NCERT Solutions for Miscellaneous Exercise Chapter 4 Class 12 - Determinants

Question:2 Without expanding the determinant, prove that
$\begin{vmatrix} a &a^2 &bc \\ b& b^2 &ca \\ c & c^2 &ab \end{vmatrix}= \begin{vmatrix} 1 &a^2 &a^3 \\ 1 &b^2 &b^3 \\ 1 & c^2 &c^3 \end{vmatrix}$

Answer:

We have the

$L.H.S. = \begin{vmatrix} a &a^2 &bc \\ b& b^2 &ca \\ c & c^2 &ab \end{vmatrix}$

Multiplying rows with a, b, and c respectively.

$R_{1} \rightarrow aR_{1}, R_{2} \rightarrow bR_{2},\ and\ R_{3} \rightarrow cR_{3}$

we get;

$= \frac{1}{abc} \begin{vmatrix} a^2 &a^3 &abc \\ b^2& b^3 &abc \\ c^2& c^3 & abc \end{vmatrix}$

$= \frac{1}{abc}.abc \begin{vmatrix} a^2 &a^3 & 1\\ b^2& b^3 &1 \\ c^2& c^3 & 1 \end{vmatrix}$ $[after\ taking\ out\ abc\ from\ column\ 3].$

$= \begin{vmatrix} a^2 &a^3 & 1\\ b^2& b^3 &1 \\ c^2& c^3 & 1 \end{vmatrix} = \begin{vmatrix} 1&a^2 & a^3\\ 1& b^2 &b^3 \\ 1& c^2 & c^3 \end{vmatrix}$ $[Applying\ C_{1}\leftrightarrow C_{3}\ and\ C_{2} \leftrightarrow C_{3}]$

= R.H.S.

Hence proved. L.H.S. =R.H.S.

Question:3 Evaluate $\begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta &-\sin \alpha \\ -\sin \beta & \cos \beta &0 \\ \sin \alpha \cos \beta &\sin \alpha \sin \beta & \cos \alpha \end{vmatrix}$.

Answer:

Given determinant $\begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta &-\sin \alpha \\ -\sin \beta & \cos \beta &0 \\ \sin \alpha \cos \beta &\sin \alpha \sin \beta & \cos \alpha \end{vmatrix}$;

$= \cos \alpha \cos \beta \begin{vmatrix} \cos \beta &0 \\ \sin \alpha \sin \beta & \cos \alpha \end{vmatrix} - \cos \alpha \sin \beta \begin{vmatrix} -\sin \beta & 0 \\ \sin \alpha \cos \beta & \cos \alpha \end{vmatrix} -\sin \alpha \begin{vmatrix} -\sin \beta &\cos \beta \\ \sin \alpha \cos \beta& \sin \alpha \sin \beta \end{vmatrix}$$= \cos \alpha \cos \beta (\cos \beta \cos \alpha -0 )- \cos \alpha \sin \beta (-\cos \alpha\sin \beta- 0) -\sin \alpha (-\sin \alpha\sin^2\beta - \sin \alpha \cos^2 \beta)$

$= \cos^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta +\sin^2 \alpha \sin^2\beta + \sin^2 \alpha \cos^2 \beta$

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

$= \cos^2 \alpha(1) +\sin^2 \alpha(1) = 1$.

Question 4: If \( A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \), find \( (AB)^{-1} \).
Answer:

We know from the identity that:

$(AB)^{-1} = B^{-1}A^{-1}$

Then we can find easily,

Given

Given
$A^{-1} = \begin{bmatrix} 3 & -1 & 1 \ -15 & 6 & -5 \ 5 & -2 & 2 \end{bmatrix}$,
$B = \begin{bmatrix} 1 & 2 & -2 \ -1 & 3 & 0 \ 0 & -2 & 1 \end{bmatrix}$

Then we have to basically find the $B^{-1}$ matrix.

