AP Inter 2nd Year Maths Exercise 4a Solutions

Referring to the AP Inter 2nd Year Maths Study Material Chapter 4 Determinants Exercise 4a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Determinants Solutions Exercise 4a

Question 1.
Evaluate \(\left|\begin{array}{cc}
2 & 4 \\
-5 & -1
\end{array}\right|\)
Solution:
|A| = \(\left|\begin{array}{cc}
2 & 4 \\
-5 & -1
\end{array}\right|\) = 2(-1) – 4(-5) = -2 + 20 = 18 [∵ \(\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|\) = ad – bc]

Question 2.
Evaluate \(\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
Solution:
\(\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\) = (cos θ)(cos θ) – (-sin θ)(sin θ) = cos²θ + sin²θ = 1

Question 3.
Find the determinant of \(\left[\begin{array}{cc}
2 & 1 \\
1 & -5
\end{array}\right]\)
Solution:
det A = ad – bc = 2(-5) – 1(1) = -10 – 1 = -11

Question 4.
Find the determinant of \(\left[\begin{array}{cc}
4 & 5 \\
-6 & 2
\end{array}\right]\)
Solution:
det A = ad – bc = 4(2) – 5(-6) = 8 + 30 = 38

Question 5.
Find the determinant of \(\left[\begin{array}{cc}
i & 0 \\
0 & -i
\end{array}\right]\)
Solution:
det A = i(-i) – 0 = -i² = -(-1) = 1

Question 6.
Evaluate \(\left|\begin{array}{cc}
x^2-x+1 & x-1 \\
x+1 & x+1
\end{array}\right|\)
Solution:
\(\left|\begin{array}{cc}
x^2-x+1 & x-1 \\
x+1 & x+1
\end{array}\right|\) = (x² – x + 1)(x + 1) – (x – 1)(x + 1) [∵ \(\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|\) = ad – bc]
= x³ + x² – x² + x – x + 1 – (x² – 1)
= x³ + 1 – x² + 1
= x³ – x² + 2

Question 7.
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 = \(2\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]=\left[\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right]\) [∵ \(\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|\) = ad – bc]
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
∴ RHS = 4|A| = 4(-6) = -24
∴ |2A| = 4|A|

Question 8.
Evaluate \(\left|\begin{array}{ccc}
3 & -1 & -2 \\
0 & 0 & -1 \\
3 & -5 & 0
\end{array}\right|\)
Solution:
Let A = \(\left|\begin{array}{ccc}
3 & -1 & -2 \\
0 & 0 & -1 \\
3 & -5 & 0
\end{array}\right|\)
On expanding along the second row R₂, we get
|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

Question 9.
Evaluate \(\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\)
Solution:
Let A = \(\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\)
|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

Question 10.
Evaluate \(\left|\begin{array}{ccc}
0 & 1 & 2 \\
-1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right|\)
Solution:
Let A = \(\left|\begin{array}{ccc}
0 & 1 & 2 \\
-1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right|\)
∴ |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

Question 11.
Evaluate \(\left|\begin{array}{ccc}
2 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right| .\)
Solution:
Let A = \(\left|\begin{array}{ccc}
2 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right|\)
∴ |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 12.
If A = \(\left[\begin{array}{lll}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\), find |A|
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\)
∴ |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 = 0

Question 13.
Find the values of x, if \(\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\)
Solution:
Given that \(\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\) [∵ \(\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|\) = ad – bc]
⇒ 2 × 1 – 5 × 4 = 2x × x – 6 × 4
⇒ 2 – 20 = 2x² – 24
⇒ 2x² = 6
⇒ x² = 3
⇒ x = ±\(\sqrt{3}\)

Question 14.
Find the values of x, if \(\left|\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right|=\left|\begin{array}{cc}
x & 3 \\
2 x & 5
\end{array}\right|\)
Solution:
Given that \(\left|\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right|=\left|\begin{array}{cc}
x & 3 \\
2 x & 5
\end{array}\right|\)
⇒ 2 × 5 – 3 × 4 = x × 5 – 3 × 2x
⇒ 10 – 12 = 5x – 6x
⇒ -2 = -x
⇒ x = 2

II.

Question 1.
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: