Determinants and Cramer's Rule: Practice

Multiply down the main diagonal, then subtract the anti-diagonal product.

$\begin{vmatrix} 5 & 2 \\ 1 & 3 \end{vmatrix} = 5 \times 3 - 2 \times 1 = 13$

The determinant multiplies down the main diagonal and subtracts the product up the anti-diagonal.

$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$

A zero entry just makes one product zero; the rule is unchanged.

$\begin{vmatrix} 6 & 0 \\ 5 & 2 \end{vmatrix} = 6 \times 2 - 0 \times 5 = 12$

Around a square array, the bars denote the determinant, a signed number, not the absolute value. Here it is even negative.

$\begin{vmatrix} 2 & 3 \\ 5 & 1 \end{vmatrix} = 2 \times 1 - 3 \times 5 = -13$

The coefficient determinant uses only the four coefficients on the left-hand sides.

$D = \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 2 \times 4 - 3 \times 1 = 5$

For $D_x$, replace the $x$-column (the coefficients $2$ and $1$) with the constants $8$ and $9$.

$D_x = \begin{vmatrix} 8 & 3 \\ 9 & 4 \end{vmatrix} = 8 \times 4 - 3 \times 9 = 32 - 27 = 5$

Cramer's rule divides the swapped-column determinant by the coefficient determinant.

$x = \frac{D_x}{D} = \frac{5}{5} = 1$

Use the $x$ part of Cramer's rule, which needs $D_x$ and $D$.

$x = \frac{D_x}{D} = \frac{12}{4} = 3$

For $y$, divide $D_y$ by $D$.

$y = \frac{D_y}{D} = \frac{-8}{4} = -2$

Main-diagonal product minus anti-diagonal product.

$\begin{vmatrix} 7 & 2 \\ 3 & 1 \end{vmatrix} = 7 \times 1 - 2 \times 3 = 7 - 6 = 1$

The two rows are proportional, so the two diagonal products are equal and cancel.

$\begin{vmatrix} 5 & 5 \\ 2 & 2 \end{vmatrix} = 5 \times 2 - 5 \times 2 = 0$

The swapped-column determinant goes on top; the coefficient determinant goes on the bottom.

$x = \frac{D_x}{D}$