So, Given matrix $B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$

$|B| = 1(3-0) -2(-1-0)-2(2-0) = 3+2-4 = 1 \neq 0$

Hence its inverse $B^{-1}$ exists;

Now, as we know that

$B^{-1} = \frac{1}{|B|} adjB$

So, calculating cofactors of B,

$B_{11} = (-1)^{1+1}(3-0) = 3$ $B_{12} = (-1)^{1+2}(-1-0) = 1$

$B_{13} = (-1)^{1+3}(2-0) = 2$ $B_{21} = (-1)^{2+1}(2-4) = 2$

$B_{22} = (-1)^{2+2}(1-0) = 1$ $B_{23} = (-1)^{2+3}(-2-0) = 2$

$B_{31} = (-1)^{3+1}(0+6) = 6$ $B_{32} = (-1)^{3+2}(0-2) = 2$

$B_{33} = (-1)^{3+3}(3+2) = 5$

$adjB = \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}$

$B^{-1} = \frac{1}{|B|} adjB = \frac{1}{1} \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}$

Now, We have both $A^{-1}$ as well as $B^{-1}$ ;

Putting in the relation we know; $(AB)^{-1} = B^{-1}A^{-1}$

$(AB)^{-1}=\frac{1}{1} \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$

$= \begin{bmatrix} 9-30+30 &-3+12-12 &3-10+12 \\ 3-15+10&-1+6-4 &1-5+4 \\ 6-30+25 &-2+12-10 &2-10+10 \end{bmatrix}$

$= \begin{bmatrix} 9 &-3 &5 \\ -2&1 &0 \\ 1 &0 &2\end{bmatrix}$

Question 5(i): Let $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$. Verify that $100[\operatorname{adj} A]^{-1} = \operatorname{adj}(A^{-1})$.
Answer:

Given that $A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$;

So, let us assume that $A^{-1} = B$ matrix and $adjA = C$ then;

$|A| = 1(15-1) -2(10-1) +1(2-3) = 14-18-1 = -5 \neq 0$

Hence its inverse exists;

$A^{-1} = \frac{1}{|A|} adjA$ or $B = \frac{1}{|A|}C$;

so, we now calculate the value of $adjA$

Cofactors of A;

$A_{11}= (-1)^{1+1}(15-1) = 14$ $A_{12}= (-1)^{1+2}(10-1) = -9$

$A_{13}= (-1)^{1+3}(2-3) = -1$ $A_{21}= (-1)^{2+1}(10-1) = -9$

$A_{22}= (-1)^{2+2}(5-1) = 4$ $A_{23}= (-1)^{2+3}(1-2) = 1$

$A_{31}= (-1)^{3+1}(2-3) = -1$ $A_{32}= (-1)^{3+2}(1-2) = 1$

$A_{33}= (-1)^{3+3}(3-4) = -1$

$\Rightarrow adjA =C= \begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

$A^{-1} =B= \frac{1}{|A|} adjA = \frac{1}{-5}\begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

Finding the inverse of C;

$|C| = 14(-4-1)+9(9+1)-1(-9+4) = -70+90+5 = 25 \neq 0$

Hence its inverse exists;

$C^{-1} = \frac{1}{|C|}adj C$

Now, finding the $adjC$;

$C_{11}= (-1)^{1+1}(-4-1) = -5$ $C_{12}= (-1)^{1+2}(9+1) = -10$

$C_{13}= (-1)^{1+3}(-9+4) = -5$ $C_{21}= (-1)^{2+1}(9+1) = -10$

$C_{22}= (-1)^{2+2}(-14-1) = -15$ $C_{23}= (-1)^{2+3}(14-9) = -5$

$C_{31}= (-1)^{3+1}(-9+4) = -5$ $C_{32}= (-1)^{3+2}(14-9) = -5$

$C_{33}= (-1)^{3+3}(56-81) = -25$

$adjC = \begin{bmatrix} -5 &-10 &-5 \\ -10& -15 & -5\\ -5& -5& -25 \end{bmatrix}$

$C^{-1} = \frac{1}{|C|}adjC = \frac{1}{25}\begin{bmatrix} -5 &-10 &-5 \\ -10& -15 & -5\\ -5& -5& -25 \end{bmatrix} = \begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

or $L.H.S. = C^{-1} = [adjA]^{-1} = \begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

Now, finding the R.H.S.

