Q.Compute $|A|$ if $A=\begin{bmatrix}3 & 4 & 1\\ 0 & -1 & 2\\ 5 & -2 & 6\end{bmatrix}$

To every square matrix $A=[a_{ij}]$ of order $n$ we associate a single number, the determinant, written $\det A$ or $|A|$. Determinants are defined only for square matrices; the matrix itself is a representation, the determinant is a value derived from it.

• Order 1: $A=[a]$, so $|A|=a$.
• Order 2: $A=\begin{pmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}$, so $|A| = a_{11}a_{22}-a_{12}a_{21}$ (product of the main diagonal minus product of the other diagonal).
• Order 3: for $A=\begin{pmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{pmatrix}$, we first define, for each entry $a_{ij}$, its minor $M_{ij}$ — the order-2 determinant left after deleting row $i$ and column $j$ — and its cofactor $A_{ij}=(-1)^{i+j}M_{ij}$, a signed minor.

Laplace expansion (Result 7.1/7.2). The determinant equals the sum of the products of the entries of any one row (or column) with their corresponding cofactors — e.g. expanding along row 1: $|A| = a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}$. This value is the same no matter which row or column is chosen for the expansion. For easiest hand computation, expand along the row/column with the most zeros. …