8 & 3\end{vmatrix}$

$\displaystyle \Rightarrow 2x^{2}-40=18-40$

$\displaystyle \Rightarrow 2x^{2}=18$

$\displaystyle \Rightarrow x^{2}=9$

$\displaystyle \Rightarrow x=\pm3$

$\\$

$\displaystyle \textbf{Question 11: } \text{Find the integral value of } x \text{ if } \begin{vmatrix}x^{2} & x & 1\\ 0 & 2 & 1\\ 3 & 1 & 4\end{vmatrix} = 28$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: }\begin{vmatrix}x^{2} & x & 1\\ 0 & 2 & 1\\ 3 & 1 & 4\end{vmatrix}=28$

$\displaystyle \Rightarrow x^{2}(8-1)-x(0-3)+1(0-6)$

$\displaystyle \Rightarrow 8x^{2}-x^{2}+3x-6=28$

$\displaystyle \Rightarrow 7x^{2}+3x-6=28$

$\displaystyle \Rightarrow 7x^{2}+3x-34=0$

$\displaystyle \Rightarrow (7x+17)(x-2)=0$

$\displaystyle \Rightarrow x=2$

$\displaystyle \text{Integral value of }x\text{ is }2.\ \text{Thus,}$

$\displaystyle x=-\frac{17}{7}\text{ is not an integer.}$

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$\displaystyle \textbf{Question 12: } \text{For what value of } x \text{ is singular? }$

$\displaystyle \text{(i) } A=\begin{bmatrix}1+x & 7\\ 3-x & 8\end{bmatrix}$ $\displaystyle \text{(ii) } A=\begin{bmatrix}x-1 & 1 & 1\\ 1 & x-1 & 1\\ 1 & 1 & x-1\end{bmatrix}$