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  4. If the inverse of the matrix A=[3 4 5 2 -1 8 5 -2 7] is B, then BT=

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Q. If the inverse of the matrix $A=\begin{bmatrix}3 & 4 & 5 \\ 2 & -1 & 8 \\ 5 & -2 & 7\end{bmatrix}$ is $B$, then $B^{T}=$

AP EAMCETAP EAMCET 2019

Solution:

Given, $A=\begin{bmatrix}3 & 4 & 5 \\ 2 & -1 & 8 \\ 5 & -2 & 7\end{bmatrix}$, then
$|A|=\begin{bmatrix}3 & 4 & 5 \\ 2 & -1 & 8 \\ 5 & -2 & 7\end{bmatrix}$
$=3(-7+16)-4(14-40)+5(-4+5)$
$=27+104+5=136 \neq 0$
Thus, $|A| \neq 0$ and therefore $A^{-1}$ exists.
$\therefore A_{11}=\left|\begin{array}{cc}-1 & 8 \\ -2 & 7\end{array}\right|=9, A_{12}=-\left|\begin{array}{cc}2 & 8 \\ 5 & 7\end{array}\right|=26$
$A_{13}=\begin{vmatrix}2 & -1 \\ 5 & -2\end{vmatrix}=1$
$A_{21}=-\begin{vmatrix}4 & 5 \\ -2 & 7\end{vmatrix}=-38, A_{22}=\begin{vmatrix}3 & 5 \\ 5 & 7\end{vmatrix}=-4$
$A_{23}=-\begin{vmatrix}3 & 4 \\ 5 & -2\end{vmatrix}=26$
$A_{31}=\begin{vmatrix}4 & 5 \\ -1 & 8\end{vmatrix}=37, A_{32}=-\begin{vmatrix}3 & 5 \\ 2 & 8\end{vmatrix}=-14$,
$A_{33}=\begin{vmatrix}3 & 4 \\ 2 & -1\end{vmatrix}=-11$
$\therefore \text{adj}(A)=\begin{bmatrix}9 & 26 & 1 \\ -38 & -4 & 26 \\ 37 & -14 & -11\end{bmatrix}^{'}=\begin{bmatrix}9 & -38 & 37 \\ 26 & -4 & -14 \\ 1 & 26 & -11\end{bmatrix}$
Hence,
$A^{-1}=\frac{1}{|A|}(\text{adj} A)=\frac{1}{136}\begin{bmatrix}9 & -38 & 37 \\ 26 & -4 & -14 \\ 1 & 26 & -11\end{bmatrix}=B$
$ \therefore B^{T} =\frac{1}{136}\begin{bmatrix}9 & -38 & 37 \\ 26 & -4 & -14 \\ 1 & 26 & -11\end{bmatrix}^{'} $
$=\frac{1}{136}\begin{bmatrix}9 & 26 & 1 \\ -38 & -4 & 26 \\ 37 & -14 & -11\end{bmatrix}$