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In the inverse of the matrix $A=\left[\begin{array}{ccc}-1 & -3 & -2 \\ 0 & 1 & 2 \\ 3 & 4 & 5\end{array}\right]$ is $A^{-1}=$ $\left[\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right]$, then $a_1+c_2+b_3=$
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Verified Answer
The correct answer is:
$\frac{2}{3}$
Cofactors of $A$ :
$$
\begin{array}{lll}
C_{11}=-3 & C_{12}=6 & C_{13}=-3 \\
C_{21}=7 & C_{22}=1 & C_{23}=-5 \\
C_{31}=-4 & C_{32}=2 & C_{33}=-1
\end{array}
$$
$\begin{array}{ll}\therefore & \text { adj. } A\left[\begin{array}{ccc}-3 & 7 & -4 \\ 6 & 1 & 2 \\ -3 & -5 & -1\end{array}\right] \\ \therefore & |A|=-1(-3)-3(6)-3(-2)=-9 \\ \therefore & A^{-1}=\frac{-1}{9}\left[\begin{array}{ccc}-3 & 7 & -4 \\ 6 & 1 & 2 \\ -3 & -5 & -1\end{array}\right] \\ \therefore & a_1+c_2+b_3=\frac{3}{9}+\frac{5}{9}-\frac{2}{9}=\frac{2}{3} .\end{array}$
$$
\begin{array}{lll}
C_{11}=-3 & C_{12}=6 & C_{13}=-3 \\
C_{21}=7 & C_{22}=1 & C_{23}=-5 \\
C_{31}=-4 & C_{32}=2 & C_{33}=-1
\end{array}
$$
$\begin{array}{ll}\therefore & \text { adj. } A\left[\begin{array}{ccc}-3 & 7 & -4 \\ 6 & 1 & 2 \\ -3 & -5 & -1\end{array}\right] \\ \therefore & |A|=-1(-3)-3(6)-3(-2)=-9 \\ \therefore & A^{-1}=\frac{-1}{9}\left[\begin{array}{ccc}-3 & 7 & -4 \\ 6 & 1 & 2 \\ -3 & -5 & -1\end{array}\right] \\ \therefore & a_1+c_2+b_3=\frac{3}{9}+\frac{5}{9}-\frac{2}{9}=\frac{2}{3} .\end{array}$
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