Linear Algebra Examples

$5 x - y = - 2$ , $53 x - 8 y = 0$

Represent the system of equations in matrix format.

$[\begin{array}{c} 5 & - 1 \\ 53 & - 8 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 2 \\ 0 \end{array}]$

The determinant of $[\begin{array}{c} 5 & - 1 \\ 53 & - 8 \end{array}]$ is $13$ .

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

$D = d e t e r m i n a n t [\begin{array}{c} 5 & - 1 \\ 53 & - 8 \end{array}] = | \begin{array}{c} 5 & - 1 \\ 53 & - 8 \end{array} |$

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

$D = (5) (- 8) - 53 \cdot - 1$

Simplify the determinant.

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Simplify each term.

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Multiply $5$ by $- 8$ .

$D = - 40 - 53 \cdot - 1$

Multiply $- 53$ by $- 1$ .

$D = - 40 + 53$

Add $- 40$ and $53$ .

$D = 13$

The determinant of $[\begin{array}{c} - 2 & - 1 \\ 0 & - 8 \end{array}]$ is $16$ .

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

$D_{x} = d e t e r m i n a n t [\begin{array}{c} - 2 & - 1 \\ 0 & - 8 \end{array}] = | \begin{array}{c} - 2 & - 1 \\ 0 & - 8 \end{array} |$

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

$D_{x} = (- 2) (- 8) + 0 \cdot - 1$

Simplify the determinant.

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Simplify each term.

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Multiply $- 2$ by $- 8$ .

$D_{x} = 16 + 0 \cdot - 1$

Multiply $0$ by $- 1$ .

$D_{x} = 16 + 0$

Add $16$ and $0$ .

$D_{x} = 16$

The determinant of $[\begin{array}{c} 5 & - 2 \\ 53 & 0 \end{array}]$ is $106$ .

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

$D_{y} = d e t e r m i n a n t [\begin{array}{c} 5 & - 2 \\ 53 & 0 \end{array}] = | \begin{array}{c} 5 & - 2 \\ 53 & 0 \end{array} |$

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

$D_{y} = (5) (0) - 53 \cdot - 2$

Simplify the determinant.

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Simplify each term.

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Multiply $5$ by $0$ .

$D_{y} = 0 - 53 \cdot - 2$

Multiply $- 53$ by $- 2$ .

$D_{y} = 0 + 106$

Add $0$ and $106$ .

$D_{y} = 106$

Remove the extra parentheses from the expression $\frac{16}{13}$ .

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Remove the parentheses from the numerator.

$x = \frac{16}{13}$

Remove the parentheses from the denominator.

$x = \frac{16}{13}$

Remove the extra parentheses from the expression $\frac{106}{13}$ .

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Remove the parentheses from the numerator.

$y = \frac{106}{13}$

Remove the parentheses from the denominator.

$y = \frac{106}{13}$

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

$x = \frac{16}{13}, y = \frac{106}{13}$

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