1   Evaluate the determinants in Exercise $1$ and $2.$ $ \begin{vmatrix} 2 & 4 \\[0.3em] -5 & -1 \end{vmatrix}$

Solution :

$ \begin{vmatrix} 2 & 4 \\[0.3em] -5 & -1 \end{vmatrix}$ =$2(-1)-4(-5) =-2+20=18$

2   Evaluate the determinants in Exercise $1$ and $2.$$\\$ $(i) \begin{vmatrix} \cos \theta & -\sin \theta\\[0.3em] \sin \theta & \cos \theta \end{vmatrix}$ $\ \ \ \ $ $(ii) \begin{vmatrix} x^2-x+1 & x-1\\[0.3em] x+1 & x+1 \end{vmatrix}$

Solution :

$(ii) \begin{vmatrix} x^2-x+1 & x-1\\[0.3em] x+1 & x+1 \end{vmatrix}$$\\$ $=(x^2-x+1)(x+1)-(x-1)(x+1)\\ =x^3-x^2+x+x^2-x+1-(x^2-1)\\ =x^3+1-x^2+1\\ =x^3-x^2+2$

$(i) \begin{vmatrix} \cos \theta & -\sin \theta\\[0.3em] \sin \theta & \cos \theta \end{vmatrix}$ = $(\cos \theta )(\cos \theta ) -(-\sin \theta)(\sin \theta ) =\cos^2\theta + \sin ^\theta =1$

3   If $ A=\begin{bmatrix} 1 &2 \\[0.3em] 4 &2 \end{bmatrix}$ ,then show that $ |2A|=4|A|$

Solution :

The given matrix is $A=\begin{bmatrix} 1 &2 \\[0.3em] 4 &2 \end{bmatrix}$$\\$ $\therefore 2A =2\begin{bmatrix} 1 &2 \\[0.3em] 4 &2 \end{bmatrix}=\begin{bmatrix} 2&4 \\[0.3em] 8&4 \end{bmatrix}\\ L.H.S:|2A|=\begin{vmatrix} 2&4 \\[0.3em] 8& 4 \end {vmatrix} \\ =2*4-4*8\\ =8-32=-24$$\\$ Now , $|A|=\begin{bmatrix} 1& 2 \\[0.3em] 4 &2 \end {bmatrix} =1*2-2*4=2*8=-6\\ R>H>S:4|A|=4*(-6) =-24\\ \therefore L.H.S.=\therefore R.H.S$

4   $A=\begin{bmatrix} 1 &0 &1 \\[0.3em] 0& 1& 2 \\[0.3em] 0 & 0 & 4 \end{bmatrix}$ , then show that $|3A|=27|A|.$

Solution :

The given matrix is $A=\begin {bmatrix} 1&0 &1 \\[0.3em] 0& 1& 2 \\[0.3em] 0 & 0 & 4 \end{bmatrix}$.$\\$ It can be observed that in the first column, two entries are zero. Thus, we expand along the first column $(C 1 )$ for easier calculation.$\\$ $ |A|=1\begin{vmatrix} 1 & 2\\[0.3em] 0 & 4 \end{vmatrix}$ $ -0 \begin{vmatrix} 0 &1 \\[0.3em] 0 &4 \end{vmatrix}$ $ +0 \begin{vmatrix} 0 & 1 \\[0.3em] 1 &2 \end{vmatrix} $ $=1(4-0)-0+0=4\\ \therefore 27|A|=27(4)=108 .....(i)$ $\\$ Now , $3A=3\begin{bmatrix} 1&0 & 1 \\[0.3em] 0 & 1 & 2 \\[0.3em] 0 & 0 & 4 \end {bmatrix}$ $\\$ $\therefore |3A|=3 \begin{vmatrix} 3 & 6\\[0.3em] 0 & 12 \end{vmatrix}$ $-0 \begin{vmatrix} 0 & 3 \\[0.3em] 0& 12 \end{vmatrix}$ $+0 \begin{vmatrix} 0 & 3 \\[0.3em] 3 & 6 \end{vmatrix}$$ \\$$ =3(36-0)=3(36)=108 ....(ii)$$\\$ From equations (i) and (ii), we have: $|3 A = 27 A|$$\\$ Hence, the given result is proved.

