Linear Algebra Examples

$4 x - y = - 4$ , $6 x - y = 0$

Represent the system of equations in matrix format.

$[\begin{array}{c} 4 & - 1 \\ 6 & - 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 4 \\ 0 \end{array}]$

The determinant of $[\begin{array}{c} 4 & - 1 \\ 6 & - 1 \end{array}]$ is $2$ .

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

$D = d e t e r m i n a n t [\begin{array}{c} 4 & - 1 \\ 6 & - 1 \end{array}] = | \begin{array}{c} 4 & - 1 \\ 6 & - 1 \end{array} |$

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

$D = (4) (- 1) - 6 \cdot - 1$

Simplify the determinant.

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Simplify each term.

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Multiply $4$ by $- 1$ to get $- 4$ .

$D = - 4 - 6 \cdot - 1$

Multiply $- 6$ by $- 1$ to get $6$ .

$D = - 4 + 6$

Add $- 4$ and $6$ to get $2$ .

$D = 2$

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

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

$D_{x} = d e t e r m i n a n t [\begin{array}{c} - 4 & - 1 \\ 0 & - 1 \end{array}] = | \begin{array}{c} - 4 & - 1 \\ 0 & - 1 \end{array} |$

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $- 4$ by $- 1$ to get $4$ .

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

Multiply $0$ by $- 1$ to get $0$ .

$D_{x} = 4 + 0$

Add $4$ and $0$ to get $4$ .

$D_{x} = 4$

The determinant of $[\begin{array}{c} 4 & - 4 \\ 6 & 0 \end{array}]$ is $24$ .

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

$D_{y} = d e t e r m i n a n t [\begin{array}{c} 4 & - 4 \\ 6 & 0 \end{array}] = | \begin{array}{c} 4 & - 4 \\ 6 & 0 \end{array} |$

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

$D_{y} = (4) (0) - 6 \cdot - 4$

Simplify the determinant.

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Simplify each term.

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Multiply $4$ by $0$ to get $0$ .

$D_{y} = 0 - 6 \cdot - 4$

Multiply $- 6$ by $- 4$ to get $24$ .

$D_{y} = 0 + 24$

Add $0$ and $24$ to get $24$ .

$D_{y} = 24$

Find the value of $x$ by Cramer's Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{4}{2}$ .

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Remove the extra parentheses from the expression $\frac{4}{2}$ .

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Remove the parentheses from the numerator.

$x = \frac{4}{2}$

Remove the parentheses from the denominator.

$x = \frac{4}{2}$

Divide $4$ by $2$ to get $2$ .

$x = 2$

Find the value of $y$ by Cramer's Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{24}{2}$ .

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Remove the extra parentheses from the expression $\frac{24}{2}$ .

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Remove the parentheses from the numerator.

$y = \frac{24}{2}$

Remove the parentheses from the denominator.

$y = \frac{24}{2}$

Divide $24$ by $2$ to get $12$ .

$y = 12$

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

$x = 2, y = 12$

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