NCERT Solutions for Miscellaneous Exercise Chapter 4 Class 12 - Determinants

As the name suggests miscellaneous exercise consists of mixed questions from all other exercises of the chapter. In NCERT solutions for Class 12 Maths chapter 4 miscellaneous exercise, you will get questions like solving determinants using cofactor expansion, solving determinants using properties, solving system of linear equation, checking the consistency of the system of linear equations, etc. In this Class 12 Maths chapter 4 miscellaneous solutions, you will get some difficult questions as compared to previous exercises. So, If you are not able to solve these questions at first by yourself, you don't need to be a worry. Over 95% of the questions in the board exams are not asked from Class 12 Maths chapter 4 miscellaneous exercise. You can Check Class 12 Maths chapter 4 miscellaneous exercise solutions in this article. Check here for NCERT Solutions.

Also, see

Determinants Exercise 4.1
Determinants Exercise 4.2
Determinants Exercise 4.3
Determinants Exercise 4.4
Determinants Exercise 4.5
Determinants Exercise 4.6

Determinants Miscellaneous Exercise

Question:1 Prove that the determinant $\begin{vmatrix} x & \sin \theta &\cos \theta \\ -\sin \theta &-x & 1\\ \cos \theta &1 &x \end{vmatrix}$ is independent of $\theta$ .

Answer:

Calculating the determinant value of $\begin{vmatrix} x & \sin \theta &\cos \theta \\ -\sin \theta &-x & 1\\ \cos \theta &1 &x \end{vmatrix}$ ;

$= x\begin{bmatrix} -x &1 \\ 1& x \end{bmatrix}-\sin \Theta\begin{bmatrix} -\sin \Theta &1 \\ \cos \Theta& x \end{bmatrix} + \cos \Theta \begin{bmatrix} -\sin \Theta &-x \\ \cos \Theta& 1 \end{bmatrix}$

$= x(-x^2-1)-\sin \Theta (-x\sin \Theta-\cos \Theta)+\cos\Theta(-\sin \Theta+x\cos\Theta)$

$= -x^3-x+x\sin^2 \Theta+ \sin \Theta\cos \Theta-\cos\Theta\sin \Theta+x\cos^2\Theta$

$= -x^3-x+x(\sin^2 \Theta+\cos^2\Theta)$

$= -x^3-x+x = -x^3$

Clearly, the determinant is independent of $\Theta$ .

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 $a,b$ and $c$ are real numbers, and

$\Delta =\begin{vmatrix} b+c & c+a &a+b \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}=0$

Show that either $a+b+c=0$ or $a=b=c$

Answer:

We have given $\Delta =\begin{vmatrix} b+c & c+a &a+b \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}=0$

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

$\Delta =\begin{vmatrix} 2(a+b+c) & 2(a+b+c) &2(a+b+c) \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}$

Taking out common factor 2(a+b+c) from the first row;

$\Delta =2(a+b+c)\begin{vmatrix} 1 & 1 &1 \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}$

Now, applying the column transformations; $C_{1}\rightarrow C_{1} - C_{2}\ and\ C_{2} \rightarrow C_{2}- C_{3}$

we have;

$=2(a+b+c)\begin{vmatrix} 0 & 0 &1 \\ c-b &a-c &b+c \\ a-c & b-a & c+a \end{vmatrix}$

$=2(a+b+c)[(c-b)(b-a)-(a-c)^2]$

$=2(a+b+c)[ab+bc+ca-a^2-b^2-c^2]$

and given that the determinant is equal to zero. i.e., $\triangle = 0$ ;

$(a+b+c)[ab+bc+ca-a^2-b^2-c^2] = 0$

So, either $(a+b+c) = 0$ or $[ab+bc+ca-a^2-b^2-c^2] = 0$ .

we can write $[ab+bc+ca-a^2-b^2-c^2] = 0$ as;

$\Rightarrow -2ab-2bc-2ca+2a^2+2b^2+2c^2 =0$

$\Rightarrow (a-b)^2+(b-c)^2+(c-a)^2 =0$

$\because (a-b)^2,(b-c)^2,\ and\ (c-a)^2$ are non-negative.

Hence $(a-b)^2= (b-c)^2=(c-a)^2 = 0$ .

we get then $a=b=c$

Therefore, if given $\triangle$ = 0 then either $(a+b+c) = 0$ or $a=b=c$ .

