1.

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)


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