Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.5 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.5

Question 1.
(left[begin{array}{ll}1 & 2 \3 & 4 end{array}right])
Solution:
A₁₁ = (-1)¹⁺¹ M₁₁ = 4
A₁₂ = (-1)¹⁺² M₁₂ = – 3
A₂₁ = (-1)²⁺¹ M₂₁ = – 2
A₂₂ = (-1)²⁺² M₂₂ = 1
Let A = (left[begin{array}{ll}1 & 2 \3 & 4 end{array}right]). So, adj A = (left[begin{array}{ll}
A_{11} & A_{12} \
A_{21} & A_{22}
end{array}right])‘
= (left[begin{array}{cc}
4 & -3 \
-2 & 1
end{array}right]^{prime}) = (left[begin{array}{cc} 4 & -2 \ -3 & 1 end{array}right])

Question 2.
(left[begin{array}{ccc} 1 & -1 & 2 \ 2 & 3 & 5 \ -2 & 0 & 1 end{array}right])
Solution:

Question 3.
(left[begin{array}{ll} 2 & 3 \ -4 & -6 end{array}right])
Solution:

Question 4.
(left[begin{array}{ccc} 1 & -1 & 2 \ 3 & 0 & -2 \ 1 & 0 & 3 end{array}right])
Solution:

Question 5.
(left[begin{array}{ll} 2 & -2 \ 4 & 3 end{array}right])
Solution:
Let A = (left[begin{array}{ll} 2 & -2 \ 4 & 3 end{array}right])

Question 6.
(left[begin{array}{ll} -1 & 5 \ -3 & 2 end{array}right])
Solution:
Let A = (left[begin{array}{ll} -1 & 5 \ -3 & 2 end{array}right])

Question 7.
(left[begin{array}{lll}
1 & 2 & 3 \
0 & 2 & 4 \
0 & 0 & 5
end{array}right])
Solution:

Question 8.
(left[begin{array}{lll}
1 & 0 & 0 \
3 & 3 & 0 \
5 & 2 & -1
end{array}right])
Solution:

Question 9.
(left[begin{array}{lll}
2 & 1 & 3 \
4 & -1 & 0 \
-7 & 2 & 1
end{array}right])
Solution:

Question 10.
(left[begin{array}{lll}
1 & -1 & 2 \
0 & 2 & -3 \
3 & -2 & 4
end{array}right])
Solution:

Question 11.
(left[begin{array}{ccc}
1 & 0 & 0 \
0 & cos alpha & sin alpha \
0 & sin alpha & -cos alpha
end{array}right])
Solution:

Question 12.
Let A = (left[begin{array}{ll} 3 & 7 \ 2 & 5 end{array}right]) and B = (left[begin{array}{ll} 6 & 8 \ 7 & 9 end{array}right]), verify that (AB)^-1 = B^-1A^-1.
Solution:

Question 13.
If A = (left[begin{array}{ll} 3 & 1 \ -1 & 2 end{array}right]), show that A² – 5A + 7I = O. Hence, find A^-1.
Solution:
A = (left[begin{array}{ll} 3 & 1 \ -1 & 2 end{array}right])
∴ A² = (left[begin{array}{ll} 3 & 1 \ -1 & 2 end{array}right]) (left[begin{array}{ll} 3 & 1 \ -1 & 2 end{array}right]) = (left[begin{array}{cc} 9-1 & 3+2 \ -3-2 & -1+4 end{array}right])
= (left[begin{array}{ll} 8 & 5 \ -5 & 3 end{array}right])
∴ A² – 5A + 7I = (left[begin{array}{ll} 8 & 5 \ -5 & 3 end{array}right]) – 5(left[begin{array}{ll} 3 & 1 \ -1 & 2 end{array}right]) + 7(left[begin{array}{ll} 1 & 0 \ 0 & 1 end{array}right])
= (left[begin{array}{ll} 8 & 5 \ -5 & 3 end{array}right]) + (left[begin{array}{ll} -15 & -5 \ 5 & -10 end{array}right]) + (left[begin{array}{ll} 7 & 0 \ 0 & 7 end{array}right])
= (left[begin{array}{cc} 8-15+7 & 5-5+0 \-5+5+0 & 3-10+7 end{array}right]) = (left[begin{array}{ll} 0 & 0 \ 0 & 0 end{array}right]) = O.
∴ A² – 5A + 7I = O.
Multiplying by A^-1, we get
(A^-1A)A – 5^-1A + 7A^-1 I = O
⇒ IA – 5I + 7A^-1 = O
∴ 7A^-1 = 5I – IA = 5I – A
= 5(left[begin{array}{ll} 1 & 0 \ 0 & 1 end{array}right]) – (left[begin{array}{ll} 3 & 1 \ -1 & 2 end{array}right]) = (left[begin{array}{ll} 5 & 0 \ 0 & 5 end{array}right]) – (left[begin{array}{ll} 3 & 1 \ -1 & 2 end{array}right])
= (left[begin{array}{ll} 5 & 0 \ 0 & 5 end{array}right]) – (left[begin{array}{ll} 3 & 1 \ -1 & 2 end{array}right])
= (left[begin{array}{ll} 5-3 & 0-1 \ 0+1 & 5-2 end{array}right]) = (left[begin{array}{ll} 2 & -1 \ 1 & 3 end{array}right])
∴ A^-1 = (frac { 1 }{ 7 })(left[begin{array}{ll} 2 & -1 \ 1 & 3 end{array}right]).

