Solve the Matrix Equation [[-3,-14],[2,9]]x=[[8,-4],[0,-4]]

$[\begin{array}{c} - 3 & - 14 \\ 2 & 9 \end{array}] x = [\begin{array}{c} 8 & - 4 \\ 0 & - 4 \end{array}]$

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 $[\begin{array}{c} - 3 & - 14 \\ 2 & 9 \end{array}]$ .

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

$determinant [\begin{array}{c} - 3 & - 14 \\ 2 & 9 \end{array}] = | \begin{array}{c} - 3 & - 14 \\ 2 & 9 \end{array} |$

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

$(- 3) (9) - 2 \cdot - 14$

Simplify the determinant.

Simplify each term.

Multiply $- 3$ by $9$ .

$- 27 - 2 \cdot - 14$

Multiply $- 2$ by $- 14$ .

$- 27 + 28$

Add $- 27$ and $28$ .

$1$

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

$\frac{1}{1} [\begin{array}{c} 9 & - (- 14) \\ - (2) & - 3 \end{array}]$

Simplify each element in the matrix.

Rearrange $- (- 14)$ .

$\frac{1}{1} [\begin{array}{c} 9 & 14 \\ - (2) & - 3 \end{array}]$

Rearrange $- (2)$ .

$\frac{1}{1} [\begin{array}{c} 9 & 14 \\ - 2 & - 3 \end{array}]$

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

$[\begin{array}{c} \frac{1}{1} \cdot 9 & \frac{1}{1} \cdot 14 \\ \frac{1}{1} \cdot - 2 & \frac{1}{1} \cdot - 3 \end{array}]$

Simplify each element in the matrix.

Rearrange $\frac{1}{1} \cdot 9$ .

$[\begin{array}{c} 9 & \frac{1}{1} \cdot 14 \\ \frac{1}{1} \cdot - 2 & \frac{1}{1} \cdot - 3 \end{array}]$

Rearrange $\frac{1}{1} \cdot 14$ .

$[\begin{array}{c} 9 & 14 \\ \frac{1}{1} \cdot - 2 & \frac{1}{1} \cdot - 3 \end{array}]$

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

$[\begin{array}{c} 9 & 14 \\ - 2 & \frac{1}{1} \cdot - 3 \end{array}]$

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

$[\begin{array}{c} 9 & 14 \\ - 2 & - 3 \end{array}]$

Step 2

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

$[\begin{array}{c} 9 & 14 \\ - 2 & - 3 \end{array}] \cdot ([\begin{array}{c} - 3 & - 14 \\ 2 & 9 \end{array}] x) = [\begin{array}{c} 9 & 14 \\ - 2 & - 3 \end{array}] \cdot [\begin{array}{c} 8 & - 4 \\ 0 & - 4 \end{array}]$

Step 3

Simplify the left side of the equation.

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

$[\begin{array}{c} 9 \cdot - 3 + 14 \cdot 2 & 9 \cdot - 14 + 14 \cdot 9 \\ - 2 \cdot - 3 - 3 \cdot 2 & - 2 \cdot - 14 - 3 \cdot 9 \end{array}] x = [\begin{array}{c} 9 & 14 \\ - 2 & - 3 \end{array}] \cdot [\begin{array}{c} 8 & - 4 \\ 0 & - 4 \end{array}]$

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

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] x = [\begin{array}{c} 9 & 14 \\ - 2 & - 3 \end{array}] \cdot [\begin{array}{c} 8 & - 4 \\ 0 & - 4 \end{array}]$

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

$x = [\begin{array}{c} 9 & 14 \\ - 2 & - 3 \end{array}] \cdot [\begin{array}{c} 8 & - 4 \\ 0 & - 4 \end{array}]$

Step 4

Simplify the right side of the equation.

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

$x = [\begin{array}{c} 9 \cdot 8 + 14 \cdot 0 & 9 \cdot - 4 + 14 \cdot - 4 \\ - 2 \cdot 8 - 3 \cdot 0 & - 2 \cdot - 4 - 3 \cdot - 4 \end{array}]$

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

$x = [\begin{array}{c} 72 & - 92 \\ - 16 & 20 \end{array}]$

The matrix is in the most simplified form.

$x = [\begin{array}{c} 72 & - 92 \\ - 16 & 20 \end{array}]$

Solve the Matrix Equation [[-3,-14],[2,9]]x=[[8,-4],[0,-4]]

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