Solve the Matrix Equation [[-1,2],[-6,10]]z=[[6],[22]]

$[\begin{array}{c} - 1 & 2 \\ - 6 & 10 \end{array}] z = [\begin{array}{c} 6 \\ 22 \end{array}]$

Step 1

Find the inverse matrix.

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 $[\begin{array}{c} - 1 & 2 \\ - 6 & 10 \end{array}]$ .

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

$determinant [\begin{array}{c} - 1 & 2 \\ - 6 & 10 \end{array}] = | \begin{array}{c} - 1 & 2 \\ - 6 & 10 \end{array} |$

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

$(- 1) (10) + 6 \cdot 2$

Simplify the determinant.

Simplify each term.

Multiply $- 1$ by $10$ .

$- 10 + 6 \cdot 2$

Multiply $6$ by $2$ .

$- 10 + 12$

Add $- 10$ and $12$ .

$2$

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

$\frac{1}{2} [\begin{array}{c} 10 & - (2) \\ - (- 6) & - 1 \end{array}]$

Simplify each element in the matrix.

Rearrange $- (2)$ .

$\frac{1}{2} [\begin{array}{c} 10 & - 2 \\ - (- 6) & - 1 \end{array}]$

Rearrange $- (- 6)$ .

$\frac{1}{2} [\begin{array}{c} 10 & - 2 \\ 6 & - 1 \end{array}]$

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

$[\begin{array}{c} \frac{1}{2} \cdot 10 & \frac{1}{2} \cdot - 2 \\ \frac{1}{2} \cdot 6 & \frac{1}{2} \cdot - 1 \end{array}]$

Simplify each element in the matrix.

Rearrange $\frac{1}{2} \cdot 10$ .

$[\begin{array}{c} 5 & \frac{1}{2} \cdot - 2 \\ \frac{1}{2} \cdot 6 & \frac{1}{2} \cdot - 1 \end{array}]$

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

$[\begin{array}{c} 5 & - 1 \\ \frac{1}{2} \cdot 6 & \frac{1}{2} \cdot - 1 \end{array}]$

Rearrange $\frac{1}{2} \cdot 6$ .

$[\begin{array}{c} 5 & - 1 \\ 3 & \frac{1}{2} \cdot - 1 \end{array}]$

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

$[\begin{array}{c} 5 & - 1 \\ 3 & - \frac{1}{2} \end{array}]$

Step 2

Assuming that $X$ is the matrix to solve for, multiply the inverse matrix by both sides of the equation.

$[\begin{array}{c} 5 & - 1 \\ 3 & - \frac{1}{2} \end{array}] \cdot ([\begin{array}{c} - 1 & 2 \\ - 6 & 10 \end{array}] x) = [\begin{array}{c} 5 & - 1 \\ 3 & - \frac{1}{2} \end{array}] \cdot [\begin{array}{c} 6 \\ 22 \end{array}]$

Step 3

Simplify the left side of the equation.

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

$[\begin{array}{c} 5 \cdot - 1 - 1 \cdot - 6 & 5 \cdot 2 - 1 \cdot 10 \\ 3 \cdot - 1 - \frac{1}{2} \cdot - 6 & 3 \cdot 2 - \frac{1}{2} \cdot 10 \end{array}] x = [\begin{array}{c} 5 & - 1 \\ 3 & - \frac{1}{2} \end{array}] \cdot [\begin{array}{c} 6 \\ 22 \end{array}]$

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

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] x = [\begin{array}{c} 5 & - 1 \\ 3 & - \frac{1}{2} \end{array}] \cdot [\begin{array}{c} 6 \\ 22 \end{array}]$

Multiplying the identity matrix by any matrix $x$ is matrix $x$ .

$x = [\begin{array}{c} 5 & - 1 \\ 3 & - \frac{1}{2} \end{array}] \cdot [\begin{array}{c} 6 \\ 22 \end{array}]$

Step 4

Simplify the right side of the equation.

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

$x = [\begin{array}{c} 5 \cdot 6 - 1 \cdot 22 \\ 3 \cdot 6 - \frac{1}{2} \cdot 22 \end{array}]$

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

$x = [\begin{array}{c} 8 \\ 7 \end{array}]$

The matrix is in the most simplified form.

$x = [\begin{array}{c} 8 \\ 7 \end{array}]$

Solve the Matrix Equation [[-1,2],[-6,10]]z=[[6],[22]]

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