PSEB 12th Class Maths Solutions Chapter 4 Determinants Ex 4.1

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 4 Determinants Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1

Direction (1 – 2): Evaluate the determinants.

Question 1.
\(\left|\begin{array}{cc}
2 & 4 \\
-5 & -1
\end{array}\right|\)
Solution.
Let A = \(\left|\begin{array}{cc}
2 & 4 \\
-5 & -1
\end{array}\right|\)
∴ |A| = 2(- 1) – 4(- 5) = – 2 + 20 = 18.

Question 2.
(i) \(\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)

(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 θ) (cos θ) – (- sin θ)(sin θ)
= cos² θ + sin² θ = 1

(ii) \(\left|\begin{array}{cc}
x^{2}-x+1 & x-1 \\
x+1 & x+1
\end{array}\right|\) = (x² – x + 1) (x + 1) – (x – 1) (x + 1)
= x³ – x² + x + x² – x + 1 – (x² – 1)
= x³ + 1 – x² + 1
= x³ – x² + 2.

Question 3.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\) then show that [2A] = 4|A|
Solution.
The given matrix is A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\)
2A = \(2\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]=\left[\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right]\)
L.H.S.= |2A|
= \(\left|\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right|\)
= 2 × 4 – 4 × 8
= 8 – 32 = – 24

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

Question 4.
If A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right]\),then show that |3A| = 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 the matrix A along the first column (C₁) for finding |A|.

From Eqs. (j) and (ii), we have,
|3A| = 27 |A|
Hence, the given result is proved.

Question 5.
Evaluate the following 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.
(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| = \(0\left|\begin{array}{cc}
-1 & -2 \\
-5 & 0
\end{array}\right|+0\left|\begin{array}{cc}
3 & -2 \\
3 & 0
\end{array}\right|-(-1)\left|\begin{array}{cc}
3 & -1 \\
3 & -5
\end{array}\right|\)
= – 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
|A| = \(3\left|\begin{array}{cc}
1 & -2 \\
3 & 1
\end{array}\right|+4\left|\begin{array}{cc}
1 & -2 \\
2 & 1
\end{array}\right|+5\left|\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right|\)

= 3 (1 + 6) + 4 (1 + 4) + 5 (3 – 2)
= 3(7) + 4(5) + 5(1)
= 21 + 20 + 5 = 46.

(iii) \(\left|\begin{array}{ccc}
0 & 1 & 2 \\
-1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right|\)
By expanding along the first row, we have
|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) \(\left[\begin{array}{ccc}
2 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right]\)
By expanding along the first row, we have
|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 = \(\left[\begin{array}{ccc}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\)

By expanding along the first row, we have

|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
= 6 – 6 = 0.

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

(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) Given, \(\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\)

On expending both determinants, we get
⇒ 2 × 1 – 5 × 4 = 2 × x – 6 × 4
⇒ 2 – 20 = 2x² – 24
⇒ 2x² = 6
⇒ x² = 3
⇒ x = ±√3.

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

On expending both determinants, we get
⇒ 2 × 5 – 3 × 4 = x × 5 – 3 × 2x
⇒ 10 – 12 = 5x – 6x
⇒ – 2 = – x
⇒ x = 2.

Question 8.
If \(\left|\begin{array}{cc}
x & 2 \\
18 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & 2 \\
18 & 6
\end{array}\right|\), then x is equal to
(A) 6
(B) ± 6
(C) – 6
(D) 0
Solution.
Given, \(\left|\begin{array}{cc}
x & 2 \\
18 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & 2 \\
18 & 6
\end{array}\right|\)

On expending both determinants, we get
⇒ x² – 36 = 36 – 36
⇒ x² – 36 = 0
⇒ x² – 36 = 0
⇒ x = ± 6.
Hence, the correct answer is (B).