Linear Algebra Examples

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

Combine the similar matrices with each others.

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

Simplify each element of the matrix $[\begin{array}{c} - 1 + 1 & - 1 + 0 \\ 2 + 0 & - 2 + 1 \end{array}]$ .

Tap for more steps...

Combine the same size matrices $[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$ and $[\begin{array}{c} - 1 & - 1 \\ 2 & - 2 \end{array}]$ by adding the corresponding elements of each.

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

Simplify element $0, 0$ of the matrix.

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

Simplify element $0, 1$ of the matrix.

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

Simplify element $1, 0$ of the matrix.

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

Simplify element $1, 1$ of the matrix.

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

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

Tap for more steps...

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

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

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

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

Simplify the determinant.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $0$ by $- 1$ to get $0$ .

$0 - 2 \cdot - 1$

Multiply $- 2$ by $- 1$ to get $2$ .

$0 + 2$

Add $0$ and $2$ to get $2$ .

$2$

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

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

Simplify each element of the matrix $[\begin{array}{c} - 1 & - (- 1) \\ - (2) & 0 \end{array}]$ .

Tap for more steps...

Simplify element $0, 1$ by multiplying $- (- 1)$ to get $1$ .

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

Simplify element $1, 0$ by multiplying $- (2)$ to get $- 2$ .

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

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

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

Simplify each element of the matrix $[\begin{array}{c} \frac{1}{2} \cdot - 1 & \frac{1}{2} \cdot 1 \\ \frac{1}{2} \cdot - 2 & \frac{1}{2} \cdot 0 \end{array}]$ .

Tap for more steps...

Simplify element $0, 0$ by multiplying $\frac{1}{2} \cdot - 1$ to get $- \frac{1}{2}$ .

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

Simplify element $0, 1$ by multiplying $\frac{1}{2} \cdot 1$ to get $\frac{1}{2}$ .

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

Simplify element $1, 0$ by multiplying $\frac{1}{2} \cdot - 2$ to get $- 1$ .

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

Simplify element $1, 1$ by multiplying $\frac{1}{2} \cdot 0$ to get $0$ .

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

Enter YOUR Problem