Find the Inverse [[2,-1],[1,0]]

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

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

Find the determinant of $[\begin{array}{c} 2 & - 1 \\ 1 & 0 \end{array}]$ .

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

$determinant [\begin{array}{c} 2 & - 1 \\ 1 & 0 \end{array}] = | \begin{array}{c} 2 & - 1 \\ 1 & 0 \end{array} |$

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

$(2) (0) - 1 \cdot - 1$

Simplify the determinant.

Simplify each term.

Multiply $2$ by $0$ .

$0 - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$0 + 1$

Add $0$ and $1$ .

$1$

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

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

Simplify each element in the matrix.

Rearrange $- (- 1)$ .

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

Rearrange $- (1)$ .

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

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

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

Simplify each element in the matrix.

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

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

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

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

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

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

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

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

Find the Inverse [[2,-1],[1,0]]

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