Linear Algebra Examples

- 0.9 x - 0.8 y = - 0.5

$- 0.9 x - 0.8 y = - 0.5$ ,

0.08 (y + 0.5) = - 0.09 x

$0.08 (y + 0.5) = - 0.09 x$

Step 1

Find the

A X = B

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

[\begin{array}{c} - 0.9 & - 0.8 \\ 0.09 & 0.08 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 0.5 \\ - 0.04 \end{array}]

$[\begin{array}{c} - 0.9 & - 0.8 \\ 0.09 & 0.08 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 0.5 \\ - 0.04 \end{array}]$

Step 2

Find the inverse of the coefficient matrix.

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The inverse of a

2 \times 2

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

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

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

| A |

$| A |$ is the determinant of

A

$A$ .

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

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

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

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

Find the determinant of

[\begin{array}{c} - 0.9 & - 0.8 \\ 0.09 & 0.08 \end{array}]

$[\begin{array}{c} - 0.9 & - 0.8 \\ 0.09 & 0.08 \end{array}]$ .

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

determinant [\begin{array}{c} - 0.9 & - 0.8 \\ 0.09 & 0.08 \end{array}] = | \begin{array}{c} - 0.9 & - 0.8 \\ 0.09 & 0.08 \end{array} |

$determinant [\begin{array}{c} - 0.9 & - 0.8 \\ 0.09 & 0.08 \end{array}] = | \begin{array}{c} - 0.9 & - 0.8 \\ 0.09 & 0.08 \end{array} |$

The determinant of a

2 \times 2

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

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

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

(- 0.9) (0.08) - 0.09 \cdot - 0.8

$(- 0.9) (0.08) - 0.09 \cdot - 0.8$

Simplify the determinant.

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Simplify each term.

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Multiply

- 0.9

$- 0.9$ by

0.08

$0.08$ .

- 0.072 - 0.09 \cdot - 0.8

$- 0.072 - 0.09 \cdot - 0.8$

Multiply

- 0.09

$- 0.09$ by

- 0.8

$- 0.8$ .

- 0.072 + 0.072

$- 0.072 + 0.072$

- 0.072 + 0.072

$- 0.072 + 0.072$

Add

- 0.072

$- 0.072$ and

0.072

$0.072$ .

0

$0$

0

$0$

0

$0$

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

\frac{1}{0} [\begin{array}{c} 0.08 & - (- 0.8) \\ - (0.09) & - 0.9 \end{array}]

$\frac{1}{0} [\begin{array}{c} 0.08 & - (- 0.8) \\ - (0.09) & - 0.9 \end{array}]$

Simplify each element in the matrix.

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Rearrange

- (- 0.8)

$- (- 0.8)$ .

\frac{1}{0} [\begin{array}{c} 0.08 & 0.8 \\ - (0.09) & - 0.9 \end{array}]

$\frac{1}{0} [\begin{array}{c} 0.08 & 0.8 \\ - (0.09) & - 0.9 \end{array}]$

Rearrange

- (0.09)

$- (0.09)$ .

\frac{1}{0} [\begin{array}{c} 0.08 & 0.8 \\ - 0.09 & - 0.9 \end{array}]

$\frac{1}{0} [\begin{array}{c} 0.08 & 0.8 \\ - 0.09 & - 0.9 \end{array}]$

\frac{1}{0} [\begin{array}{c} 0.08 & 0.8 \\ - 0.09 & - 0.9 \end{array}]

$\frac{1}{0} [\begin{array}{c} 0.08 & 0.8 \\ - 0.09 & - 0.9 \end{array}]$

Multiply

\frac{1}{0}

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

[\begin{array}{c} \frac{1}{0} \cdot 0.08 & \frac{1}{0} \cdot 0.8 \\ \frac{1}{0} \cdot - 0.09 & \frac{1}{0} \cdot - 0.9 \end{array}]

$[\begin{array}{c} \frac{1}{0} \cdot 0.08 & \frac{1}{0} \cdot 0.8 \\ \frac{1}{0} \cdot - 0.09 & \frac{1}{0} \cdot - 0.9 \end{array}]$

Rearrange

\frac{1}{0} \cdot 0.08

$\frac{1}{0} \cdot 0.08$ .

[\begin{array}{c} Undefined & \frac{1}{0} \cdot 0.8 \\ \frac{1}{0} \cdot - 0.09 & \frac{1}{0} \cdot - 0.9 \end{array}]

$[\begin{array}{c} Undefined & \frac{1}{0} \cdot 0.8 \\ \frac{1}{0} \cdot - 0.09 & \frac{1}{0} \cdot - 0.9 \end{array}]$

Since the matrix is undefined, it cannot be solved.

Undefined

$Undefined$

Undefined