Question:5 Solve the equation

$\begin{vmatrix} x+a & x &x \\ x &x+a &x\\ x & x & x+a \end{vmatrix}=0; a\neq 0$

Answer:

Given determinant $\begin{vmatrix} x+a & x &x \\ x &x+a &x\\ x & x & x+a \end{vmatrix}=0; a\neq 0$

Applying the row transformation; $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ we have;

$\begin{vmatrix} 3x+a & 3x+a &3x+a \\ x &x+a &x\\ x & x & x+a \end{vmatrix} =0$

Taking common factor (3x+a) out from first row.

$(3x+a)\begin{vmatrix} 1 & 1 &1 \\ x &x+a &x\\ x & x & x+a \end{vmatrix} =0$

Now applying the column transformations; $C_{1} \rightarrow C_{1}-C_{2}$ and $C_{2} \rightarrow C_{2}-C_{3}$ .

we get;

$(3x+a)\begin{vmatrix} 0 & 0 &1 \\ -a &a &x\\ 0 & -a & x+a \end{vmatrix} =0$

$=(3x+a)(a^2)=0$ as $a^2 \neq 0$ ,

or $3x+a=0$ or $x= -\frac{a}{3}$

Question:6 Prove that $\begin{vmatrix} a^2 &bc &ac+a^2 \\ a^2+ab & b^2 & ac\\ ab &b^2+bc &c^2 \end{vmatrix}=4a^2b^2c^2$ .

Answer:

Given matrix $\begin{vmatrix} a^2 &bc &ac+a^2 \\ a^2+ab & b^2 & ac\\ ab &b^2+bc &c^2 \end{vmatrix}$

Taking common factors a,b and c from the column $C_{1}, C_{2}, and\ C_{3}$ respectively.

we have;

$\triangle = abc\begin{vmatrix} a &c &a+c \\ a+b & b & a\\ b &b+c &c \end{vmatrix}$

Applying $R_{2} \rightarrow R_{2}-R_{1}\ and\ R_{3} \rightarrow R_{3} - R_{1}$ , we have;

$\triangle = abc\begin{vmatrix} a &c &a+c \\ b & b-c &-c\\ b-a &b &-a \end{vmatrix}$

Then applying $R_{2} \rightarrow R_{2}+R_{1}$ , we get;

$\triangle = abc\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ b-a &b &-a \end{vmatrix}$

Applying $R_{3} \rightarrow R_{3}+R_{2}$ , we have;

$\triangle = abc\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ 2b &2b &0 \end{vmatrix} = 2ab^2c\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ 1 &1 &0 \end{vmatrix}$

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

$\triangle = 2ab^2c\begin{vmatrix} a &c-a &a+c \\ a+b & -a &a\\ 1 &0 &0 \end{vmatrix}$

So we can now expand the remaining determinant along $R_{3}$ we have;

$\triangle = 2ab^2c\left [ a(c-a)+a(a+c) \right ]$

$= 2ab^2c\left [ ac-a^2+a^2+ac) \right ] = 2ab^2c\left [ 2ac \right ]$

$= 4a^2b^2c^2$

Hence proved.

Question:7 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 $A^-^1=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$ and $B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$

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:8(i) Let $A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$ . Verify that,

$\dpi{100} [adj A]^-^1 = adj (A^-^1)$

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:8(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 $adjB$ ;

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

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

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

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

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

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

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

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

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

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

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

Question:9 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:10 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:11 Using properties of determinants, prove that

$\begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ \beta & \beta ^2 &\gamma +\alpha \\ \gamma &\gamma ^2 &\alpha +\beta \end{vmatrix}=(\beta -\gamma )(\gamma -\alpha )(\alpha -\beta )(\alpha +\beta +\gamma )$

Answer:

Given determinant $\triangle = \begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ \beta & \beta ^2 &\gamma +\alpha \\ \gamma &\gamma ^2 &\alpha +\beta \end{vmatrix}$

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

$\triangle = \begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ \beta -\alpha & \beta ^2 - \alpha^2 &\alpha - \beta \\ \gamma-\alpha &\gamma ^2-\alpha^2 &\alpha -\gamma \end{vmatrix}$

$= (\beta-\alpha)(\gamma-\alpha)\begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ 1 & \beta + \alpha &-1 \\ 0 &\gamma-\beta &0\end{vmatrix}$

Expanding the remaining determinant;

$= (\beta-\alpha)(\gamma-\alpha)[-(\gamma-\beta)(-\alpha-\beta-\gamma)]$

$= (\beta-\alpha)(\gamma-\alpha)(\gamma-\beta)(\alpha+\beta+\gamma)$

$=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)$

hence the given result is proved.

