Question

If the inverse of the matrix $A=\left[\begin{array}{ccc}3& 4& 5\\ 2& -1& 8\\ 5& -2& 7\end{array}\right]$ is $B$, then $B^T=$

• A. $\frac{1}{136}\left[\begin{array}{ccc}9& 26& 1\\ -38& -4& 26\\ 37& -14& -11\end{array}\right]$
• B. $\frac{1}{136}\left[\begin{array}{ccc}9& -38& 37\\ 26& -4& -14\\ 1& 26& -11\end{array}\right]$
• C. $\frac{1}{136}\left[\begin{array}{ccc}9& 26& 1\\ 37& -14& -11\\ -38& -4& 26\end{array}\right]$
• D. $\frac{1}{136}\left[\begin{array}{ccc}9& 1& 26\\ -38& 26& -4\\ 37& -11& -14\end{array}\right]$

Answer 1

Answer: (A)

Given, $A=\left[\begin{array}{ccc}3& 4& 5\\ 2& -1& 8\\ 5& -2& 7\end{array}\right]$, then
$|A|=\left[\begin{array}{ccc}3& 4& 5\\ 2& -1& 8\\ 5& -2& 7\end{array}\right]$
$=3(-7+16)-4(14-40)+5(-4+5)$
$=27+104+5=136\ne 0$
Thus, $|A|\ne 0$ and therefore $A^{-1}$ exists.
$\therefore A_{11}=\left|\begin{array}{cc}-1& 8\\ -2& 7\end{array}\right|=9,A_{12}=-\left|\begin{array}{cc}2& 8\\ 5& 7\end{array}\right|=26$
$A_{13}=\left|\begin{array}{cc}2& -1\\ 5& -2\end{array}\right|=1$
$A_{21}=-\left|\begin{array}{cc}4& 5\\ -2& 7\end{array}\right|=-38,A_{22}=\left|\begin{array}{cc}3& 5\\ 5& 7\end{array}\right|=-4$
$A_{23}=-\left|\begin{array}{cc}3& 4\\ 5& -2\end{array}\right|=26$
$A_{31}=\left|\begin{array}{cc}4& 5\\ -1& 8\end{array}\right|=37,A_{32}=-\left|\begin{array}{cc}3& 5\\ 2& 8\end{array}\right|=-14$,
$A_{33}=\left|\begin{array}{cc}3& 4\\ 2& -1\end{array}\right|=-11$
$\therefore \text{adj}(A)={\left[\begin{array}{ccc}9& 26& 1\\ -38& -4& 26\\ 37& -14& -11\end{array}\right]}^{{}^{\prime }}=\left[\begin{array}{ccc}9& -38& 37\\ 26& -4& -14\\ 1& 26& -11\end{array}\right]$
Hence,
$A^{-1}=\frac{1}{|A|}(\text{adj}A)=\frac{1}{136}\left[\begin{array}{ccc}9& -38& 37\\ 26& -4& -14\\ 1& 26& -11\end{array}\right]=B$
$\therefore B^T=\frac{1}{136}{\left[\begin{array}{ccc}9& -38& 37\\ 26& -4& -14\\ 1& 26& -11\end{array}\right]}^{{}^{\prime }}$
$=\frac{1}{136}\left[\begin{array}{ccc}9& 26& 1\\ -38& -4& 26\\ 37& -14& -11\end{array}\right]$