Linear Algebra Examples

$[\begin{array}{c} - 2 & 0 \\ - 5 & - 4 \end{array}] - [\begin{array}{c} - 4 & 0 \\ 2 & 2 \end{array}]$

Step 1

Subtract the corresponding elements.

$[\begin{array}{c} - 2 + 4 & 0 - 0 \\ - 5 - 2 & - 4 - 2 \end{array}]$

Step 2

Simplify each element.

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Add $- 2$ and $4$ .

$[\begin{array}{c} 2 & 0 - 0 \\ - 5 - 2 & - 4 - 2 \end{array}]$

Subtract $0$ from $0$ .

$[\begin{array}{c} 2 & 0 \\ - 5 - 2 & - 4 - 2 \end{array}]$

Subtract $2$ from $- 5$ .

$[\begin{array}{c} 2 & 0 \\ - 7 & - 4 - 2 \end{array}]$

Subtract $2$ from $- 4$ .

$[\begin{array}{c} 2 & 0 \\ - 7 & - 6 \end{array}]$

Step 3

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 $A$ .

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

Step 4

Find the determinant of $[\begin{array}{c} 2 & 0 \\ - 7 & - 6 \end{array}]$ .

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

$determinant [\begin{array}{c} 2 & 0 \\ - 7 & - 6 \end{array}] = | \begin{array}{c} 2 & 0 \\ - 7 & - 6 \end{array} |$

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

$(2) (- 6) + 7 \cdot 0$

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $- 6$ .

$- 12 + 7 \cdot 0$

Multiply $7$ by $0$ .

$- 12 + 0$

Add $- 12$ and $0$ .

$- 12$

Step 5

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

$\frac{1}{- 12} [\begin{array}{c} - 6 & - (0) \\ - (- 7) & 2 \end{array}]$

Step 6

Simplify each element in the matrix.

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Rearrange $- (0)$ .

$\frac{1}{- 12} [\begin{array}{c} - 6 & 0 \\ - (- 7) & 2 \end{array}]$

Rearrange $- (- 7)$ .

$\frac{1}{- 12} [\begin{array}{c} - 6 & 0 \\ 7 & 2 \end{array}]$

Step 7

Multiply $\frac{1}{- 12}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{- 12} \cdot - 6 & \frac{1}{- 12} \cdot 0 \\ \frac{1}{- 12} \cdot 7 & \frac{1}{- 12} \cdot 2 \end{array}]$

Step 8

Simplify each element in the matrix.

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Rearrange $\frac{1}{- 12} \cdot - 6$ .

$[\begin{array}{c} \frac{1}{2} & \frac{1}{- 12} \cdot 0 \\ \frac{1}{- 12} \cdot 7 & \frac{1}{- 12} \cdot 2 \end{array}]$

Rearrange $\frac{1}{- 12} \cdot 0$ .

$[\begin{array}{c} \frac{1}{2} & 0 \\ \frac{1}{- 12} \cdot 7 & \frac{1}{- 12} \cdot 2 \end{array}]$

Rearrange $\frac{1}{- 12} \cdot 7$ .

$[\begin{array}{c} \frac{1}{2} & 0 \\ - \frac{7}{12} & \frac{1}{- 12} \cdot 2 \end{array}]$

Rearrange $\frac{1}{- 12} \cdot 2$ .

$[\begin{array}{c} \frac{1}{2} & 0 \\ - \frac{7}{12} & - \frac{1}{6} \end{array}]$

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