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1.

If A is a square matrix of order n, prove that |A adj A| = |A|n.

Answer»

A adj A| = |A|n

LHS |A adj A|

A|.|adj A|

|A|.|A|n–1

|A|n–1+1

|A|n = RHS

Hence, LHS = RHS
Discussion
2.

Find the adjoint of the Matrices.\(\begin{bmatrix}-3&5\\2&4\end{bmatrix}\)Verify that (adj A) A=|A| I=A (adj A) for the above matrices.

Answer»

A = \(\begin{bmatrix}-3&5\\2&4\end{bmatrix}\)

Cofactors of A are 

C11 = 4 

C12 = – 2 

C21 = – 5 

C22 = – 3 

Since, adj A = \(\begin{bmatrix}C_{11}&C_{12}\\2_{21}&C_{22}\end{bmatrix}^T\)

(adj A) = \(\begin{bmatrix}4&-2\\-5&-3\end{bmatrix}^T\)

\(=\begin{bmatrix}4&-5\\-2&-3\end{bmatrix}^T\)

Now, (adj A)A \(=\begin{bmatrix}4&-5\\-2&-3\end{bmatrix}\)\(\begin{bmatrix}-3&5\\2&4\end{bmatrix}\)\(=\begin{bmatrix}-12&-10&20&-20\\6&-6&-10&-12\end{bmatrix}\)

(adj A)A \(=\begin{bmatrix}-22&0\\0&-22\end{bmatrix}\)

And, |A|.I =\(\begin{bmatrix}-3&5\\2&4\end{bmatrix}\)\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)=(-22)\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)\(=\begin{bmatrix}-22&0\\0&-22\end{bmatrix}\)

Also, A(adj A) = \(\begin{bmatrix}-3&5\\2&4\end{bmatrix}\)\(\begin{bmatrix}4&-5\\-2&-3\end{bmatrix}\)\(=\begin{bmatrix}-12&-10&20&-20\\6&-6&-10&-12\end{bmatrix}\)

A(adj A) \(=\begin{bmatrix}-22&0\\0&-22\end{bmatrix}\)

Hence, (adj A)A = |A|.I = A.(adj A)
Discussion
3.

Find the adjoint of the Matrices.\(\begin{bmatrix}a&b\\c&d\end{bmatrix}\)Verify that (adj A) A=|A| I=A (adj A) for the above matrices.

Answer»

A = \(\begin{bmatrix} a&b \\ c&d \end{bmatrix}\)

Cofactors of A are

C11 = d

C12 = – c

C21 = – b

C22 = a

Since, adj A =\(\begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T\)

(adj A) = \(\begin{bmatrix}d&-c\\-b&a\end{bmatrix}^T\)

\(=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\)

Now, (adj A)A \(=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\)\(\begin{bmatrix} a&b \\ c&d \end{bmatrix}\)\(=\begin{bmatrix}ad&-bc&bd&-bd\\-ac&+ac&-bc&+ad\end{bmatrix}\)

(adj A)A =\(\begin{bmatrix}ad&-bc&0\\0&ad&-bc\end{bmatrix}\)

And, |A|.I =\(\begin{bmatrix} a&b \\ c&d \end{bmatrix}\)\(\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}\) = (ad - bc)\(\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}\)= \(\begin{bmatrix}ad&-bc&0\\0&ad&-bc\end{bmatrix}\) 

Also, A(adj A) = \(\begin{bmatrix} a&b \\ c&d \end{bmatrix}\)\(\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\)\(=\begin{bmatrix}ad&-bc&bd&-bd\\-ac&+ac&-bc&+ad\end{bmatrix}\)

Hence, (adj A)A = |A|.I = A.(adj A)
Discussion
4.

Find the adjoint of the Matrices and Verify that (adj A) A = |A| I = A (adj A)\(\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}\)Verify that (adj A) A=|A| I=A (adj A) for the above matrices.

Answer»

A = \(\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}\)

Cofactors of A are: 

C11 = – 3 C21 = 2 C31 = 2 

C12 = 2 C22 = – 3 C23 = 2 

C13 = 2 C23 = 2 C33 = – 3

adj A =\(\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{33}\end{bmatrix}^T\)

\(\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}\)

Now, (adj A).A =\(\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}\)\(\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}\)

\(=\begin{bmatrix}-3+4+4&-6+2+4&-6+4+2\\2-3+4&4-3+4&4-6+2\\2+4-6&4+2-6&4+4-3\end{bmatrix}\)

\(=\begin{bmatrix}5&0&5\\0&5&0\\0&0&5\end{bmatrix}\)

Also, |A|.I =\(\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}\)\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)\(=(-3+4+4)\)\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)\(=\begin{bmatrix}5&0&5\\0&5&0\\0&0&5\end{bmatrix}\)

Then, A.(adj A) =\(\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}\)\(\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}\)

\(=\begin{bmatrix}-3+4+4&-6+2+4&-6+4+2\\2-3+4&4-3+4&4-6+2\\2+4-6&4+2-6&4+4-3\end{bmatrix}\)

\(=\begin{bmatrix}5&0&5\\0&5&0\\0&0&5\end{bmatrix}\)

Since, (adj A).A = |A|.I = A(adj A)
Discussion
5.

Find the adjoint of the Matrices.\(\begin{bmatrix}1&tan\frac{a}{2}\\-tan\frac{a}{2}&1\end{bmatrix}\)Verify that (adj A) A=|A| I=A (adj A) for the above matrices.

