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.