Linear Algebra Examples

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

Step 1

Subtract the corresponding elements.

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

Step 2

Simplify each element.

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Step 2.1

Subtract $1$ from $2$ .

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

Step 2.2

Subtract $2$ from $4$ .

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

Step 2.3

Subtract $3$ from $6$ .

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

Step 2.4

Subtract $4$ from $8$ .

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

Step 3

The inverse of a $2 \times 2$ matrix can be found using the formula $\frac{1}{a d - b c} [\begin{array}{c} d & - b \\ - c & a \end{array}]$ where $a d - b c$ is the determinant.

Step 4

Find the determinant.

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Step 4.1

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

$1 \cdot 4 - 3 \cdot 2$

Step 4.2

Simplify the determinant.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply $4$ by $1$ .

$4 - 3 \cdot 2$

Step 4.2.1.2

Multiply $- 3$ by $2$ .

$4 - 6$

Step 4.2.2

Subtract $6$ from $4$ .

$- 2$

Step 5

Since the determinant is non-zero, the inverse exists.

Step 6

Substitute the known values into the formula for the inverse.

$\frac{1}{- 2} [\begin{array}{c} 4 & - 2 \\ - 3 & 1 \end{array}]$

Step 7

Move the negative in front of the fraction.

$- \frac{1}{2} [\begin{array}{c} 4 & - 2 \\ - 3 & 1 \end{array}]$

Step 8

Multiply $- \frac{1}{2}$ by each element of the matrix.

$[\begin{array}{c} - \frac{1}{2} \cdot 4 & - \frac{1}{2} \cdot - 2 \\ - \frac{1}{2} \cdot - 3 & - \frac{1}{2} \cdot 1 \end{array}]$

Step 9

Simplify each element in the matrix.

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Step 9.1

Cancel the common factor of $2$ .

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Step 9.1.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$[\begin{array}{c} \frac{- 1}{2} \cdot 4 & - \frac{1}{2} \cdot - 2 \\ - \frac{1}{2} \cdot - 3 & - \frac{1}{2} \cdot 1 \end{array}]$

Step 9.1.2

Factor $2$ out of $4$ .

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

Step 9.1.3

Cancel the common factor.

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

Step 9.1.4

Rewrite the expression.

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

Step 9.2

Multiply $- 1$ by $2$ .

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

Step 9.3

Cancel the common factor of $2$ .

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Step 9.3.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Step 9.3.2

Factor $2$ out of $- 2$ .

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

Step 9.3.3

Cancel the common factor.

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

Step 9.3.4

Rewrite the expression.

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

Step 9.4

Multiply $- 1$ by $- 1$ .

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

Step 9.5

Multiply $- \frac{1}{2} \cdot - 3$ .

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Step 9.5.1

Multiply $- 3$ by $- 1$ .

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

Step 9.5.2

Combine $3$ and $\frac{1}{2}$ .

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

Step 9.6

Multiply $- 1$ by $1$ .

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

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