Precalculus Examples

$x + 3 y = 6$ , $- x - 3 y + 6 = 0$

Step 1

Subtract $6$ from both sides of the equation.

$x + 3 y = 6$

$- x - 3 y = - 6$

Step 2

Represent the system of equations in matrix format.

$[\begin{array}{c} 1 & 3 \\ - 1 & - 3 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 6 \\ - 6 \end{array}]$

Step 3

Find the determinant of the coefficient matrix $[\begin{array}{c} 1 & 3 \\ - 1 & - 3 \end{array}]$ .

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Step 3.1

Write $[\begin{array}{c} 1 & 3 \\ - 1 & - 3 \end{array}]$ in determinant notation.

$| \begin{array}{c} 1 & 3 \\ - 1 & - 3 \end{array} |$

Step 3.2

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$1 \cdot - 3 - (- 1 \cdot 3)$

Step 3.3

Simplify the determinant.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Multiply $- 3$ by $1$ .

$- 3 - (- 1 \cdot 3)$

Step 3.3.1.2

Multiply $- (- 1 \cdot 3)$ .

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Step 3.3.1.2.1

Multiply $- 1$ by $3$ .

$- 3 - - 3$

Step 3.3.1.2.2

Multiply $- 1$ by $- 3$ .

$- 3 + 3$

Step 3.3.2

Add $- 3$ and $3$ .

$0$

$D = 0$

Step 4

Since the determinant is $0$ , the system cannot be solved using Cramer's Rule.