Linear Algebra Examples

$5 x - y = - 2$ , $3 x - 4 y = 0$

Represent the system of equations in matrix format.

$[\begin{array}{c} 5 & - 1 \\ 3 & - 4 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 2 \\ 0 \end{array}]$

The determinant of $[\begin{array}{c} 5 & - 1 \\ 3 & - 4 \end{array}]$ is $- 17$ .

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These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} 5 & - 1 \\ 3 & - 4 \end{array}] = | \begin{array}{c} 5 & - 1 \\ 3 & - 4 \end{array} |$

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

$D = (5) (- 4) - 3 \cdot - 1$

Simplify the determinant.

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Simplify each term.

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Multiply $5$ by $- 4$ .

$D = - 20 - 3 \cdot - 1$

Multiply $- 3$ by $- 1$ .

$D = - 20 + 3$

Add $- 20$ and $3$ .

$D = - 17$

The determinant of $[\begin{array}{c} - 2 & - 1 \\ 0 & - 4 \end{array}]$ is $8$ .

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These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} - 2 & - 1 \\ 0 & - 4 \end{array}] = | \begin{array}{c} - 2 & - 1 \\ 0 & - 4 \end{array} |$

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

$D_{x} = (- 2) (- 4) + 0 \cdot - 1$

Simplify the determinant.

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Simplify each term.

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Multiply $- 2$ by $- 4$ .

$D_{x} = 8 + 0 \cdot - 1$

Multiply $0$ by $- 1$ .

$D_{x} = 8 + 0$

Add $8$ and $0$ .

$D_{x} = 8$

The determinant of $[\begin{array}{c} 5 & - 2 \\ 3 & 0 \end{array}]$ is $6$ .

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These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} 5 & - 2 \\ 3 & 0 \end{array}] = | \begin{array}{c} 5 & - 2 \\ 3 & 0 \end{array} |$

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

$D_{y} = (5) (0) - 3 \cdot - 2$

Simplify the determinant.

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Simplify each term.

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Multiply $5$ by $0$ .

$D_{y} = 0 - 3 \cdot - 2$

Multiply $- 3$ by $- 2$ .

$D_{y} = 0 + 6$

Add $0$ and $6$ .

$D_{y} = 6$

Find the value of $x$ by Cramer's Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{8}{- 17}$ .

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Remove parentheses.

$x = \frac{8}{- 17}$

Move the negative in front of the fraction.

$x = - \frac{8}{17}$

Find the value of $y$ by Cramer's Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{6}{- 17}$ .

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Remove parentheses.

$y = \frac{6}{- 17}$

Move the negative in front of the fraction.

$y = - \frac{6}{17}$

The solution to the system of equations using Cramer's Rule.

$x = - \frac{8}{17}, y = - \frac{6}{17}$

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