Q.1: Find the adjoint of each of the matrices
\(\begin{bmatrix} 2 & 4 \\ 5 & 7 \end{bmatrix}\)
Sol:
Suppose A = \(\begin{bmatrix} 2 & 4 \\ 5 & 7 \end{bmatrix}\)
For X = \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\\\)
Then adj (X) = \(\begin{bmatrix} d & – b \\ – c & a \end{bmatrix}\)
We have,
A 11 = 7, A12 = – 4, A21 = -5, A22 = 2
Therefore, adj (A) = \(\begin{bmatrix} 7 & – 4 \\ – 5 & 2 \end{bmatrix}\)
Q.2: Find the adjoint of each of the matrices
\(\begin{bmatrix} 1 & – 1 & 2 \\ 2 & 3 & 5 \\ – 2 & 0 & 1 \end{bmatrix}\)
Sol:
Suppose, D = \(\begin{bmatrix} 1 & – 1 & 2 \\ 2 & 3 & 5 \\ – 2 & 0 & 1 \end{bmatrix}\)
We have,
\(D_{11} = \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = 3 – 0 = 3 \\ D_{12} = – \begin{vmatrix} 2 & 5 \\ – 2 & 1 \end{vmatrix} = – (2 + 10) = – 12 \\ D_{13} = \begin{vmatrix} 2 & 3 \\ – 2 & 0 \end{vmatrix} = 0 + 6 = 6 \\\) \(D_{21} = – \begin{vmatrix} – 1 & 2 \\ 0 & 1 \end{vmatrix} = – (- 1 – 0) = 1 \\ D_{22} = \begin{vmatrix} 1 & 2 \\ – 2 & 1 \end{vmatrix} = 1 + 4 = 5 \\ D_{23} = – \begin{vmatrix} 1 & – 1 \\ – 2 & 0 \end{vmatrix} = – (0 – 2) = 2 \\\) \(D_{31} = \begin{vmatrix} – 1 & 2 \\ 3 & 5 \end{vmatrix} = – 5 – 6 = – 11 \\ D_{32} = – \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = – (5 – 4) = – 1 \\ D_{33} = \begin{vmatrix} 1 & – 1 \\ 2 & 3 \end{vmatrix} = 3 + 2 = 5\\\)Hence, adj D = \(\\\begin{bmatrix} D ^{11} & D ^{12} & D ^{13} \\ D ^{21} & D ^{22} & D ^{23} \\ D ^{31} & D ^{32} & D ^{33} \end{bmatrix} = \begin{bmatrix} 3 & 1 & – 11 \\ – 12 & 5 & – 1 \\ 6 & 2 & 5 \end{bmatrix}\)
Q.3: Prove whether D (adj D) = (adj D) D = \(\left | D \right | I\)
\(D = \begin{bmatrix} 2 & 3 \\ – 4 & – 6 \end{bmatrix}\)
Sol:
\(D = \begin{bmatrix} 2 & 3 \\ – 4 & – 6 \end{bmatrix}\)We have,
\(\left | D \right | = – 12 – (- 12) = – 12 + 12 = 0 \\ \left | D \right | I = 0 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)D 11 = – 6, D 12 = 4, D 21 = – 3, D 22 = 2
adj D = \(\begin{bmatrix} – 6 & – 3 \\ 4 & 2 \end{bmatrix}\)
Now,
D (adj D) = \(\begin{bmatrix} 2 & 3 \\ – 4 & – 6 \end{bmatrix} \begin{bmatrix} – 6 & – 3 \\ 4 & 2 \end{bmatrix} \\ = \begin{bmatrix} – 12 + 12 & – 6 + 6 \\ 24 – 24 & 12 – 12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
And,
(adj D) D = \(\begin{bmatrix} – 6 & – 3 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ – 4 & – 6 \end{bmatrix} \\ = \begin{bmatrix} – 12 + 12 & – 18 + 18 \\ 8 – 8 & 12 – 12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
Thus, D (adj D) = (adj D) D = \(\left | D \right | I\)
Q.4: Obtain the inverse of the matrices if it exists
\(\begin{bmatrix} 2 & – 2 \\ 4 & 3 \end{bmatrix}\)
Sol:
Suppose D = \(\begin{bmatrix} 2 & – 2 \\ 4 & 3 \end{bmatrix}\)
We have,
\(\left | D \right | = 6 + 8 = 14\)Now,
D 11 = 3, D 12 = – 4, D 21 = 2, D 22 = 2
adj D = \(\begin{bmatrix} 3 & 2 \\ – 4 & 2 \end{bmatrix} \\ D ^{- 1} = \frac{1}{\left | D \right |} = \frac{1}{14}\; \begin{bmatrix} 3 & 2 \\ – 4 & 2 \end{bmatrix}\)
Q.5: Obtain the inverse of the matrices if it exists
\(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}\)
Sol:
Suppose D = \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}\)
We have,
\(\left | D \right | = 1 (10 – 0) – 2 (0 – 0) + 3 (0 – 0) = 10\)Now, D 11 = 10 – 0 = 10, D 12 = – (0 – 0) = 0, D 13 = 0 – 0 = 0, D 21 = – (10 – 0) = – 10, D 22 = 5 – 0 = 5, D 23 = – (0 – 0) = 0, D 31 = 8 – 6 = 2, D 32 = – (4 – 0) = – 4, D 33 = 2 – 0 = 2
adj D = \(\begin{bmatrix} 10 & – 10 & 2 \\ 0 & 5 & – 4 \\ 0 & 0 & 2 \end{bmatrix} \\\)
D-1 = \(\\\frac{1}{\left | D \right |} adj D = \frac{1}{10} \begin{bmatrix} 10 & – 10 & 2 \\ 0 & 5 & – 4 \\ 0 & 0 & 2 \end{bmatrix}\)
Q.6: Find the inverse of each of the matrices (if it exists).