$adj (A^{-1}) = adj B$

$A^{-1} =B= \begin{bmatrix} \frac{-14}{5} &&\frac{9}{5} &&\frac{1}{5} \\ \\ \frac{9}{5}&& \frac{-4}{5}&& \frac{-1}{5}\\ \\ \frac{1}{5}&& \frac{-1}{5} &&\frac{1}{5}\end{bmatrix}$

Cofactors of B;

$B_{11}= (-1)^{1+1}(\frac{-4}{25}-\frac{1}{25}) = \frac{-1}{5}$

$B_{12}= (-1)^{1+2}(\frac{9}{25}+\frac{1}{25}) =- \frac{2}{5}$

$B_{13}= (-1)^{1+3}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{21}= (-1)^{2+1}(\frac{9}{25}+\frac{1}{25}) = -\frac{2}{5}$

$B_{22}= (-1)^{2+2}(\frac{-14}{25}-\frac{1}{25}) = \frac{-3}{5}$

$B_{23}= (-1)^{2+3}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{31}= (-1)^{3+1}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{32}= (-1)^{3+2}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{33}= (-1)^{3+3}(\frac{56}{25}-\frac{81}{25}) = -1$

$R.H.S. = adjB = adj(A^{-1}) =\begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

Hence L.H.S. = R.H.S. proved.

Question 5(ii):Let $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$, verify that $(A^{-1})^{-1} = A$.

Answer:

Given that $A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$;

So, let us assume that $A^{-1} = B$

$|A| = 1(15-1) -2(10-1) +1(2-3) = 14-18-1 = -5 \neq 0$

Hence its inverse exists;

$A^{-1} = \frac{1}{|A|} adjA$ or $B = \frac{1}{|A|}C$;

so, we now calculate the value of $adjA$

Cofactors of A;

$A_{11}= (-1)^{1+1}(15-1) = 14$ $A_{12}= (-1)^{1+2}(10-1) = -9$

$A_{13}= (-1)^{1+3}(2-3) = -1$ $A_{21}= (-1)^{2+1}(10-1) = -9$

$A_{22}= (-1)^{2+2}(5-1) = 4$ $A_{23}= (-1)^{2+3}(1-2) = 1$

$A_{31}= (-1)^{3+1}(2-3) = -1$ $A_{32}= (-1)^{3+2}(1-2) = 1$

$A_{33}= (-1)^{3+3}(3-4) = -1$

$\Rightarrow adjA =C= \begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

$A^{-1} =B= \frac{1}{|A|} adjA = \frac{1}{-5}\begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix} = \begin{bmatrix} \frac{-14}{5} &&\frac{9}{5} &&\frac{1}{5} \\ \\ \frac{9}{5} && \frac{-4}{5} &&\frac{-1}{5} \\ \\ \frac{1}{5} &&\frac{-1}{5} &&\frac{1}{5} \end{bmatrix}$

Finding the inverse of B ;

$|B| = \frac{-14}{5}(\frac{-4}{25}-\frac{1}{25})-\frac{9}{5}(\frac{9}{25}+\frac{1}{25})+\frac{1}{5}(\frac{-9}{25}+\frac{4}{25})$

$= \frac{70}{125}-\frac{90}{125}-\frac{5}{125} = \frac{-25}{125} = \frac{-1}{5} \neq 0$

Hence its inverse exists;

$B^{-1} = \frac{1}{|B|}adj B$

Now, finding the $adjB$;

$A^{-1} =B= \frac{1}{|A|} adjA = \frac{1}{-5}\begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix} = \begin{bmatrix} \frac{-14}{5} &&\frac{9}{5} &&\frac{1}{5} \\ \\ \frac{9}{5} && \frac{-4}{5} &&\frac{-1}{5} \\ \\ \frac{1}{5} &&\frac{-1}{5} &&\frac{1}{5} \end{bmatrix}$

$B_{11}= (-1)^{1+1}(\frac{-4}{25}-\frac{1}{25}) = \frac{-1}{5}$ $B_{12}= (-1)^{1+2}(\frac{9}{25}+\frac{1}{25}) = \frac{-2}{5}$

$B_{13}= (-1)^{1+3}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$ $B_{21}= (-1)^{2+1}(\frac{9}{25}+\frac{1}{25}) = -\frac{2}{5}$

$B_{22}= (-1)^{2+2}(\frac{-14}{25}-\frac{1}{25}) = \frac{-3}{5}$ $B_{23}= (-1)^{2+3}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{31}= (-1)^{3+1}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{32}= (-1)^{3+2}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{33}= (-1)^{3+3}(\frac{56}{25}-\frac{81}{25}) = \frac{-25}{25} =-1$

