OP Malhotra Class 12 Maths Solutions Chapter 5 Determinants Ex 5(b)

Effective OP Malhotra Class 12 Solutions Chapter 5 Determinants Ex 5(b) can help bridge the gap between theory and application.

S Chand Class 12 ICSE Maths Solutions Chapter 5 Determinants Ex 5(b)

Question 1.
Write minors and cofactors of elements of determinant \(\left|\begin{array}{ll}
a & c \\
b & d
\end{array}\right|\).
Solution:
Let |A|
∴ M₁₁ = d ; M₁₂ = b;
M₂₁ = c & M₂₂ = a
Thus factors are :
A₁₁ = (-1)¹⁺¹ M₁₁ = M₁₁ = d ;
A₁₂ = (-1)¹⁺² M₁₂ = – M₁₂ = – b ;
A₂₁ = (-1)²⁺¹ M₂₁ = – M₂₁ = – c
A₂₂ = (-1)²⁺² M₂₂ = M₂₂ = a

Question 2.
Write down the minors of – 2 and 4 in
\(\left|\begin{array}{rrr}
2 & 1 & 1 \\
1 & -2 & -3 \\
3 & 2 & 4
\end{array}\right|\)
Solution:
Minor of – 2 = M₂₂ = \(\left|\begin{array}{ll}
2 & 1 \\
3 & 4
\end{array}\right|\)
= 8 – 3 = 5
and Minor of 4 = M₃₃ = \(\left|\begin{array}{cc}
2 & 1 \\
1 & -2
\end{array}\right|\)
= 2 x (- 2) – 1 x 1
= – 4 – 1 = – 5

Question 3.
Write down the co-factors of 3 and – 2 of the determinant \(\left|\begin{array}{rrr}
1 & 0 & -2 \\
3 & -1 & 2 \\
4 & 5 & 6
\end{array}\right|\)
Solution:

Question 4.
(i) Write the co-factors of elements of the second row of the determinant \(\left|\begin{array}{rrr}
1 & 2 & 3 \\
-4 & 3 & 6 \\
2 & -7 & 9
\end{array}\right|\)
(ii) Using the cofactors of elements of second row evaluate ∆ = \(\left|\begin{array}{lll}
5 & 2 & 8 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right|\)
Solution:

Question 5.
Write the minors and co-factors of each element of the first column of the following determinants and evaluate the determinant in each case.
(i) \(\left|\begin{array}{cc}
5 & 20 \\
0 & -1
\end{array}\right|\)
(ii) \(\left|\begin{array}{lll}
1 & a & b c \\
1 & b & c d \\
1 & c & a b
\end{array}\right|\)
(iii) \(\left|\begin{array}{lll}
0 & 2 & 6 \\
1 & 5 & 0 \\
3 & 7 & 1
\end{array}\right|\)
Solution:
(i) Let ∆ = \(\left|\begin{array}{cc}
5 & 20 \\
0 & -1
\end{array}\right|\)
The element of C₁ are 5 & 0
i.e. C₁₁ = 5 ; C₂₁ = 0
∴ M₁₁ = – 1 ; M₂₁ = 20
Thus, A₁₁ = (- 1)¹⁺¹ M₁₁ = M₁₁ = – 1 ;
A₂₁ = (- 1)²⁺¹ M₂₁ = – M₂₁ = – 20
∴ ∆ = C₁₁A₁₁ + C₂₁A₂₁
= 5 x (- 1) + 0 x (- 20) = – 5