\(\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos \alpha & sin \alpha \\ 0 & sin \alpha & – cos \alpha \end{bmatrix}\)
Sol:
Let A = \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos \alpha & sin \alpha \\ 0 & sin \alpha & – cos \alpha \end{bmatrix}\)
We have,
\(\left |A \right | = 1 \left ( – cos ^{2} \alpha – sin ^{2} \alpha\right )\) \(\left |A \right | = – \left ( cos ^{2} \alpha + sin ^{2} \alpha \right ) = – 1\)Now,
\(A _{ 11 } = – cos ^{2 } \alpha – sin ^{2} \alpha = – 1 , A _{ 12 } = 0 , A _{ 13 } = 0\) \(A _{21} = 0 , A _{ 22 } = – cos \alpha , A _{ 23 } = – sin \alpha \\ A _{ 31 } = 0 , A_{32} = – sin \alpha , A_{33} = cos \alpha\)Therefore,
\(adj A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & – cos \alpha & – sin \alpha \\ 0 & – sin \alpha & cos \alpha \end{bmatrix}\)Therefore,
\(A ^{ -1 } = \frac{ 1 }{ \left | A \right |}.adj A\) = \(\begin{bmatrix} -1 & 0 & 0 \\ 0 & – cos \alpha & – sin \alpha \\ 0 & – sin \alpha & cos \alpha \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos \alpha & sin \alpha \\ 0 & sin \alpha & – cos \alpha \end{bmatrix}\)
Q.7: Let \(A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\) and \(B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}\). Verify that \(\left ( AB \right )^{ -1} = B ^{-1} A ^{ -1 }\)
Sol:
Let \(A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\)
We have ,
\(\left |A \right | = 15 – 14 = 1\)Now,
\(A _{ 11} = 5 , A _{ 12} = – 2 , A _{ 21} = – 7 , A _{ 22} = 3\)Therefore,
\(adj A = \begin{bmatrix} 5 & – 7 \\ – 2 & 3 \end{bmatrix}\)Therefore,
\(A ^{ – 1 } = \frac{1}{ \left |A \right |}. adj A = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}\)Now, let \(adj B = \begin{bmatrix} 6 & 8 \\ 7 & 6 \end{bmatrix}\)
We have ,
\(\left |B \right | = 54 – 56 = – 2\)Therefore,
\(adj B = \begin{bmatrix} 9 & – 8 \\ – 7 & 6 \end{bmatrix}\)Therefore,
\(B ^{ – 1 } = \frac{1}{ \left |B \right |}. adj B = 0.5 \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}\) \(\begin{bmatrix} \frac{ – 9}{2} & 4 \\ \frac{7}{2} & – 3 \end{bmatrix}\)Now,
\(B^{-1}A^{-1} = \begin{bmatrix} \frac{-9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix}\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}\) \(B^{-1}A^{-1} = \begin{bmatrix} \frac{-45}{2} – 8 & \frac{63}{2} + 12 \\ \frac{35}{2} + 6 & \frac{-49}{2} – 9 \end{bmatrix}\begin{bmatrix} \frac{ 61}{2} & \frac{87}{2} \\ \frac{47}{2} & \frac{ -67}{2} \end{bmatrix}\)Then,
\(\\\boldsymbol{\Rightarrow }\) \(AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\begin{bmatrix} 6 & 7 \\ 7 & 9 \end{bmatrix}\\\)
\(\\\boldsymbol{\Rightarrow }\) \(AB = \begin{bmatrix} 18 + 49 & 24 + 63 \\ 12 + 35 & 16 + 45 \end{bmatrix}\\\)
\(\\\boldsymbol{\Rightarrow }\) \(AB = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}\\\)
Therefore, we have \(AB = \begin{bmatrix} 18 + 49 & 24 + 63 \\ 12 + 35 & 16 + 45 \end{bmatrix}\)
Also,
\(adj (AB) = \begin{bmatrix} 61 & – 87 \\ – 47 & 67 \end{bmatrix}\)Therefore, \(\left ( AB \right )^{ -1 } = \frac{1}{ \left |AB \right |} adj\left ( AB \right ) = \frac{-1}{2}\begin{bmatrix} 61 & – 87 \\ – 47 & 67 \end{bmatrix}\) . . . . . . . . . . . . . . . (2)
\(= \begin{bmatrix} \frac{61}{2} & \frac{87}{2}\\ \frac{47}{2} & \frac{-67}{2} \end{bmatrix}\)From (1) and (2), we have: (AB)−1 = B−1 A -1 Hence, the given result is proved.
