Solve the Matrix Equation [[-1,1],[5,-2]]c=[[4],[-26]]

$[\begin{array}{c} - 1 & 1 \\ 5 & - 2 \end{array}] c = [\begin{array}{c} 4 \\ - 26 \end{array}]$

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

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

$determinant [\begin{array}{c} - 1 & 1 \\ 5 & - 2 \end{array}] = | \begin{array}{c} - 1 & 1 \\ 5 & - 2 \end{array} |$

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

$(- 1) (- 2) - 5 \cdot 1$

Simplify the determinant.

Simplify each term.

Multiply $- 1$ by $- 2$ .

$2 - 5 \cdot 1$

Multiply $- 5$ by $1$ .

$2 - 5$

Subtract $5$ from $2$ .

$- 3$

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

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

Simplify each element in the matrix.

Rearrange $- (1)$ .

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

Rearrange $- (5)$ .

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

Multiply $\frac{1}{- 3}$ by each element of the matrix.

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

Simplify each element in the matrix.

Rearrange $\frac{1}{- 3} \cdot - 2$ .

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

Rearrange $\frac{1}{- 3} \cdot - 1$ .

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

Rearrange $\frac{1}{- 3} \cdot - 5$ .

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

Rearrange $\frac{1}{- 3} \cdot - 1$ .

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

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

$[\begin{array}{c} \frac{2}{3} & \frac{1}{3} \\ \frac{5}{3} & \frac{1}{3} \end{array}] \cdot ([\begin{array}{c} - 1 & 1 \\ 5 & - 2 \end{array}] x) = [\begin{array}{c} \frac{2}{3} & \frac{1}{3} \\ \frac{5}{3} & \frac{1}{3} \end{array}] \cdot [\begin{array}{c} 4 \\ - 26 \end{array}]$

Simplify the left side of the equation.

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

$[\begin{array}{c} \frac{2}{3} \cdot - 1 + \frac{1}{3} \cdot 5 & \frac{2}{3} \cdot 1 + \frac{1}{3} \cdot - 2 \\ \frac{5}{3} \cdot - 1 + \frac{1}{3} \cdot 5 & \frac{5}{3} \cdot 1 + \frac{1}{3} \cdot - 2 \end{array}] x = [\begin{array}{c} \frac{2}{3} & \frac{1}{3} \\ \frac{5}{3} & \frac{1}{3} \end{array}] \cdot [\begin{array}{c} 4 \\ - 26 \end{array}]$

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

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] x = [\begin{array}{c} \frac{2}{3} & \frac{1}{3} \\ \frac{5}{3} & \frac{1}{3} \end{array}] \cdot [\begin{array}{c} 4 \\ - 26 \end{array}]$

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

$x = [\begin{array}{c} \frac{2}{3} & \frac{1}{3} \\ \frac{5}{3} & \frac{1}{3} \end{array}] \cdot [\begin{array}{c} 4 \\ - 26 \end{array}]$

Simplify the right side of the equation.

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

$x = [\begin{array}{c} \frac{2}{3} \cdot 4 + \frac{1}{3} \cdot - 26 \\ \frac{5}{3} \cdot 4 + \frac{1}{3} \cdot - 26 \end{array}]$

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

$x = [\begin{array}{c} - 6 \\ - 2 \end{array}]$

The matrix is in the most simplified form.

$x = [\begin{array}{c} - 6 \\ - 2 \end{array}]$

Solve the Matrix Equation [[-1,1],[5,-2]]c=[[4],[-26]]

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