(ii) Let ∆ = \(\left|\begin{array}{lll}
1 & a & b c \\
1 & b & c d \\
1 & c & a b
\end{array}\right|\)
The element of C₁₁ are ; a₁₁ = 1 ; a₂₁= 1 ; a₂₁ = 1;
M₁₁ = \(\frac { 1 }{ 2 }\) = ab² – c²a = a(b² – c²)
M₂₁ = \(\frac { 1 }{ 2 }\) = a²b – bc² = b(a² – c²)
M₃₁ = \(\frac { 1 }{ 2 }\) = a²c – b²c = c(a² – b²)
∴ Cofactors of first column are ;
A₁₁ = (- 1)¹⁺¹ M₁₁ = M₁₁ = a(b² – c²)
A₂₁ = (- 1)²⁺¹ M₂₁ = – M₂₁ = – b(a² – c²)
A₃₁ = (- 1)³⁺¹ M₃₁ = M₃₁ = c(a² – b²)
∴ Value of determinant
= a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
= a(b² – c²) – b(a² – c²) + c(a² – b²)
= ab{b – a) + bc(c – b) + ca(a – c)

(iii) Let ∆ = \(\left|\begin{array}{lll}
0 & 2 & 6 \\
1 & 5 & 0 \\
3 & 7 & 1
\end{array}\right|\)
The element of C₁ are ; a₁₁ = 0 ; a₂₁= 1 ; a₃₁ = 3;
The minors of C₁ are ;
M₁₁ = \(\left|\begin{array}{ll}
5 & 0 \\
7 & 1
\end{array}\right|\) = 5 – 0 = 5;
M₂₁ = \(\left|\begin{array}{ll}
2 & 6 \\
7 & 1
\end{array}\right|\) = 2 – 42 = – 40
and M₃₁ = \(\left|\begin{array}{ll}
2 & 6 \\
5 & 0
\end{array}\right|\) = 0 – 30 = – 30
The cofactors of C, are ;
A₁₁ = (-1)¹⁺¹ M₁₁ = + M₁₁ = 5 ;
A₃₁ = (-1)³⁺¹ M₃₁ = M₃₁ = – 30 ;
A₂₁ = (-1)²⁺¹ M₂₁ = – M₂₁ = – 40
Thus, Value of acicrminant
= a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
= 0 x 5 + 1 x 40 + 3 x (- 30) = – 50

Question 6.
Evaluate \(\left|\begin{array}{rrr}
5 & 1 & 0 \\
2 & 3 & -1 \\
-3 & 2 & 0
\end{array}\right|\)
Solution:
Let ∆ = \(\left|\begin{array}{rrr}
5 & 1 & 0 \\
2 & 3 & -1 \\
-3 & 2 & 0
\end{array}\right|\);
expanding along C₃
= 0 x \(\left|\begin{array}{cc}
2 & 3 \\
-3 & 2
\end{array}\right|+1\left|\begin{array}{cc}
5 & 1 \\
-3 & 2
\end{array}\right|+0\left|\begin{array}{cc}
5 & 1 \\
2 & 3
\end{array}\right|\)
= 0 + 1(10 + 3) + 0 (15 – 2) = 13

Question 7.
Find the value of the determinants
(i) \(\left|\begin{array}{rrr}
1 & 3 & 5 \\
2 & 6 & 10 \\
31 & 11 & 38
\end{array}\right|\)
(ii) \(\left|\begin{array}{ccc}
x+\lambda & x & x \\
x & x+\lambda & x \\
x & x & x+\lambda
\end{array}\right|\)
(iii) \(\left|\begin{array}{ccc}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right|\)
Solution:
(i) Let ∆ = \(\left|\begin{array}{rrr}
1 & 3 & 5 \\
2 & 6 & 10 \\
31 & 11 & 38
\end{array}\right|\);
expanding along R₁

Question 8.
Expand the determinants by minors of the given row or column
(i) \(\left|\begin{array}{rrr}
1 & 3 & -1 \\
2 & 1 & 4 \\
6 & 1 & 1
\end{array}\right|\); column 1
(ii) \(\left|\begin{array}{rrr}
2 & -1 & 4 \\
3 & 0 & 1 \\
2 & 1 & -1
\end{array}\right|\); column 2
(iii) \(\left|\begin{array}{rrr}
5 & 1 & -1 \\
2 & 3 & -1 \\
4 & 2 & 3
\end{array}\right|\); row 2
Solution:

Question 9.
Solve for x :
(i) \(\left|\begin{array}{lll}
x & 0 & 0 \\
2 & 1 & 3 \\
0 & 1 & 4
\end{array}\right|\) = 3
(ii) \(\left|\begin{array}{rrr}
x^2 & x & 1 \\
0 & 2 & 1 \\
3 & 1 & 4
\end{array}\right|\) = 28
Solution:
(i) Given \(\left|\begin{array}{lll}
x & 0 & 0 \\
2 & 1 & 3 \\
0 & 1 & 4
\end{array}\right|\) = 3
expanding along R₁, we have
\(x\left|\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right|-0\left|\begin{array}{ll}
2 & 3 \\
0 & 4
\end{array}\right|+0\left|\begin{array}{ll}
2 & 1 \\
0 & 2
\end{array}\right|\) = 3
⇒ x(4 – 3) – 0(8 – 0) + 0(2 – 0) = 3
⇒ – x = 3

(ii) Given \(\left|\begin{array}{rrr}
x^2 & x & 1 \\
0 & 2 & 1 \\
3 & 1 & 4
\end{array}\right|\) = 28
expanding along C₁
\(x^2\left|\begin{array}{ll}
2 & 1 \\
1 & 4
\end{array}\right|-0\left|\begin{array}{ll}
x & 1 \\
1 & 4
\end{array}\right|+3\left|\begin{array}{ll}
x & 1 \\
2 & 1
\end{array}\right|\) = 28
x² (8 – 1) – 0(4x – 1) + 3(x – 2) = 28
⇒ 7x² + 3x – 6 – 28 = 0
⇒ 7x² + 3x – 34 = 0
⇒ (x – 2) (7x + 17) = 0
⇒ x = 2, \(\frac { -17 }{ 7 }\)

Question 10.
Show that
\(\left|\begin{array}{rrr}
1 & a & b \\
-a & 1 & c \\
-b & -c & 1
\end{array}\right|\) = 1 + a² + b² + c².
Solution:
L.H.S. = \(\left|\begin{array}{rrr}
1 & a & b \\
-a & 1 & c \\
-b & -c & 1
\end{array}\right|\);
expanding along R₁
= \(1\left|\begin{array}{cc}
1 & c \\
-c & 1
\end{array}\right|-a\left|\begin{array}{ll}
-a & c \\
-b & 1
\end{array}\right|+b\left|\begin{array}{cc}
-a & 1 \\
-b & -c
\end{array}\right|\)
= = 1(1 + c²) – a(- a + bc) + b(ac + b)
= 1 + c² + a² + b² = R.H.S

Question 11.
Expand the simplify the following :
(i) \(\left|\begin{array}{ccc}
1 & x & y \\
0 & \cos x & \sin y \\
0 & \sin x & \cos y
\end{array}\right|\)
(ii) \(\left|\begin{array}{rrr}
0 & \tan \theta & 1 \\
1 & -\sec \theta & 0 \\
\sec \theta & \tan \theta & 1
\end{array}\right|\)
(iii) \(\left|\begin{array}{ccc}
\sin \theta & 1 & 0 \\
0 & \cos \phi & -\cos \theta \\
\sin \phi & 0 & 1
\end{array}\right|\)
Solution:

Question 12.
If one root of is x = – 9, find the other roots.
Solution:
Given
expanding along R₁
\(7\left|\begin{array}{ll}
x & 2 \\
3 & 7
\end{array}\right|-6\left|\begin{array}{ll}
2 & 2 \\
x & 7
\end{array}\right|+x\left|\begin{array}{ll}
2 & x \\
x & 3
\end{array}\right|\) = 0
⇒ 7(7x – 6) – 6(14 – 2x) + x (6 – x²) = 0
⇒ 49x – 42 – 84 + 12x + 6x – x³ = 0
⇒ 67x – x³ – 126 = 0
⇒ x³ – 67x + 126 = 0
⇒ (x – 2)(x² + 2x – 63) = 0
⇒ (x – 2)(x – 7)(x + 9) = 0
⇒ x = 2, 7, – 9
Since one of the root of given eqn be x = – 9
Thus, the other two roots are 2 & 7.