5   Evaluate the determinants $\\$ $ (i) \begin{vmatrix} 3 & -1 & -2 \\[0.3em] 0 & 0 & -1 \\[0.3em] 3 & -5 & 0 \end{vmatrix}$ $ \ \ \ $ $ (ii) \begin{vmatrix} 3 & -4 & 5 \\[0.3em] 1 & 1 & -2 \\[0.3em] 2 & 3 & 1 \end{vmatrix}$ $ \ \ \ $ $(iii) \begin{vmatrix} 0 & 1 & 2 \\[0.3em] -1 & 0 & -3 \\[0.3em] -2 & 3 & 0\end{vmatrix}$ $(iv) \begin{vmatrix} 2 & -1 & -2 \\[0.3em] 0 & 2 & -1 \\[0.3em] 3 & -5 & 0 \end{vmatrix}$

Solution :

(ii)Let $ \begin{vmatrix} 3 & -4 & 5 \\[0.3em] 1 & 1 & -2 \\[0.3em] 2 & 3 & 1 \end{vmatrix}$$\\$ By expanding along the first row, we have:$\\$ $|A| =3\begin{vmatrix} 1 & -2 \\[0.3em] 3 & 1\end{vmatrix}$ $+4\begin{vmatrix} 1 & -2 \\[0.3em] 2 & 1 \end{vmatrix}$ $+5\begin{vmatrix} 1 & 1 \\[0.3em] 2 & 3 \end{vmatrix} $$\\$ $ =3(1+6)+4(1+4)+5(3-2) \\ =3(7)+4(5)+5(1)\\ 21+20+5=46$

(iv) Let $ A=\begin{vmatrix} 2 & -1 & -2 \\[0.3em] 0 & 2 & -1 \\[0.3em] 3 & -5 & 0 \end{vmatrix}$$\\$ By expanding along the first column, we have:$\\$ $|A|=2\begin{vmatrix} 2 & -1\\[0.3em] -5 & 0\end{vmatrix} -0\begin{vmatrix} -1 & -2 \\[0.3em] -5 &0 \end{vmatrix}+3\begin{vmatrix} -1 & -2 \\[0.3em] 2 & -1 \end{vmatrix} $$\\$ $ =2(0-5) -0+3(1+4)\\ =-10+15=5$

$ (i) \begin{vmatrix} 3 & -1 & -2 \\[0.3em] 0 & 0 & -1 \\[0.3em] 3 & -5 & 0 \end{vmatrix}$$\\$ It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.$\\$ $|A|=-0\begin{vmatrix} -1 & -2 \\[0.3em] -5 & 0\end{vmatrix}$ $ +0\begin{vmatrix} 3 & -2 \\[0.3em] 3 & 0\end{vmatrix}$ $ -(-1)\begin{vmatrix} 3 & -1 \\[0.3em] 3 & -5\end{vmatrix}$ $=(-15+3) =-12$

(iii) Let $A=\begin{vmatrix} 0 & 1 & 2 \\[0.3em] -1 & 0 & -3 \\[0.3em] -2 & 3 & 0\end{vmatrix}$$\\$ By expanding along the first row, we have:$\\$ $|A|=0\begin{vmatrix} 0 & -3 \\[0.3em] 3 & 0\end{vmatrix} -1\begin{vmatrix} -1 & -3 \\[0.3em] -2 & 0\end{vmatrix} +2\begin{vmatrix} -1 & 0\\[0.3em] -2 & 3 \end{vmatrix}$$\\$ $ =0-1(0-6)+2(-3-0)\\ =-1(-6)+2(-3)\\ =6-6=0$

6   If $A=\begin{vmatrix} 1&1&-2\\[0.3em] 2&1&-3\\[0.3em] 5&4&-9 \end{vmatrix}$,find $|A|$

Solution :