$adjB = \begin{bmatrix} \frac{-1}{5} &&\frac{-2}{5} &&\frac{-1}{5} \\ \\\frac{-2}{5}&& \frac{-3}{5} && \frac{-1}{5}\\ \\ \frac{-1}{5}&& \frac{-1}{5}&& -1 \end{bmatrix}$

$B^{-1} = \frac{1}{|B|}adjB = \frac{-5}{1}\begin{bmatrix} \frac{-1}{5} &&\frac{-2}{5} &&\frac{-1}{5} \\ \\\frac{-2}{5}&& \frac{-3}{5} && \frac{-1}{5}\\ \\ \frac{-1}{5}&& \frac{-1}{5}&& -1 \end{bmatrix}= \begin{bmatrix} 1 &2 &1 \\ 2& 3& 1\\ 1 & 1 &5 \end{bmatrix}$

$L.H.S. = B^{-1} = (A^{-1})^{-1} = \begin{bmatrix} 1 &2 &1 \\ 2& 3& 1\\ 1 & 1 &5 \end{bmatrix}$

$R.H.S. = A = \begin{bmatrix} 1 &2 &1 \\ 2& 3& 1\\ 1& 1 &5 \end{bmatrix}$

Hence proved L.H.S. =R.H.S..

Question:6 Evaluate $\begin{vmatrix} x & y &x+y \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Answer:

We have determinant $\triangle = \begin{vmatrix} x & y &x+y \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Applying row transformations; $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ , we have then;

$\triangle = \begin{vmatrix} 2(x+y) & 2(x+y) &2(x+y) \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Taking out the common factor 2(x+y) from the row first.

$= 2(x+y)\begin{vmatrix} 1 & 1 &1 \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Now, applying the column transformation; $C_{1} \rightarrow C_{1} - C_{2}$ and $C_{2} \rightarrow C_{2} - C_{1}$ we have ;

$= 2(x+y)\begin{vmatrix} 0 & 0 &1 \\ -x & y &x \\ y & x-y & y \end{vmatrix}$

Expanding the remaining determinant;

$= 2(x+y)(-x(x-y)-y^2) = 2(x+y)[-x^2+xy-y^2]$

$= -2(x+y)[x^2-xy+y^2] = -2(x^3+y^3)$.

Question:7 Evaluate $\begin{vmatrix} 1 & x &y \\ 1 &x+y &y \\ 1 &x &x+y \end{vmatrix}$

Answer:

We have determinant $\triangle = \begin{vmatrix} 1 & x &y \\ 1 &x+y &y \\ 1 &x &x+y \end{vmatrix}$

Applying row transformations; $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{2} \rightarrow R_{2}-R_{3}$ then we have then;

$\triangle = \begin{vmatrix} 0 & -y &0 \\ 0 &y &-x \\ 1 &x &x+y \end{vmatrix}$

Taking out the common factor -y from the row first.

$\triangle = -y\begin{vmatrix} 0 & 1 &0 \\ 0 &y &-x \\ 1 &x &x+y \end{vmatrix}$

Expanding the remaining determinant;

$-y[1(-x-o)] = xy$

Question:8 Solve the system of equations

$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$

$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

Answer:

We have a system of equations;

$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$

$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

So, we will convert the given system of equations in a simple form to solve the problem by the matrix method;

Let us take, $\frac{1}{x} = a$, $\frac{1}{y} = b\ and\ \frac{1}{z} = c$

Then we have the equations;

$2a +3b+10c = 4$

$4a-6b+5c =1$

$6a+9b-20c = 2$

We can write it in the matrix form as $AX =B$ , where

$A= \begin{bmatrix} 2 &3 &10 \\ 4& -6 & 5\\ 6 & 9 & -20 \end{bmatrix} , X = \begin{bmatrix} a\\b \\c \end{bmatrix}\ and\ B = \begin{bmatrix} 4\\1 \\ 2 \end{bmatrix}.$

Now, Finding the determinant value of A;

$|A| = 2(120-45)-3(-80-30)+10(36+36)$

$=150+330+720$

$=1200 \neq 0$

Hence we can say that A is non-singular $\therefore$ its invers exists;

Finding cofactors of A;

$A_{11} = 75$ , $A_{12} = 110$, $A_{13} = 72$

$A_{21} = 150$, $A_{22} = -100$, $A_{23} = 0$

$A_{31} =75$, $A_{31} =30$, $A_{33} =-24$

$\therefore$ as we know $A^{-1} = \frac{1}{|A|}adjA$

$= \frac{1}{1200} \begin{bmatrix} 75 &150 &75 \\ 110& -100& 30\\ 72&0 &-24 \end{bmatrix}$

Now we will find the solutions by relation $X = A^{-1}B$.

