Exercise 4.1 (Revised) - Chapter 4 Determinants class 12 ncert solutions Maths - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Chapter 4 - Determinants NCERT Solutions Class 12 Maths | Step-by-Step Solutions & Explanations

Ex 4.1 Question 1:
Evaluate the determinants in Exercises 1 and 2.
$
\left|\begin{array}{cc}
2 & 4 \\
-5 & -1
\end{array}\right|
$

Answer
$
\left|\begin{array}{cc}
2 & 4 \\
-5 & -1
\end{array}\right|=2(-1)-4(-5)=-2+20=18
$

Ex 4.1 Question 2:
Evaluate the determinants in Exercises 1 and 2.
(i) $\left|\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right|_{(\mathrm{ii})}\left|\begin{array}{cc}x^2-x+1 & x-1 \\ x+1 & x+1\end{array}\right|$

Answer
(i) $\left|\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right|=(\cos \theta)(\cos \theta)-(-\sin \theta)(\sin \theta)=\cos ^2 \theta+\sin ^2 \theta=1$
$
\begin{aligned}
& \left|\begin{array}{cc}
x^2-x+1 & x-1 \\
x+1 & x+1
\end{array}\right| \\
\text { (ii) } & \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
\end{aligned}
$

Ex 4.1 Question 3:
$
A=\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right] \text {, then show that }|2 A|=4|A|
$

Answer

The given matrix is
$
A=\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right] .
$
$
\begin{aligned}
& \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] \\
& \therefore \text { 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
\end{aligned}
$

Now, $|A|=\left|\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right|=1 \times 2-2 \times 4=2-8=-6$
$
\therefore \text { R.H.S. }=4|A|=4 \times(-6)=-24
$
$\therefore$ L.H.S. $=$ R.H.S.

Ex 4.1 Question 4:
$
\mathrm{A}=\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right] \text {, then show that }|3 A|=27|A| \text {. }
$

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

The given matrix is
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.
$
\begin{aligned}
& |\mathrm{A}|=1\left|\begin{array}{ll}
1 & 2 \\
0 & 4
\end{array}\right|-0\left|\begin{array}{ll}
0 & 1 \\
0 & 4
\end{array}\right|+0\left|\begin{array}{ll}
0 & 1 \\
1 & 2
\end{array}\right|=1(4-0)-0+0=4 \\
& \therefore 27|\mathrm{~A}|=27(4)=108
\end{aligned}
$

Now, $3 A=3\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]=\left[\begin{array}{ccc}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12\end{array}\right]$
$
\begin{aligned}
\therefore|3 \mathrm{~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)=3(36)=108
\end{aligned} .
$

From equations (i) and (ii), we have:
$
|3 A|=27|A|
$

Hence, the given result is proved.

Ex 4.1 Question 5:
Evaluate the determinants
(i) $\left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|_{\text {(iii) }}\left|\begin{array}{ccc}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{array}\right|$
(ii) $\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|_{\text {(iv) }}\left[\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right]$

Answer

(i) Let

$
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.
$
|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]
$

By expanding along the first row, we have:
$
\begin{aligned}
|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
\end{aligned}
$

(iii) Let
$
A=\left[\begin{array}{ccc}
0 & 1 & 2 \\
-1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right] \text {. }
$

By expanding along the first row, we have:
$
\begin{aligned}
|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
\end{aligned}
$

(iv) Let
$
A=\left[\begin{array}{ccc}
2 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right] .
$

By expanding along the first column, we have:
$
\begin{aligned}
|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
\end{aligned}
$

Ex 4.1 Question 6:
$
A=\left[\begin{array}{lll}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right] \text {, find }|A| \text {. }
$

Answer
$
A=\left[\begin{array}{lll}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right] .
$

By expanding along the first row, we have:
$
\begin{aligned}
|A| & =1\left|\begin{array}{ll}
1 & -3 \\
4 & -9
\end{array}\right|-1\left|\begin{array}{ll}
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 \\
& =6-6 \\
& =0
\end{aligned}
$

Ex 4.1 Question 7:
Find values of $x$, if
(i) $\left|\begin{array}{ll}2 & 4 \\ 2 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$ (ii) $\left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \\ 2 x & 5\end{array}\right|$

Answer
$
\begin{aligned}
& \text { (i) }\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right| \\
& \Rightarrow 2 \times 1-5 \times 4=2 x \times x-6 \times 4 \\
& \Rightarrow 2-20=2 x^2-24 \\
& \Rightarrow 2 x^2=6 \\
& \Rightarrow x^2=3 \\
& \Rightarrow x= \pm \sqrt{3}
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) }\left|\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right|=\left|\begin{array}{cc}
x & 3 \\
2 x & 5
\end{array}\right| \\
& \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
\end{aligned}
$

Ex 4.1 Question 8:
If $\left|\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right|$, then $x$ is equal to
(A) 6 (B) $\pm 6$ (C) -6 (D) 0

Answer
Answer: B
$
\begin{aligned}
& \left|\begin{array}{cc}
x & 2 \\
18 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & 2 \\
18 & 6
\end{array}\right| \\
& \Rightarrow x^2-36=36-36 \\
& \Rightarrow x^2-36=0 \\
& \Rightarrow x^2=36 \\
& \Rightarrow x= \pm 6
\end{aligned}
$

Hence, the correct answer is B.