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Find the inverse of the following matrices.

$A=\left(\begin{array}{ll}{3} & {2} \\ {3} & {4}\end{array}\right)$

$\Rightarrow det(A)=12-6=6 \neq 0$

$A^{-1}=\frac{1}{6}\left(\begin{array}{cc}{4} & {-2} \\ {-3} & {3}\end{array}\right)$

$A^{-1}=\left(\begin{array}{cc}{4 / 6} & {-2 / 6} \\ {-3 / 6} & {3 / 6}\end{array}\right)$

$A=\left(\begin{array}{lll}{3} & {1} & {2} \\ {2} & {4} & {6} \\ {3} & {5} & {4}\end{array}\right)$

$\Rightarrow det(A)=-36$

$A^{-1}=\frac{1}{det(A)} \cdot adj(A)$

$adj(A) = [C_{ij}]^{T}$

$\left[C_{i j}\right]$

$\left(\begin{array}{ccc}{+} & {-} & {+} \\ {-} & {+} & {-} \\ {+} & {-} & {+}\end{array}\right)$

$C_{11}=+\left|\begin{array}{ll}{4}&{6} \\ {5}&{4}\end{array}\right|, C_{12}=-\left|\begin{array}{ll}{2} & {6} \\ {3} & {4}\end{array}\right|, \quad C_{13}=+\left|\begin{array}{ll}{2} & {4} \\ {3} & {5}\end{array}\right|$

$=16-30=-14 \quad=-(8-18)=10 \quad 10-12=-2$

$c_{21}=-\left|\begin{array}{cc}{1} & {2} \\ {5} & {4}\end{array}\right|, \quad c_{22}=+\left|\begin{array}{cc}{3} & {2} \\ {3} & {4}\end{array}\right|, \quad C_{23}=-\left|\begin{array}{cc}{3} & {1} \\ {3} & {5}\end{array}\right|$

$-(4-10)=6 \quad 12-6=6 \quad-(15-3)=-12$

$c_{31}=+\left|\begin{array}{cc}{1} & {2} \\ {4} & {6}\end{array}\right|, \quad c_{32}=-\left|\begin{array}{cc}{3} & {2} \\ {2} & {6}\end{array}\right|, \quad c_{33}=+\left|\begin{array}{cc}{3} & {1} \\ {2} & {4}\end{array}\right|$

$6-8=-2 \quad -(18-4)=-14 \quad 12 - 2=16$

$C_{i j}=\left(\begin{array}{ccc}{-14} & {10} & {-2} \\ {6} & {6} & {-12} \\ {-2} & {-14} & {10}\end{array}\right)$

$\left[C_{i j}\right]^{T}=\left(\begin{array}{ccc}{-14} & {6} & {-2} \\ {10} & {6} & {-14} \\ {-2} & {-12} & {10}\end{array}\right)=a d j(A)$

$A^{-1}=\frac{1}{-36}\left(\begin{array}{ccc}{-14} & {6} & {-2} \\ {10} & {6} & {-14} \\ {-2} & {-12} & {10}\end{array}\right)$

$A^{-1}=\left(\begin{array}{ccc}{14 / 36} & {-6 / 36} & {2 / 36} \\ {-10 / 36} & {-6 / 36} & {14 / 36} \\ {2 / 36} & {-12 / 36} & {-10 / 36}\end{array}\right)$