Q.8: If \(A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\), show that A 2 – 5 A + 7 I = 0. Hence find A – 1
Sol:
\(A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)A 2 = A. A = \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)\(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\) = \(A = \begin{bmatrix} 9 – 1 & 3 +2 \\ -3 – 2 & – 1 + 4 \end{bmatrix}\) = \(\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}\)
Therefore,
A 2 – 5 A + 7 I = 0
= \(\\\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}\) – 5 \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\) + 7 \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\\\)
= \(\\\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}\) – \(\begin{bmatrix} 15 & 5 \\ – 5 & 10 \end{bmatrix}\) + \(\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\\\)
= \(\\\begin{bmatrix} – 7 & 0 \\ 0 & – 7 \end{bmatrix}\) – \(\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\) = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\\\)
Therefore, A 2 – 5 A + 7 I = 0
Therefore, A.A – 5 A = – 7I
\(\boldsymbol{\Rightarrow }\) A.A ( A -1 ) – 5A A -1 = – 7I A -1 . . . . . . . . . . . . . . . . . [ post – multiplying by A -1 as \(\left | A \right | \neq 0\) ]
\(\boldsymbol{\Rightarrow }\) A. ( A A -1 ) – 5I = – 7 A -1
\(\boldsymbol{\Rightarrow }\) AI – 5I = – 7 A -1
\(\boldsymbol{\Rightarrow }\) A -1 \(= – \frac{ 1 }{ 7 }\left ( A – 5I \right )\)
\(\boldsymbol{\Rightarrow }\) A -1 \(= – \frac{ 1 }{ 7 }\left ( 5I – A \right )\)
=\(\\\frac{1}{7} \left ( \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} – \begin{bmatrix} 3 & 1 \\ – 1 & 2 \end{bmatrix} \right ) = \frac{1}{ 7 } \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}\)
Therefore,
A -1 = \(\frac{1}{ 7 } \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}\)
Q.9: Let A be a nonsingular square matrix of order 3 × 3. Then \(\left | adj A \right |\) is equal to
(A) \(\left | adj A \right |\)
(B) \(\left | adj A \right |^{2}\)
(C) \(\left | adj A \right |^{ 3 }\)
(D) 3 \(\left | adj A \right |\)
Sol:
The correct option is B
\(\left ( adj A \right )A = \left | A \right |I = \begin{bmatrix} \left | A \right | & 0 & 0 \\ 0 & \left | A \right | & 0 \\ 0 & 0 & \left | A \right | \end{bmatrix}\)\(\boldsymbol{\Rightarrow }\) \(\left ( adj A \right )A = \begin{bmatrix} \left | A \right | & 0 & 0 \\ 0 & \left | A \right | & 0 \\ 0 & 0 & \left | A \right | \end{bmatrix}\)
\(\boldsymbol{\Rightarrow }\) \(\left ( adj A \right )\left |A \right | = \left | A \right |^{3} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \left | A \right |^{3} \left (I \right )\)
Therefore, \(\left |\left ( adj A \right ) \right | = \left | A \right |^{2}\)
Hence, the correct Sol: is B.
