# GSEB Solutions Class 12 Maths Chapter 4 Determinants Ex 4.5

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.