Linear Algebra Examples

$20 = 8 y + 11 x$ ,

$11 x + 8 y = 79$

Step 1

Find the

$A X = B$ from the system of equations.

$[\begin{array}{c} - 11 & - 8 \\ 11 & 8 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 20 \\ 79 \end{array}]$

Step 2

Find the inverse of the coefficient matrix.

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The inverse of a

$2 \times 2$ matrix can be found using the formula

$\frac{1}{| A |} [\begin{array}{c} d & - b \\ - c & a \end{array}]$ where

$| A |$ is the determinant of

$A$ .

$A = [\begin{array}{c} a & b \\ c & d \end{array}]$ then

$A^{- 1} = \frac{1}{| A |} [\begin{array}{c} d & - b \\ - c & a \end{array}]$

Find the determinant of

$[\begin{array}{c} - 11 & - 8 \\ 11 & 8 \end{array}]$ .

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

$determinant [\begin{array}{c} - 11 & - 8 \\ 11 & 8 \end{array}] = | \begin{array}{c} - 11 & - 8 \\ 11 & 8 \end{array} |$

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$(- 11) (8) - 11 \cdot - 8$

Simplify the determinant.

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Simplify each term.

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Multiply

$- 11$ by

$8$ .

$- 88 - 11 \cdot - 8$

Multiply

$- 11$ by

$- 8$ .

$- 88 + 88$

Add

$- 88$ and

$88$ .

$0$

Substitute the known values into the formula for the inverse of a matrix.

$\frac{1}{0} [\begin{array}{c} 8 & - (- 8) \\ - (11) & - 11 \end{array}]$

Simplify each element in the matrix.

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Rearrange

$- (- 8)$ .

$\frac{1}{0} [\begin{array}{c} 8 & 8 \\ - (11) & - 11 \end{array}]$

Rearrange

$- (11)$ .

$\frac{1}{0} [\begin{array}{c} 8 & 8 \\ - 11 & - 11 \end{array}]$

Multiply

$\frac{1}{0}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{0} \cdot 8 & \frac{1}{0} \cdot 8 \\ \frac{1}{0} \cdot - 11 & \frac{1}{0} \cdot - 11 \end{array}]$

Rearrange

$\frac{1}{0} \cdot 8$ .

$[\begin{array}{c} Undefined & \frac{1}{0} \cdot 8 \\ \frac{1}{0} \cdot - 11 & \frac{1}{0} \cdot - 11 \end{array}]$

Since the matrix is undefined, it cannot be solved.

$Undefined$

Undefined