Linear Algebra Examples

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

Step 1

To evaluate a square matrix to a positive integer power $n$ , multiply $n$ copies of the matrix.

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

Step 2

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

Tap for more steps...

Step 2.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is $2 \times 2$ and the second matrix is $2 \times 2$ .

Step 2.2

Multiply each row in the first matrix by each column in the second matrix.

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

Step 2.3

Simplify each element of the matrix by multiplying out all the expressions.

$[\begin{array}{c} - 5 & 6 \\ - 9 & 10 \end{array}] \cdot [\begin{array}{c} 1 & - 2 \\ 3 & - 4 \end{array}]$

Step 3

Multiply $[\begin{array}{c} - 5 & 6 \\ - 9 & 10 \end{array}] [\begin{array}{c} 1 & - 2 \\ 3 & - 4 \end{array}]$ .

Tap for more steps...

Step 3.1

Step 3.2

Multiply each row in the first matrix by each column in the second matrix.

$[\begin{array}{c} - 5 \cdot 1 + 6 \cdot 3 & - 5 \cdot - 2 + 6 \cdot - 4 \\ - 9 \cdot 1 + 10 \cdot 3 & - 9 \cdot - 2 + 10 \cdot - 4 \end{array}]$

Step 3.3

Simplify each element of the matrix by multiplying out all the expressions.

$[\begin{array}{c} 13 & - 14 \\ 21 & - 22 \end{array}]$

Step 4

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 5

Find the determinant.

Tap for more steps...

Step 5.1

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

$13 \cdot - 22 - 21 \cdot - 14$

Step 5.2

Simplify the determinant.

Tap for more steps...

Step 5.2.1

Simplify each term.

Tap for more steps...

Step 5.2.1.1

Multiply $13$ by $- 22$ .

$- 286 - 21 \cdot - 14$

Step 5.2.1.2

Multiply $- 21$ by $- 14$ .

$- 286 + 294$

Step 5.2.2

Add $- 286$ and $294$ .

$8$

Step 6

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

Step 7

Substitute the known values into the formula for the inverse.

$\frac{1}{8} [\begin{array}{c} - 22 & 14 \\ - 21 & 13 \end{array}]$

Step 8

Multiply $\frac{1}{8}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{8} \cdot - 22 & \frac{1}{8} \cdot 14 \\ \frac{1}{8} \cdot - 21 & \frac{1}{8} \cdot 13 \end{array}]$

Step 9

Simplify each element in the matrix.

Tap for more steps...

Step 9.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 9.1.1

Factor $2$ out of $8$ .

$[\begin{array}{c} \frac{1}{2 (4)} \cdot - 22 & \frac{1}{8} \cdot 14 \\ \frac{1}{8} \cdot - 21 & \frac{1}{8} \cdot 13 \end{array}]$

Step 9.1.2

Factor $2$ out of $- 22$ .

$[\begin{array}{c} \frac{1}{2 \cdot 4} \cdot (2 \cdot - 11) & \frac{1}{8} \cdot 14 \\ \frac{1}{8} \cdot - 21 & \frac{1}{8} \cdot 13 \end{array}]$

Step 9.1.3

Cancel the common factor.

$[\begin{array}{c} \frac{1}{2 \cdot 4} \cdot (2 \cdot - 11) & \frac{1}{8} \cdot 14 \\ \frac{1}{8} \cdot - 21 & \frac{1}{8} \cdot 13 \end{array}]$

Step 9.1.4

Rewrite the expression.

$[\begin{array}{c} \frac{1}{4} \cdot - 11 & \frac{1}{8} \cdot 14 \\ \frac{1}{8} \cdot - 21 & \frac{1}{8} \cdot 13 \end{array}]$

Step 9.2

Combine $\frac{1}{4}$ and $- 11$ .

$[\begin{array}{c} \frac{- 11}{4} & \frac{1}{8} \cdot 14 \\ \frac{1}{8} \cdot - 21 & \frac{1}{8} \cdot 13 \end{array}]$

Step 9.3

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{11}{4} & \frac{1}{8} \cdot 14 \\ \frac{1}{8} \cdot - 21 & \frac{1}{8} \cdot 13 \end{array}]$

Step 9.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 9.4.1

Factor $2$ out of $8$ .

$[\begin{array}{c} - \frac{11}{4} & \frac{1}{2 (4)} \cdot 14 \\ \frac{1}{8} \cdot - 21 & \frac{1}{8} \cdot 13 \end{array}]$

Step 9.4.2

Factor $2$ out of $14$ .

$[\begin{array}{c} - \frac{11}{4} & \frac{1}{2 \cdot 4} \cdot (2 \cdot 7) \\ \frac{1}{8} \cdot - 21 & \frac{1}{8} \cdot 13 \end{array}]$

Step 9.4.3

Cancel the common factor.

$[\begin{array}{c} - \frac{11}{4} & \frac{1}{2 \cdot 4} \cdot (2 \cdot 7) \\ \frac{1}{8} \cdot - 21 & \frac{1}{8} \cdot 13 \end{array}]$

Step 9.4.4

Rewrite the expression.

$[\begin{array}{c} - \frac{11}{4} & \frac{1}{4} \cdot 7 \\ \frac{1}{8} \cdot - 21 & \frac{1}{8} \cdot 13 \end{array}]$

Step 9.5

Combine $\frac{1}{4}$ and $7$ .

$[\begin{array}{c} - \frac{11}{4} & \frac{7}{4} \\ \frac{1}{8} \cdot - 21 & \frac{1}{8} \cdot 13 \end{array}]$

Step 9.6

Combine $\frac{1}{8}$ and $- 21$ .

$[\begin{array}{c} - \frac{11}{4} & \frac{7}{4} \\ \frac{- 21}{8} & \frac{1}{8} \cdot 13 \end{array}]$

Step 9.7

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{11}{4} & \frac{7}{4} \\ - \frac{21}{8} & \frac{1}{8} \cdot 13 \end{array}]$

Step 9.8

Combine $\frac{1}{8}$ and $13$ .

$[\begin{array}{c} - \frac{11}{4} & \frac{7}{4} \\ - \frac{21}{8} & \frac{13}{8} \end{array}]$