Question 14.
For the matrix A = (left[begin{array}{ll} 3 & 2 \ 1 & 1 end{array}right]), find the numbers a and b such that A² + aA + bI = O. Hence, find A^-1.
Solution:
A² = A x A = (left[begin{array}{ll} 3 & 2 \ 1 & 1 end{array}right]) (left[begin{array}{ll} 3 & 2 \ 1 & 1 end{array}right])
= (left[begin{array}{ll} 9+2 & 6+2 \ 3+1 & 2+1 end{array}right]) = (left[begin{array}{ll} 11 & 8 \ 4 & 3 end{array}right]).
Now, A² + aA + bI = O
∴ (left[begin{array}{ll} 11 & 8 \ 4 & 3 end{array}right]) + (left[begin{array}{ll} 3a & 2a \ a & a end{array}right]) + (left[begin{array}{ll} b & 0 \ 0 & b end{array}right]) = (left[begin{array}{ll} 0 & 0 \ 0 & 0 end{array}right])
or (left[begin{array}{cc} 11+3 a+b & 8+2 a \ 4+a & 3+a+b end{array}right]) = (left[begin{array}{ll} 0 & 0 \ 0 & 0 end{array}right])
So, 4 + a = 0, ∴ a = – 4, 3 + a + b = 0 or 3 – 4 + b = 0 ⇒ b = 1.
Also, 11 + 3a + b = 11 – 12 + 1 = 0
and 8 + 2a = 0.
Thus, for a = – 4, b = 1, A² + aA + bI = O.
Putting a = – 4 and b = 1
in A² + aA + bI = O, we get A² – 4A + I = O.
Multiplying by A^-1, we get
(A^-1A)A – 4(A^-1A) + A^-1I = O
or A – 4I + A^-1 = O or A^-1 = 4I – A
⇒ A^-1 = 4(left[begin{array}{ll} 1 & 0 \ 0 & 1 end{array}right]) – (left[begin{array}{ll} 3 & 2 \ 1 & 1 end{array}right]) = (left[begin{array}{ll} 4 & 0 \ 0 & 4 end{array}right]) – (left[begin{array}{ll} 3 & 2 \ 1 & 1 end{array}right])
= (left[begin{array}{ll} 1 & -2 \ -1 & 3 end{array}right])

Question 15.
For the matrix A = (left[begin{array}{ccc} 1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3 end{array}right]) show that A³ – 6A² + 5A + 11I₃ = O. Hence, find A^-1.
Solution:

Question 16.
If A = (left[begin{array}{ccc} 2 & -1 & 1 \ 1 & 2 & -1 \ 1 & -1 & 2 end{array}right]), verify that A³ – 6A² + 9A – 4I = O. Hence, find A^-1.
Solution:

Question 17.
Let A be a non-singular square matrix of order 3 x 3. Then, | adj A | is equal to
(A) |A|
(B) |A|²
(C) |A|³
(D) 3|A|
Solution:

Sum of products of elements of a row and their corresponding cofactors = |A| and sum of products of elements of a row arid cofactor of another row = 0.
∴ |A adj A| = |A| | adj A | = (left|begin{array}{ccc}
mathrm{A} & 0 & 0 \
0 & |mathrm{~A}| & 0 \
0 & 0 & mathrm{~A}
end{array}right|)
= |A|³
∴ Dividing by |A|, | adj A | = | A |².
So, part (B) is the correct answer.

Question 18.
If A is an invertible matrix of order 2, then det (A^-1) ¡s equal to
(A) det |A|
(B) (frac { 1 }{ det(A) })
(C) 1
(D) 0
Solution:
|A| ≠ 0 ⇒ A^-1 ⇒ A^-1 = 1.
∴ |AA^-1A| = |I| = 1.
or |A| |A^-1| = 1
∴ |A^-1| = (frac { 1 }{ |A| }).
∴ Part (B) is the correct answer.