Class 12 ncert solution math exercise 4.1

  EXERCISE 4.1 ( Determinants )

Evaluate the determinant:(Class 12 ncert solution math exercise 4.1)

Question 1:  \left|\begin{array}{cc}2 & 4 \\ -5 & -1\end{array}\right|

Solution: Let |A|=\left|\begin{array}{cc}2 & 4 \\ -5 & -1\end{array}\right|

Hence,

|A| =\left|\begin{array}{cc} 2 & 4 \\ -5 & -1 \end{array}\right|

=2(-1)-4(-5)

=-2+20

=18

Question 2: Evaluate the determinants:

(i) \begin{vmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{vmatrix}

(ii) \left|\begin{array}{cc}x^{2}-x+1 & x-1 \\ x+1 & x+1\end{array}\right|

Solution: (i) \left|\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right|
=(\cos \theta)(\cos \theta)-(-\sin \theta)(\sin \theta)
=\cos ^{2} \theta+\sin ^{2} \theta
=1

(ii) \left|\begin{array}{cc} x^{2}-x+1 & x-1 \\ x+1 & x+1 \end{array}\right|
=\left(x^{2}-x+1\right)(x+1)-(x-1)(x+1)
=x^{3}-x^{2}+x+x^{2}-x+1-\left(x^{2}-1\right)
=x^{3}+1-x^{2}+1
=x^{3}-x^{2}+2

Question 3: If A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right], then show that |2 A|=4|A|

Solution: The given matrix is A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right]

Therefore,

2 A =2\left[\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right]
=\left[\begin{array}{ll} 2 & 4 \\ 8 & 4 \end{array}\right]

Hence,

L.H.S. =|2 A|
=\left|\begin{array}{ll} 2 & 4 \\ 8 & 4 \end{array}\right|
=2 \times 4-4 \times 8
=8-32
=-24

Now,

|A| =\left|\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right|=1 \times 2-2 \times 4
=2-8
=-6

Therefore,

R.H.S. =4|A|
=4(-6)
=-24

Thus, |2 A|=4|A| proved.

Question 4: If A=\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right] , then show that |3 A|=27|A|

Solution:The given matrix is

A=\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right]

It can be observed that in the first column, two entries are zero. Thus, we expand along the first column \left(C_{1}\right) for easier calculation.

|A| =1\left|\begin{array}{ll} 1 & 2 \\ 0 & 4 \end{array}\right|-0\left|\begin{array}{ll} 0 & 1 \\ 1 & 4 \end{array}\right|+0\left|\begin{array}{ll} 0 & 1 \\ 1 & 2 \end{array}\right|

=1(4-0)-0+0
=4

Therefore,

27|A| =27|4|

=108

Now,

3 A=3\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right]

=\left[\begin{array}{ccc} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{array}\right]

Therefore,

|3 A| =3\left|\begin{array}{lc} 3 & 6 \\ 0 & 12 \end{array}\right|-0\left|\begin{array}{cc} 0 & 3 \\ 0 & 12 \end{array}\right|+0\left|\begin{array}{ll} 0 & 3 \\ 3 & 6 \end{array}\right|

=3(36-0)
=36(36)
=108

From equations (1) and (2),

|3 A|=27|A|

Thus, |3 A|=27|A| proved.

Question 5:Evaluate the determinants

(i) \left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|

(ii) \left|\begin{array}{ccc}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{array}\right|

(iii) \left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|

(iv) \left|\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right|

Solution: A=\left|\begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array}\right|

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.

Hence,

|A| =-0\left|\begin{array}{cc} -1 & -2 \\ -5 & 0 \end{array}\right|+0\left|\begin{array}{cc} 3 & -2 \\ 3 & 0 \end{array}\right|-(-1)\left|\begin{array}{ll} 3 & -1 \\ 3 & -5 \end{array}\right|
=(-15+3)
=-12

(ii) Let A=\left|\begin{array}{ccc}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{array}\right|

Hence,

|A| =3\left|\begin{array}{cc} 1 & -2 \\ 3 & 1 \end{array}\right|+4\left|\begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array}\right|+5\left|\begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array}\right|
=3(1+6)+4(1+4)+5(3-2)
=3(7)+4(5)+5(1)
=21+20+5
=46

(iii) Let A=\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|

Hence,

|A| =0\left|\begin{array}{cc} 0 & -3 \\ 3 & 0 \end{array}\right|-1\left|\begin{array}{cc} -1 & -3 \\ -2 & 0 \end{array}\right|+2\left|\begin{array}{cc} -1 & 0 \\ -2 & 3 \end{array}\right|

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

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

=6-6=0

(iv) A=\left|\begin{array}{ccc} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right|

Hence,

|A| =2\left|\begin{array}{cc} 2 & -1 \\ -5 & 0 \end{array}\right|-0\left|\begin{array}{cc} -1 & -2 \\ -5 & 0 \end{array}\right|+3\left|\begin{array}{cc} -1 & -2 \\ 2 & -1 \end{array}\right|
=2(0-5)-0+3(1+4)
=-10+15=5

Question 6:If A=\left[\begin{array}{ccc} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{array}\right], find |A|

Solution: Let A=\left[\begin{array}{ccc} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{array}\right]

Hence,

|A| =1\left|\begin{array}{cc} 1 & -3 \\ 4 & -9 \end{array}\right|-1\left|\begin{array}{cc} 2 & -3 \\ 5 & -9 \end{array}\right|-2\left|\begin{array}{ll} 2 & 1 \\ 5 & 4 \end{array}\right|
=1(-9+12)-1(-18+15)-2(8-5)
=1(3)-1(-3)-2(3)
=3+3-6
=0

Question 7: Find the values of x, if
(i) \left|\begin{array}{cc}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|
(ii) \left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \\ 2 x & 5\end{array}\right|

Solution: (i) \left|\begin{array}{cc}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|
\Rightarrow 2\times 1-4\times 5 = 2x\times x-6\times 4
\Rightarrow 2-20 = 2x^2-24
\Rightarrow -18 +24 =2x^2
\Rightarrow x^2 = 3
\Rightarrow x =\pm\sqrt{3}

Therefore,

(ii) \left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \\ 2 x & 5\end{array}\right|

Therefore,

\Rightarrow 2 \times 5-3 \times 4=x \times 5-3 \times 2 x
\Rightarrow 10-12=5 x-6 x
\Rightarrow-2=-x
\Rightarrow x=2

Question 8: If \left|\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right|, the x is equal to
(A) 6
(B) \pm 6
(C) -6
(D) 0

Solution: \left|\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right|

Therefore,

\Rightarrow x^{2}-36=36-36
\Rightarrow x^{2}-36=0
\Rightarrow x^{2}=36
\Rightarrow x=\pm 6

Thus, the correct option is B.

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