Inverse matrices with cofactors

By the invertibility criterion a square matrix is invertible if and only if its determinant is nonzero. Since $\det A=5$, the matrix $A$ is invertible; since $\det B=0$, the matrix $B$ is not invertible, in agreement with the fact that the second row of $B$ is $\frac{3}{2}$ times the first. For the inverse of order $2$ one swaps the entries of the main diagonal, $2$ and $4$, changes the sign of the entries $3$ and $1$, and divides by the determinant. The product $A{A}^{-1}=I_2$ confirms the result.

The third column of $A$ has only one nonzero entry, so the determinant is computed by expanding along that column; the sign of the cofactor in position $(2,3)$ is $-1$, and the result is $\det A=1$. Since $\det A\ne 0$, the matrix is invertible and the formula for the inverse applies. Each cofactor $C_{ij}$ is the determinant of the matrix obtained by deleting row $i$ and column $j$, with the checkerboard sign. The cofactors are arranged in the matrix $\mathrm{cof}A$, which must be transposed, that is, the first row of $\mathrm{cof}A$ becomes the first column of $A^{-1}$. Since $\det A=1$, no division remains. The product of the first row of $A$ and the columns of $A^{-1}$ gives $(1,0,0)$, and the other rows are checked in the same way.

The determinant is computed with Sarrus’ rule: the diagonals to the right give $1\cdot 1\cdot 1+0\cdot 3\cdot 0+2\cdot 2\cdot 1=5$, those to the left give $2\cdot 1\cdot 0+1\cdot 3\cdot 1+0\cdot 2\cdot 1=3$, so $\det A=2$ and the matrix is invertible. The nine cofactors are computed with the checkerboard signs and the transpose of $\mathrm{cof}A$ is written, that is, the matrix that has $C_{1i}$, $C_{2i}$, $C_{3i}$ in row $i$. Dividing by $2$ gives $A^{-1}$. The check on the second row of $A$ and the second column of $A^{-1}$ gives $2\cdot 1+1\cdot \frac{1}{2}+3\cdot \left(-\frac{1}{2}\right)=1$, as required by the entry $(2,2)$ of $I_3$.