Question

Given a =
$\binom{ - 3 \: \: \: \: 6 }{0 \: \: \: - 9}$
and a^t is its transpose matrix . Find :

${a}^{t} - \frac{1}{3} a$

​

Answer 1

Solution :-

Given matrix ,

• $\sf [A]=\left[\begin{array}{cc}\sf -3&\sf 6\\\sf 0 & \sf -9\end{array}\right]$

We need to find ,

• $\sf A^T - \dfrac{1}{3}A$

Firstly finding $\textbf{\textsf{A ^T }}$

As we know that ,

• Transpose matrix :- Transpose of a matrix is defined as the matrix which is converted row to column and column to row .

So , now

$\to\textbf{\textsf{A ^T }}=\sf \left[\begin{array}{cc}\sf 3&\sf 0\\\sf 6 &\sf 9\end{array}\right]$

Now , $\textbf{\textsf{A ^T }}-\dfrac{\sf 1}{\sf 3}A$

Substituting the value of $\sf A^T$

$\leadsto \sf\left[ \begin{array}{cc}\sf 3&\sf 0\\\sf 6&\sf 9\end{array}\right] - \dfrac{1}{3}\left[\begin{array}{cc}\sf 3&\sf6\\\sf 0&\sf 9\end{array}\right]$

$\leadsto\sf \left[\begin{array}{cc}\sf 3&\sf 0\\\sf 6&\sf 9\end{array}\right]-\left[\begin{array}{cc}\sf 1 & \sf 2\\\sf 0 & 3 \end{array}\right]$

$\leadsto\left[\begin{array}{cc}\sf 3-1&\sf 0-2\\\sf 6-0&\sf 9-3\end{array}\right]$

$\leadsto \left[\begin{array}{cc}\textbf{\textsf{2}}&\textbf{\textsf{ - 2}}\\\textbf{\textsf{ 6}}&\textbf{\textsf{6}}\end{array}\right]$

Hence, $\sf A^T - \dfrac{1}{3}A = \left[\begin{array}{cc}\textbf{\textsf{2}}&\textbf{\textsf{ - 2}}\\\textbf{\textsf{ 6}}&\textbf{\textsf{6}}\end{array}\right]$