Linear Algebra Examples

$[\begin{array}{c} 1 & - 1 \\ - 1 & 0 \end{array}]$

Step 1

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 2

Find the determinant.

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Step 2.1

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

$1 \cdot 0 - - - 1$

Step 2.2

Simplify the determinant.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Multiply $0$ by $1$ .

$0 - - - 1$

Step 2.2.1.2

Multiply $- - - 1$ .

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Step 2.2.1.2.1

Multiply $- 1$ by $- 1$ .

$0 - 1 \cdot 1$

Step 2.2.1.2.2

Multiply $- 1$ by $1$ .

$0 - 1$

Step 2.2.2

Subtract $1$ from $0$ .

$- 1$

Step 3

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

Step 4

Substitute the known values into the formula for the inverse.

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

Step 5

Divide $1$ by $- 1$ .

$- [\begin{array}{c} 0 & 1 \\ 1 & 1 \end{array}]$

Step 6

Multiply $- 1$ by each element of the matrix.

$[\begin{array}{c} - 0 & - 1 \cdot 1 \\ - 1 \cdot 1 & - 1 \cdot 1 \end{array}]$

Step 7

Simplify each element in the matrix.

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Step 7.1

Multiply $- 1$ by $0$ .

$[\begin{array}{c} 0 & - 1 \cdot 1 \\ - 1 \cdot 1 & - 1 \cdot 1 \end{array}]$

Step 7.2

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 0 & - 1 \\ - 1 \cdot 1 & - 1 \cdot 1 \end{array}]$

Step 7.3

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 0 & - 1 \\ - 1 & - 1 \cdot 1 \end{array}]$

Step 7.4

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 0 & - 1 \\ - 1 & - 1 \end{array}]$