Linear Algebra Examples

${[\begin{array}{c} - 7 & 12 \\ - 4 & 7 \end{array}]}^{2}$

Step 1

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

$[\begin{array}{c} - 7 & 12 \\ - 4 & 7 \end{array}] \cdot [\begin{array}{c} - 7 & 12 \\ - 4 & 7 \end{array}]$

Step 2

Multiply $[\begin{array}{c} - 7 & 12 \\ - 4 & 7 \end{array}] [\begin{array}{c} - 7 & 12 \\ - 4 & 7 \end{array}]$ .

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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.

$[\begin{array}{c} - 7 \cdot - 7 + 12 \cdot - 4 & - 7 \cdot 12 + 12 \cdot 7 \\ - 4 \cdot - 7 + 7 \cdot - 4 & - 4 \cdot 12 + 7 \cdot 7 \end{array}]$

Step 2.3

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

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

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 $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$1 \cdot 1 + 0 \cdot 0$

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 $1$ by $1$ .

$1 + 0 \cdot 0$

Step 4.2.1.2

Multiply $0$ by $0$ .

$1 + 0$

Step 4.2.2

Add $1$ and $0$ .

$1$

Step 5

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

Step 6

Substitute the known values into the formula for the inverse.

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

Step 7

Divide $1$ by $1$ .

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

Step 8

Multiply $1$ by each element of the matrix.

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

Step 9

Simplify each element in the matrix.

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Step 9.1

Multiply $1$ by $1$ .

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

Step 9.2

Multiply $0$ by $1$ .

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

Step 9.3

Multiply $0$ by $1$ .

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

Step 9.4

Multiply $1$ by $1$ .

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