Q.If $A=\begin{bmatrix}1 & 2\\ 3 & 4\end{bmatrix}$, prove that $A\cdot(\text{adj }A)=(\text{adj }A)\cdot A=|A|\cdot I$

For a square matrix $A = [a_{ij}]$, first attach to every entry $a_{ij}$ its cofactor $A_{ij} = (-1)^{i+j}M_{ij}$, where $M_{ij}$ (the minor) is the determinant left after deleting row $i$ and column $j$ from $A$. Collecting all the cofactors in their original positions gives the cofactor matrix $[A_{ij}]$. The adjoint of $A$, written $\text{adj}\,A$, is defined as the transpose of this cofactor matrix: $\text{adj}\,A = [A_{ij}]^T$, so the $(i,j)$ entry of $\text{adj}\,A$ is the cofactor $A_{ji}$ (cofactor of the element in the opposite position). This extra transpose step is easy to forget but is essential — omitting it produces the cofactor matrix, not the adjoint. The adjoint's defining property comes from two facts about determinant expansion: expanding a determinant along a row using its own cofactors reproduces $|A|$ (e.g. $a_{i1}A_{i1}+a_{i2}A_{i2}+a_{i3}A_{i3}=|A|$), while expanding along a row using a different row's cofactors always g …