Linear Algebra Examples

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

Step 1

Add the corresponding elements.

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

Step 2

Simplify each element of the matrix $[\begin{array}{c} - 2 - 4 & 0 + 0 \\ - 5 + 2 & - 4 + 2 \end{array}]$ .

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Simplify $- 2 - 4$ .

$[\begin{array}{c} - 6 & 0 + 0 \\ - 5 + 2 & - 4 + 2 \end{array}]$

Simplify $0 + 0$ .

$[\begin{array}{c} - 6 & 0 \\ - 5 + 2 & - 4 + 2 \end{array}]$

Simplify $- 5 + 2$ .

$[\begin{array}{c} - 6 & 0 \\ - 3 & - 4 + 2 \end{array}]$

Simplify $- 4 + 2$ .

$[\begin{array}{c} - 6 & 0 \\ - 3 & - 2 \end{array}]$

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 $[\begin{array}{c} - 6 & 0 \\ - 3 & - 2 \end{array}]$ .

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

$determinant [\begin{array}{c} - 6 & 0 \\ - 3 & - 2 \end{array}] = | \begin{array}{c} - 6 & 0 \\ - 3 & - 2 \end{array} |$

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

$(- 6) (- 2) + 3 \cdot 0$

Simplify the determinant.

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Simplify each term.

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Multiply $- 6$ by $- 2$ .

$12 + 3 \cdot 0$

Multiply $3$ by $0$ .

$12 + 0$

Add $12$ and $0$ .

$12$

Step 5

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

$\frac{1}{12} [\begin{array}{c} - 2 & - (0) \\ - (- 3) & - 6 \end{array}]$

Step 6

Simplify each element in the matrix.

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Rearrange $- (0)$ .

$\frac{1}{12} [\begin{array}{c} - 2 & 0 \\ - (- 3) & - 6 \end{array}]$

Rearrange $- (- 3)$ .

$\frac{1}{12} [\begin{array}{c} - 2 & 0 \\ 3 & - 6 \end{array}]$

Step 7

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

$[\begin{array}{c} \frac{1}{12} \cdot - 2 & \frac{1}{12} \cdot 0 \\ \frac{1}{12} \cdot 3 & \frac{1}{12} \cdot - 6 \end{array}]$

Step 8

Simplify each element in the matrix.

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

$[\begin{array}{c} - \frac{1}{6} & \frac{1}{12} \cdot 0 \\ \frac{1}{12} \cdot 3 & \frac{1}{12} \cdot - 6 \end{array}]$

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

$[\begin{array}{c} - \frac{1}{6} & 0 \\ \frac{1}{12} \cdot 3 & \frac{1}{12} \cdot - 6 \end{array}]$

Rearrange $\frac{1}{12} \cdot 3$ .

$[\begin{array}{c} - \frac{1}{6} & 0 \\ \frac{1}{4} & \frac{1}{12} \cdot - 6 \end{array}]$

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

$[\begin{array}{c} - \frac{1}{6} & 0 \\ \frac{1}{4} & - \frac{1}{2} \end{array}]$

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