$\Rightarrow \begin{bmatrix} a\\b \\ c \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 75 &150 &75 \\ 110& -100& 30\\ 72&0 &-24 \end{bmatrix}\begin{bmatrix} 4\\1 \\ 2 \end{bmatrix}$

$= \frac{1}{1200}\begin{bmatrix} 300+150+150\\440-100+60 \\ 288+0-48 \end{bmatrix}$

$= \frac{1}{1200}\begin{bmatrix} 600\\400\\ 240 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}\\ \\ \frac{1}{3} \\ \\ \frac{1}{5} \end{bmatrix}$

Therefore we have the solutions $a = \frac{1}{2},\ b= \frac{1}{3},\ and\ c = \frac{1}{5}.$

Or in terms of x, y, and z;

$x =2,\ y =3,\ and\ z = 5$

Question 9: Choose the correct answer.

If $x$, $y$, $z$ are nonzero real numbers, then the inverse of the matrix

$A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$

is

(A) $ \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} $

(B) $ xyz \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} $

(C) $ \dfrac{1}{xyz} \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} $

(D) $ \dfrac{1}{xyz} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $

Answer:

Given Matrix $A=\begin{bmatrix} x &0 &0 \\ 0 &y &0 \\ 0 & 0 & z \end{bmatrix}$,

$|A| = x(yz-0) =xyz$

As we know,

$A^{-1} = \frac{1}{|A|}adjA$

So, we will find the $adjA$,

Determining its cofactor first,

$A_{11} = yz$ $A_{12} = 0$ $A_{13} = 0$

$A_{21} = 0$ $A_{22} = xz$ $A_{23} = 0$

$A_{31} = 0$ $A_{32} = 0$ $A_{33} = xy$

Hence $A^{-1} = \frac{1}{|A|}adjA = \frac{1}{xyz}\begin{bmatrix} yz &0 &0 \\ 0& xz & 0\\ 0& 0& xy \end{bmatrix}$

$A^{-1} = \begin{bmatrix} \frac{1}{x} &&0 &&0 \\ 0&& \frac{1}{y} && 0\\ 0&& 0&& \frac{1}{z} \end{bmatrix}$

Therefore, the correct answer is (A)

Question:10 Choose the correct answer.

Let $A=\begin{vmatrix} 1 &\sin \theta &1 \\ -\sin \theta & 1 & \sin \theta \\ -1&-\sin \theta &1 \end{vmatrix},$ where $0\leq \theta \leq 2\pi$. Then

(A)$Det(A)=0$ nbsp; (B) $Det(A)\in (2,\infty)$

(C) $Det(A)\in (2,4)$ (D)$Det(A)\in [2,4]$

Answer:

Given determinant $A=\begin{vmatrix} 1 &\sin \theta &1 \\ -\sin \theta & 1 & \sin \theta \\ -1&-\sin \theta &1 \end{vmatrix}$

$|A| = 1(1+\sin^2 \Theta) -\sin \Theta(-\sin \Theta+\sin \Theta)+1(\sin^2 \Theta +1)$

$= 1+ \sin ^2 \Theta + \sin ^2 \Theta +1$

$= 2+2\sin ^2 \Theta = 2(1+\sin^2 \Theta)$

Now, given the range of $\Theta$ from $0\leq \Theta \leq 2\pi$

$\Rightarrow 0 \leq \sin \Theta \leq 1$

$\Rightarrow 0 \leq \sin^2 \Theta \leq 1$

$\Rightarrow 1 \leq 1+\sin^2 \Theta \leq 2$

$\Rightarrow 2 \leq 2(1+\sin^2 \Theta) \leq 4$

Therefore the $|A|\ \epsilon\ [2,4]$.

Hence, the correct answer is D.