Q.10: If A is an invertible matrix of order 2, then det (A −1 ) is equal to
(A) det (A)
(B) \(\frac{ 1 }{ det \left ( A \right )}\)
(C) 1
(D) 0
Sol:
Since A is an invertible matrix , A -1 exists and A -1 = \(\frac{ 1 }{ \left | a \right |} adj A\)
As matrix A is of order 2, let \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\)
Then, \(\left |A \right | = ad – bc and adj A = \begin{bmatrix} d & -b \\ – c & a \end{bmatrix}\)
Now,
\(A^{-1} = \frac{1}{ \left | A \right |} adj A = \begin{bmatrix} \frac{d}{\left | A \right |} & \frac{-b}{ \left | A \right |}\\ \frac{-c}{ \left | A \right |} & \frac{a }{ \left | A \right |} \end{bmatrix}\)Therefore,
\(\left |A^{-1} \right | = \begin{bmatrix} \frac{d}{\left | A \right |} & \frac{-b}{ \left | A \right |}\\ \frac{-c}{ \left | A \right |} & \frac{a }{ \left | A \right |} \end{bmatrix} = \frac{1}{ \left | A \right |^{2}}\begin{bmatrix} d & – b \\ – c & a \end{bmatrix} = \frac{1}{ \left | A \right |^{2}} \left ( ad – bc \right ) = \frac{1}{ \left | A \right |^{2}} . \left |A \right | = \frac{1}{ \left | A \right |}\)Therefore,
\(det \left ( A \right )^{ -1} = \frac{ 1 }{ det \left ( A \right )}\)Hence, the correct Sol: is B.
Q.11: Suppose D = \(\begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix}\), verify that D 3 – 6 D2 + 9D – 4I = 0 and hence find D – 1
Sol:
D = \(\begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix}\\\)
D 2 = \(\\\begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix}\\\)
\(\\\begin{bmatrix} 4 + 1 + 1 & – 2 – 2 – 1 & 2 + 1 + 2 \\ – 2 – 2 – 1 & 1 + 4 + 1 & – 1 – 2 – 2 \\ 2 + 1 + 2 & – 1 – 2 – 2 & 1 + 1 + 4 \end{bmatrix} \\ = \begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix}\\\)D 3 = D 2 D = \(\\\begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix} \begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix} \\ = \begin{bmatrix} 12 + 5 + 5 & – 6 – 10 – 5 & 6 + 5 + 10 \\ – 10 – 6 – 5 & 5 + 12 + 5 & 5 – 6 – 10 \\ 10 + 5 + 6 & – 5 – 10 – 6 & 5 + 5 + 12 \end{bmatrix} \\ = \begin{bmatrix} 22 & – 21 & 21 \\ – 21 & 22 & – 21 \\ 21 & – 21 & 22 \end{bmatrix}\\\)
Now,
D 3 – 6 D2 + 9D – 4I = 0
= \(\\\begin{bmatrix} 22 & – 21 & 21 \\ – 21 & 22 & – 21 \\ 21 & – 21 & 22 \end{bmatrix} – 6 \begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix} + 9 \begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix} – 4 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\\\)
= \(\\\begin{bmatrix} 22 & – 21 & 21 \\ – 21 & 22 & – 21 \\ 21 & – 21 & 22 \end{bmatrix} – \begin{bmatrix} 36 & – 30 & 30 \\ – 30 & 36 & – 30 \\ 30 & – 30 & 36 \end{bmatrix} + \begin{bmatrix} 18 & – 9 & 9 \\ – 9 & 18 & – 9 \\ 9 & – 9 & 18 \end{bmatrix} – \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}\\\)
= \(\\ \begin{bmatrix} 40 & – 30 & 30\\ – 30 & 40 & – 30 \\ 30 & – 30 & 40 \end{bmatrix} – \begin{bmatrix} 40 & – 30 & 30\\ – 30 & 40 & – 30 \\ 30 & – 30 & 40 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\)
Hence, D 3 – 6 D2 + 9D – 4I = 0
Now,
D 3 – 6 D2 + 9D – 4I = 0
(DDD) D – 1 – 6 D + 9D – 4I = 0 [Multiplying by D – 1 on as \(\left | D \right | I\) is not equal to 0]
DD (DD– 1) – 6D (DD– 1) + 9 (DD– 1) = 4 ID– 1
DDI – 6DI + 9I = 4 D– 1
D2 – 6D + 9I = 4 D– 1
D– 1 = \(\frac{1}{4}\) (D2 – 6D + 9I)
D2 – 6D + 9I
= \(\\\begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix} – 6 \begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix} + 9 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\\)
= \(\\\begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix} – \begin{bmatrix} 12 & – 6 & 6 \\ – 6 & 12 & – 6 \\ 6 & – 6 & 12 \end{bmatrix} + \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} \\ = \begin{bmatrix} 3 & 1 & – 1 \\ 1 & 3 & 1 \\ – 1 & 1 & 3 \end{bmatrix} From\; equation\; (1), we\; have, D ^{- 1} = \frac{1}{4} \begin{bmatrix} 3 & 1 & – 1 \\ 1 & 3 & 1 \\ – 1 & 1 & 3 \end{bmatrix}\)