Question:12 Using properties of determinants, prove that

$\begin{vmatrix} x & x^2&1+px^3 \\ y& y^2& 1+py^3\\ z&z^2 & 1+pz^3 \end{vmatrix}=(1+pxyz)(x-y)(y-z)(z-x),$ where p is any scalar.

Answer:

Given the determinant $\triangle = \begin{vmatrix} x & x^2&1+px^3 \\ y& y^2& 1+py^3\\ z&z^2 & 1+pz^3 \end{vmatrix}$

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

$\triangle = \begin{vmatrix} x & x^2&1+px^3 \\ y-x& y^2-x^2& p(y^3-x^3)\\ z-x&z^2-x^2 & p(z^3-x^3) \end{vmatrix}$

Applying row transformation $R_{3} \rightarrow R_{3} - R_{2}$ we have then;

$\triangle =(y-x )(z-x)(z-y)\begin{vmatrix} x & x^2&1+px^3 \\ 1& y+x& p(y^2+x^2+xy)\\ 0&1 & p(x+y+z) \end{vmatrix}$

Now we can expand the remaining determinant to get the result;

$\triangle =(y-x )(z-x)(z-y)[(-1)(p)(xy^2+x^3+x^2y)+1+px^3+p(x+y+z)(xy)]$

$=(x-y)(y-z)(z-x)[-pxy^2-px^3-px^2y+1+px^3+px^2y+pxy^2+pxyz]$

$=(x-y)(y-z)(z-x)(1+pxyz)$

hence the given result is proved.

Question:13 Using properties of determinants, prove that

$\begin{vmatrix} 3a &-a+b &-a+c \\ -b+c &3b &-b+c \\ -c+a &-c+b &3c \end{vmatrix}=3(a+b+c)(ab+bc+ca)$

Answer:

Given determinant $\triangle = \begin{vmatrix} 3a &-a+b &-a+c \\ -b+c &3b &-b+c \\ -c+a &-c+b &3c \end{vmatrix}$

Applying the column transformation, $C_{1} \rightarrow C_{1} +C_{2}+C_{3}$ we have then;

$\triangle = \begin{vmatrix} a+b+c &-a+b &-a+c \\ a+b+c &3b &-b+c \\ a+b+c &-c+b &3c \end{vmatrix}$

Taking common factor (a+b+c) out from the column first;

$=(a+b+c) \begin{vmatrix} 1 &-a+b &-a+c \\ 1 &3b &-b+c \\1 &-c+b &3c \end{vmatrix}$

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

$\triangle=(a+b+c) \begin{vmatrix} 1 &-a+b &-a+c \\ 0 &2b+a &a-b \\0 &a-c &2c+a \end{vmatrix}$

Now we can expand the remaining determinant along $C_{1}$ we have;

$\triangle=(a+b+c) [(2b+a)(2c+a)-(a-b)(a-c)]$

$=(a+b+c) [4bc+2ab+2ac+a^2-a^2+ac+ba-bc]$

$=(a+b+c)(3ab+3bc+3ac)$

$=3(a+b+c)(ab+bc+ac)$

Hence proved.

Question:14 Using properties of determinants, prove that

$\begin{vmatrix} 1 &1+p &1+p+q \\ 2&3+2p &4+3p+2q \\ 3&6+3p & 10+6p+3q \end{vmatrix}=1$

Answer:

Given determinant $\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 2&3+2p &4+3p+2q \\ 3&6+3p & 10+6p+3q \end{vmatrix}$

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

$\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 0&1 &2+p \\ 0&3 & 7+3p \end{vmatrix}$

Now, applying another row transformation $R_{3}\rightarrow R_{3}-3R_{2}$ we have;

$\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 0&1 &2+p \\ 0&0 & 1 \end{vmatrix}$

We can expand the remaining determinant along $C_{1}$ , we have;

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

Hence the result is proved.

