NCERT Solutions Class 12 Maths Chapter 4 Exercise 4.1 - Determinants

Question:2(i) Evaluate the following determinant- $\begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta &\cos \theta \end{vmatrix}$

Answer:

The given two by two determinant is calculated as follows

$\begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} = \cos \theta (\cos \theta) - (-\sin \theta)\sin \theta = \cos^2 \theta + \sin^2 \theta = 1$

Question:2(ii) Evaluate the following determinant- $\begin{vmatrix}x^2-x+1 & x-1\\x+1 &x+1 \end{vmatrix}$

Answer:

We have determinant $\begin{vmatrix}x^2-x+1 & x-1\\x+1 &x+1 \end{vmatrix}$

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

$= (x+1)(x^2-x+1-x+1) = (x+1)(x^2-2x+2)$

$=x^3-2x^2+2x +x^2-2x+2$

$= x^3-x^2+2$

Question:3 If $A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}$ , then show that $| 2 A |=4|A|$

Answer:

Given determinant $A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}$ then we have to show that $| 2 A |=4|A|$,

So, $A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}$ then, $2A =2 \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix} = \begin{bmatrix} 2 & 4\\ 8 &4 \end{bmatrix}$

Hence we have $\left | 2A \right | = \begin{vmatrix} 2 &4 \\ 8& 4 \end{vmatrix} = 2(4) - 4(8) = -24$

So, L.H.S. = |2A| = -24

then calculating R.H.S. $4\left | A \right |$

We have,

$\left | A \right | = \begin{vmatrix} 1 &2 \\ 4& 2 \end{vmatrix} = 1(2) - 2(4) = -6$

hence R.H.S becomes $4\left | A \right | = 4\times(-6) = -24$

Therefore L.H.S. =R.H.S.

Hence proved.

Question:4 If $A =\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix}$ then show that $|3A|=27|A|$

Answer:

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

Calculating $3A =3\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix} = \begin{bmatrix} 3 &0 &3 \\ 0& 3& 6\\ 0& 0 &12 \end{bmatrix}$

So, $\left | 3A \right | = 3(3(12) - 6(0) ) - 0(0(12)-0(6)) + 3(0-0) = 3(36) = 108$

calculating $27|A|$,

$|A| = \begin{vmatrix} 1 & 0 &1 \\ 0 & 1 & 2\\ 0& 0 &4 \end{vmatrix} = 1\begin{vmatrix} 1 &2 \\ 0 & 4 \end{vmatrix} - 0\begin{vmatrix} 0 &2 \\ 0& 4 \end{vmatrix} + 1\begin{vmatrix} 0 &1 \\ 0& 0 \end{vmatrix} = 4 -0 + 0 = 4$

So, $27|A| = 27(4) = 108$

Therefore $|3A|=27|A|$.

Hence proved.

Question:5(i) Evaluate the determinants.

$\begin{vmatrix}3 &-1 &-2 \\0 &0 &-1 \\3 &-5 & 0 \end{vmatrix}$

Answer:

Given the determinant $\begin{vmatrix}3 &-1 &-2 \\0 &0 &-1 \\3 &-5 & 0 \end{vmatrix}$;

now, calculating its determinant value,

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

$= 3(0-5)+1(0+3) -2(0-0) = -15+3-0 = -12$.

Question:5(ii) Evaluate the determinants.

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

Answer:

Given determinant $\begin{vmatrix}3 &-4 &5 \\1 &1 &-2 \\2 &3 &1 \end{vmatrix}$;

Now calculating the determinant value;

$\begin{vmatrix}3 &-4 &5 \\1 &1 &-2 \\2 &3 &1 \end{vmatrix} = 3\begin{vmatrix} 1 &-2 \\ 3&1 \end{vmatrix} -(-4)\begin{vmatrix} 1 &-2 \\ 2& 1 \end{vmatrix}+5\begin{vmatrix} 1 & 1\\ 2& 3 \end{vmatrix}$

$= 3(1+6) +4(1+4) +5(3-2) = 21+20+5 = 46$.

Question:5(iii) Evaluate the determinants.

$\begin{vmatrix}0 & 1 & 2\\-1 &0 &-3 \\ -2 &3 &0 \end{vmatrix}$

Answer:

Given determinant $\begin{vmatrix}0 & 1 & 2\\-1 &0 &-3 \\ -2 &3 &0 \end{vmatrix}$;

Now calculating the determinant value;

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

$= 0 - 1(0-6)+2(-3-0) = 6 -6 =0$

Question:5(iv) Evaluate the determinants.

$\begin{vmatrix}2 &-1 &2 \\0 &2 &-1 \\3 &-5 &0 \end{vmatrix}$

Answer:

Given determinant: $\begin{vmatrix}2 &-1 &-2 \\0 &2 &-1 \\3 &-5 &0 \end{vmatrix}$,

We now calculate determinant value:

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

$=2(0-5)+1(0+3)-2(0-6) = -10+3+12 = 5$

Question:6 If $A=\begin{bmatrix}1 & 1 & -2\\ 2& 1 &-3 \\5 &4 &-9 \end{bmatrix}$ , then find $|A|$.

Answer:

Given the matrix $A=\begin{bmatrix}1 & 1 & -2\\ 2& 1 &-3 \\5 &4 &-9 \end{bmatrix}$ then,

Finding the determinant value of A;

$|A| = 1\begin{vmatrix} 1 &-3 \\ 4& -9 \end{vmatrix} -1\begin{vmatrix} 2 &-3 \\ 5& -9 \end{vmatrix}-2\begin{vmatrix} 2 &1 \\ 5& 4 \end{vmatrix}$

$= 1(-9+12)-1(-18+15)-2(8-5) =3+3-6 =0$

Question:7(i) Find values of x, if

$\begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =\begin{vmatrix}2x &4 \\6 &x \end{vmatrix}$

Answer:

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

First, we solve the determinant value of L.H.S. and equate it to the determinant value of R.H.S.,

$\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = 2(1) - 4(5) = 2 - 20 = -18$ and $\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} = 2x(x) - 4(6) = 2x^2 - 24$

So, we have then,

$-18= 2x^2-24$ or $3= x^2$ or $x= \pm \sqrt{3}$

Question:7(ii) Find values of x, if

$\begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}=\begin{vmatrix}x &3 \\2x &5 \end{vmatrix}$

Answer:

Given $\begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}=\begin{vmatrix}x &3 \\2x &5 \end{vmatrix}$;

So, we here equate both sides after calculating each side's determinant values.

L.H.S. determinant value;

$\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = 2(5) - 3(4) = 10 - 12 = -2$

Similarly R.H.S. determinant value;

$\begin{vmatrix}x &3 \\2x &5 \end{vmatrix} = 5(x) - 3(2x) = 5x - 6x =-x$

So, we have then;

$-2 = -x$ or $x =2$.

Question:8 If $\begin{vmatrix}x &2 \\18 &x \end{vmatrix}=\begin{vmatrix} 6 &2 \\ 18 &6 \end{vmatrix}$ , then $x$ is equal to

(A) $6$ (B) $\pm 6$ (C) $-6$ (D) $0$

Answer:

Solving the L.H.S. determinant ;

$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = x(x) - 2(18) = x^2 - 36$

and solving R.H.S determinant;

$\begin{vmatrix} 6 &2 \\ 18 &6 \end{vmatrix} = 36-36 = 0$

So equating both sides;

$x^2 - 36 =0$ or $x^2 = 36$ or $x = \pm 6$

Hence answer is (B).