Answer»

A = \(\begin{bmatrix}1&tan\frac{\alpha}{2}\\-tan\frac{\alpha}{2}&1\end{bmatrix}\)

Cofactors of A are

C11 = 1 

C12 = \(tan\frac{\alpha}{2}\)

C21 = \(-tan\frac{\alpha}{2}\)

C22 = 1

Since, adj A =\(\begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T\)

(adj A) =\(\begin{bmatrix}1&tan\frac{\alpha}{2}\\-tan\frac{\alpha}{2}&1\end{bmatrix}^T\)

\(=\begin{bmatrix}1&tan\frac{\alpha}{2}\\-tan\frac{\alpha}{2}&1\end{bmatrix}\)

Now, (adj A)A = \(\begin{bmatrix}1&-tan\frac{\alpha}{2}\\tan\frac{\alpha}{2}&1\end{bmatrix}\)\(\begin{bmatrix}1&tan\frac{\alpha}{2}\\-tan\frac{\alpha}{2}&1\end{bmatrix}\)

\(=\begin{bmatrix}1+tan^2\frac{\alpha}{2}&tan\frac{\alpha}{2}-tan\frac{\alpha}{2}\\tan\frac{\alpha}{2}-tan\frac{\alpha}{2}&1+tan^2\frac{\alpha}{2}\end{bmatrix}\)

(adj A)A =\(\begin{bmatrix}1+tan^2\frac{\alpha}{2}&0\\0&1+tan^2\frac{\alpha}{2}\end{bmatrix}\)

And, |A|.I =\(\begin{bmatrix}1&tan\frac{\alpha}{2}\\-tan\frac{\alpha}{2}&1\end{bmatrix}\)\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)\(=(1+tan^2\frac{\alpha}{2})\)\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

\(=\begin{bmatrix}1+tan^2\frac{\alpha}{2}&0\\0&1+tan^2\frac{\alpha}{2}\end{bmatrix}\)

Also, A(adj A) =\(\begin{bmatrix}1&tan\frac{\alpha}{2}\\-tan\frac{\alpha}{2}&1\end{bmatrix}\)\(\begin{bmatrix}1&-tan\frac{\alpha}{2}\\tan\frac{\alpha}{2}&1\end{bmatrix}\)\(=\begin{bmatrix}1+tan^2\frac{\alpha}{2}&tan\frac{\alpha}{2}-tan\frac{\alpha}{2}\\tan\frac{\alpha}{2}-tan\frac{\alpha}{2}&1+tan^2\frac{\alpha}{2}\end{bmatrix}\)

\(=\begin{bmatrix}1+tan^2\frac{\alpha}{2}&0\\0&1+tan^2\frac{\alpha}{2}\end{bmatrix}\)

Hence, (adj A)A = |A|.I = A.(adj A)
Discussion
6.

Find the adjoint of the Matrices.\(\begin{bmatrix}cos\,a&sin\,a\\sin\,a&cos\,a\end{bmatrix}\)Verify that (adj A) A=|A| I=A (adj A) for the above matrices.

Answer»

A= \(\begin{bmatrix}cos\alpha&sin\alpha\\sin\alpha&cos\alpha\end{bmatrix}\) 

Cofactors of A are

C11 = \(cos\alpha\)

C12 = \(-sin\alpha\)

C21 = \(-sin\alpha\)

C22 =\(cos\alpha\)

Since, adj A = \(\begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T\)

(adj A) =\(\begin{bmatrix}cos\alpha&-sin\alpha\\-sin\alpha&cos\alpha\end{bmatrix}^T\)

\(=\begin{bmatrix}cos\alpha&-sin\alpha\\-sin\alpha&cos\alpha\end{bmatrix}\)

Now, (adj A)A \(=\begin{bmatrix}cos\alpha&-sin\alpha\\-sin\alpha&cos\alpha\end{bmatrix}\)\(\begin{bmatrix}cos\alpha&sin\alpha\\sin\alpha&cos\alpha\end{bmatrix}\)

\(=\begin{bmatrix}-sin^2\alpha+cos^2\alpha&cos\alpha.sin\alpha-sin\alpha.cos\alpha\\-cos\alpha sin\alpha+sin\alpha coa\alpha&-sin^2\alpha+cos^2\alpha\end{bmatrix}\)

(adj A)A =\(\begin{bmatrix}cos2\alpha&0\\0&cos2\alpha\end{bmatrix}\)

And, |A|.I =\(\begin{bmatrix}cos\alpha&sin\alpha\\sin\alpha&cos\alpha\end{bmatrix}\)\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

\(=(cos^2\alpha-sin^2\alpha)\)\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

\(=\begin{bmatrix}cos^2\alpha-sin^2\alpha&0\\0&cos^2\alpha-sin^2\alpha\end{bmatrix}\)

\(=\begin{bmatrix}cos2\alpha&0\\0&cos2\alpha\end{bmatrix}\)

Also, A(adj A) =\(\begin{bmatrix}cos\alpha&sin\alpha\\sin\alpha&cos\alpha\end{bmatrix}\)\(\begin{bmatrix}cos\alpha&-sin\alpha\\-sin\alpha&cos\alpha\end{bmatrix}\)\(=\begin{bmatrix}cos^2\alpha-sin^2\alpha&0\\0&cos^2\alpha-sin^2\alpha\end{bmatrix}\)

\(=\begin{bmatrix}cos2\alpha&0\\0&cos2\alpha\end{bmatrix}\)

Hence, (adj A)A = |A|.I = A.(adj A)
Discussion