Linear Algebra Examples

$[\begin{array}{c} 2 & 4 \\ 6 & 8 \end{array}] + [\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}]$

Add the corresponding elements of $[\begin{array}{c} 2 & 4 \\ 6 & 8 \end{array}]$ to each element of $[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}]$ .

$[\begin{array}{c} 1 + 2 & 2 + 4 \\ 3 + 6 & 4 + 8 \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} 1 + 2 & 2 + 4 \\ 3 + 6 & 4 + 8 \end{array}]$ .

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Simplify $1 + 2$ .

$[\begin{array}{c} 3 & 2 + 4 \\ 3 + 6 & 4 + 8 \end{array}]$

Simplify $2 + 4$ .

$[\begin{array}{c} 3 & 6 \\ 3 + 6 & 4 + 8 \end{array}]$

Simplify $3 + 6$ .

$[\begin{array}{c} 3 & 6 \\ 9 & 4 + 8 \end{array}]$

Simplify $4 + 8$ .

$[\begin{array}{c} 3 & 6 \\ 9 & 12 \end{array}]$

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

The determinant of $[\begin{array}{c} 3 & 6 \\ 9 & 12 \end{array}]$ is $- 18$ .

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

$determinant [\begin{array}{c} 3 & 6 \\ 9 & 12 \end{array}] = | \begin{array}{c} 3 & 6 \\ 9 & 12 \end{array} |$

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

$(3) (12) - 9 \cdot 6$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $12$ .

$36 - 9 \cdot 6$

Multiply $- 9$ by $6$ .

$36 - 54$

Subtract $54$ from $36$ .

$- 18$

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

$\frac{1}{- 18} [\begin{array}{c} 12 & - (6) \\ - (9) & 3 \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} 12 & - (6) \\ - (9) & 3 \end{array}]$ .

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Rearrange $- (6)$ .

$\frac{1}{- 18} [\begin{array}{c} 12 & - 6 \\ - (9) & 3 \end{array}]$

Rearrange $- (9)$ .

$\frac{1}{- 18} [\begin{array}{c} 12 & - 6 \\ - 9 & 3 \end{array}]$

Multiply $\frac{1}{- 18}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{- 18} \cdot 12 & \frac{1}{- 18} \cdot - 6 \\ \frac{1}{- 18} \cdot - 9 & \frac{1}{- 18} \cdot 3 \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} \frac{1}{- 18} \cdot 12 & \frac{1}{- 18} \cdot - 6 \\ \frac{1}{- 18} \cdot - 9 & \frac{1}{- 18} \cdot 3 \end{array}]$ .

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Rearrange $\frac{1}{- 18} \cdot 12$ .

$[\begin{array}{c} - \frac{2}{3} & \frac{1}{- 18} \cdot - 6 \\ \frac{1}{- 18} \cdot - 9 & \frac{1}{- 18} \cdot 3 \end{array}]$

Rearrange $\frac{1}{- 18} \cdot - 6$ .

$[\begin{array}{c} - \frac{2}{3} & \frac{1}{3} \\ \frac{1}{- 18} \cdot - 9 & \frac{1}{- 18} \cdot 3 \end{array}]$

Rearrange $\frac{1}{- 18} \cdot - 9$ .

$[\begin{array}{c} - \frac{2}{3} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{- 18} \cdot 3 \end{array}]$

Rearrange $\frac{1}{- 18} \cdot 3$ .

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

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