Let $A=\begin{bmatrix} 1&1&-2\\[0.3em] 2&1&-3\\[0.3em] 5&4&-9 \end{bmatrix}$$\\$ By expanding along the first row, we have:$\\$ $A=\begin{bmatrix} 1&1&-2\\[0.3em] 2&1&-3\\[0.3em] 5&4&-9 \end{bmatrix}$$\\$ $|A|=1\begin{vmatrix} 1 & -3\\[0.3em] 4 & -9 \end{vmatrix}-1\begin{vmatrix} 2 & -3\\[0.3em] 5 & -9 \end{vmatrix}-2\begin{vmatrix} 2 & 1 \\[0.3em] 5 & 4 \end{vmatrix}$$\\$ $=1(-9+12)-1(-18+15)-2(8-5)\\ =1(3)-1(-3)-2(3)\\ =3+3-6\\ =6-6\\ =0$$\\$$\\$$\\$

7   Find values of x,if$\\$ (i)$\begin{vmatrix} 2& 4\\[0.3em] 5 & 1\end{vmatrix}=\begin{vmatrix} 2x& 4\\[0.3em] 6 & x\end{vmatrix} \ \ \ \ $(ii) $\begin{vmatrix} 2& 3\\[0.3em] 4 & 5\end{vmatrix}=\begin{vmatrix} x& 3\\[0.3em] 2x & 5\end{vmatrix}$

Solution :

(i)$\begin{vmatrix} 2& 4\\[0.3em] 5 & 1\end{vmatrix}=\begin{vmatrix} 2x& 4\\[0.3em] 6 & x\end{vmatrix} \ \ \ \ $$\\$ $\Rightarrow 2*1-5*4=2x*x-6*4\\ \Rightarrow 2-20=2x^2-24\\ \Rightarrow 2x^2=6\\ \Rightarrow x^2=3\\ \Rightarrow x=\pm3$$\\$ (ii) $\begin{vmatrix} 2& 3\\[0.3em] 4 & 5\end{vmatrix}=\begin{vmatrix} x& 3\\[0.3em] 2x & 5\end{vmatrix}$ $\\$ $\Rightarrow 2*5-3*4=x*5-3*2x\\ \Rightarrow10-12=5x-6x\\ \Rightarrow-2=-x\\ \Rightarrow x=2$$\\$

8   If$\begin{vmatrix} x & 2\\[0.3em] 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2\\[0.3em] 18 & 6 \end{vmatrix}$,then x is equal to $\\$

Solution :

$\begin{vmatrix} x & 2\\[0.3em] 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2\\[0.3em] 18 & 6 \end{vmatrix}$ $\Rightarrow x^2-36=36-36\\ \Rightarrow x^2-36=0\\ \Rightarrow x^2=36\\ \Rightarrow x=\pm6$$\\$ Hence, the correct answer is B.

9   Using the property of determinants and without expanding, prove that:$\\$ $\begin{vmatrix} x&a&x+a \\[0.3em] y & b& y+b \\[0.3em] z & c & Z+c \end{vmatrix}=0$

Solution :

$\begin{vmatrix} x&a&x+a \\[0.3em] y & b& y+b \\[0.3em] z & c & z+c \end{vmatrix}=\begin{vmatrix} x&a&x \\[0.3em] y & b& y \\[0.3em] z & c & z \end{vmatrix}+\begin{vmatrix} x&a&a \\[0.3em] y & b& b \\[0.3em] z & c & c \end{vmatrix}$$\\$ Clearly, the two determinants have two identical columns. Thus,$\\$ $=0+0=0$

10   Using the property of determinants and without expanding, prove that:$\\$ $\begin{vmatrix} a-b &b-c & c-a \\[0.3em] b-c &c-a & a-b \\[0.3em] c-a & a-b & b-c\end{vmatrix}=0$

Solution :

$\Delta = $ $\begin{vmatrix} a-b &b-c & c-a \\[0.3em] b-c &c-a & a-b \\[0.3em] c-a & a-b & b-c\end{vmatrix}=0$ $\\$ Applying $R _1 \to R _1 + R_ 2$ , we have:$\\$ $\Delta =\begin{vmatrix} a-c & b-a & c-b\\[0.3em] b-c & c-a & a-b\\[0.3em] -(a-c)& -(b-a) & -(c-b)\end{vmatrix}$$\\$ $=-\begin{vmatrix} a-c & b-a & c-b \\[0.3em] b-c & c-a & a-b \\[0.3em] a-c & b-a & c-b\end{vmatrix}$$\\$ Here, the two rows $R _1$ and $R _3$ are identical.$\\$ $\therefore \Delta =0$$\\$