Question:15 Using properties of determinants, prove that

$\begin{vmatrix} \sin \alpha &\cos \alpha &\cos (\alpha +\delta ) \\ \sin \beta & \cos \beta & \cos (\beta +\delta )\\ \sin \gamma &\cos \gamma & \cos (\gamma +\delta ) \end{vmatrix}=0$

Answer:

Given determinant $\triangle = \begin{vmatrix} \sin \alpha &\cos \alpha &\cos (\alpha +\delta ) \\ \sin \beta & \cos \beta & \cos (\beta +\delta )\\ \sin \gamma &\cos \gamma & \cos (\gamma +\delta ) \end{vmatrix}$

Multiplying the first column by $\sin \delta$ and the second column by $\cos \delta$ , and expanding the third column, we get

$\triangle =\frac{1}{\sin \delta \cos \delta} \begin{vmatrix} \sin \alpha \sin \delta &\cos \alpha\cos \delta &\cos \alpha \cos \delta -\sin \alpha \sin \delta \\ \sin \beta\sin \delta & \cos \beta\cos \delta & \cos \beta \cos \delta - \sin \beta\sin \delta \\ \sin \gamma\sin \delta &\cos \gamma\cos \delta & \cos \gamma \cos\delta- \sin \gamma\sin \delta \end{vmatrix}$

Applying column transformation, $C_{1} \rightarrow C_{1}+C_{3}$ we have then;

$\triangle =\frac{1}{\sin \delta \cos \delta} \begin{vmatrix} \cos \alpha \cos \delta &\cos \alpha\cos \delta &\cos \alpha \cos \delta -\sin \alpha \sin \delta \\ \cos \beta\cos \delta & \cos \beta\cos \delta & \cos \beta \cos \delta - \sin \beta\sin \delta \\ \cos \gamma\cos \delta &\cos \gamma\cos \delta & \cos \gamma \cos\delta- \sin \gamma\sin \delta \end{vmatrix}$

Here we can see that two columns $C_{1}\ and\ C_{2}$ are identical.

The determinant value is equal to zero. $\therefore \triangle = 0$

Hence proved.

Question:16 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:17 Choose the correct answer.

If $a,b,c,$ are in A.P, then the determinant
$\dpi{100} \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 & x+4 & x+2b\\ x+4 & x+5 &x+2c \end{vmatrix}$ is

(A) $0$ (B) $1$ (C) $x$ (D) $2x$

Answer:

Given determinant $\triangle = \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 & x+4 & x+2b\\ x+4 & x+5 &x+2c \end{vmatrix}$ and given that a, b, c are in A.P.

That means , 2b =a+c

$\triangle = \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 & x+4 & x+(a+c)\\ x+4 & x+5 &x+2c \end{vmatrix}$

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

$\triangle = \begin{vmatrix} -1 &-1 &a-c \\ x+3 & x+4 & x+(a+c)\\ 1 & 1 &c-a \end{vmatrix}$

Now, applying another row transformation, $R_{1} \rightarrow R_{1} + R_{3}$ , we have

$\triangle = \begin{vmatrix} 0 &0 &0 \\ x+3 & x+4 & x+(a+c)\\ 1 & 1 &c-a \end{vmatrix}$

Clearly we have the determinant value equal to zero;

Hence the option (A) is correct.

Question:18 Choose the correct answer.

If x, y, z are nonzero real numbers, then the inverse of matrix $A=\begin{bmatrix} x &0 &0 \\ 0 &y &0 \\ 0 & 0 & z \end{bmatrix}$ is

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

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

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:19 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) $Det(A)\in (2,\infty)$

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.

Benefits of NCERT Solutions for Class 12 Maths Chapter 4 Miscellaneous Exercise:-

Class 12 Maths chapter 4 miscellaneous solutions are designed in a very detailed manner which could be understood by an average student also.
As miscellaneous exercise questions are difficult as compared to the previous exercise, you may not be able to these questions.
You can take NCERT solutions for Class 12 Maths chapter 4 miscellaneous exercise for reference while solving miscellaneous questions.
Miscellaneous exercise chapter 4 Class 12 will check your understanding of this chapter.

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NCERT Solutions for Class 12 Maths Chapter 4
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NCERT Solutions for Class 12 Maths
NCERT Solutions for Class 12 Physics
NCERT Solutions for Class 12 Chemistry
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NCERT Exemplar Solutions for Class 12th Chemistry
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NCERT Solutions for Miscellaneous Exercise Chapter 4 Class 12 - Determinants

Determinants Miscellaneous Exercise

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