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Mathematics Question on Determinants

Find the inverse of each of the matrices(if it exists). [1−12\02−3\3−24]\begin{bmatrix}1&-1&2\\\0&2&-3\\\3&-2&4\end{bmatrix}

Answer

Let A=[1−12\02−3\3−24]\begin{bmatrix}1&-1&2\\\0&2&-3\\\3&-2&4\end{bmatrix}
By expanding along C1,we have :
IAI=1(8-6)-0+3(3-4)=2-3=-1
Now A11=8-6=2, A12=-(0+9)=-9, A13=0-6=-6
A21=-(-4+4)=0, A22=4-6=-2, A23=-(-2+3)=-1
A31=3-4=-1, A32=-(-3-0)=3, A33=2-0=2

so adj A=[20−1−9−23−6−12]\begin{bmatrix}2&0&-1\\\\-9&-2&3\\\\-6&-1&2\end{bmatrix}

so A-1=1∣A∣\frac{1}{\mid A \mid}adj A=- [20−1−9−23−6−12]\begin{bmatrix}2&0&-1\\\\-9&-2&3\\\\-6&-1&2\end{bmatrix}

=[−201\92−3\61−2]\begin{bmatrix}-2&0&1\\\9&2&-3\\\6&1&-2\end{bmatrix}