Solve the Matrix Equation [[1,-1],[3,-2]][[x],[y]]=[[5],[12]]

$[\begin{array}{c} 1 & - 1 \\ 3 & - 2 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 5 \\ 12 \end{array}]$

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

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

$determinant [\begin{array}{c} 1 & - 1 \\ 3 & - 2 \end{array}] = | \begin{array}{c} 1 & - 1 \\ 3 & - 2 \end{array} |$

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

$(1) (- 2) - 3 \cdot - 1$

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $1$ .

$- 2 - 3 \cdot - 1$

Multiply $- 3$ by $- 1$ .

$- 2 + 3$

Add $- 2$ and $3$ .

$1$

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

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

Simplify each element in the matrix.

Rearrange $- (- 1)$ .

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

Rearrange $- (3)$ .

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

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

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

Simplify each element in the matrix.

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

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

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

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

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

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

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

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

Step 2

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

$[\begin{array}{c} - 2 & 1 \\ - 3 & 1 \end{array}] \cdot ([\begin{array}{c} 1 & - 1 \\ 3 & - 2 \end{array}] x) = [\begin{array}{c} - 2 & 1 \\ - 3 & 1 \end{array}] \cdot [\begin{array}{c} 5 \\ 12 \end{array}]$

Step 3

Simplify the left side of the equation.

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

$[\begin{array}{c} - 2 \cdot 1 + 1 \cdot 3 & - 2 \cdot - 1 + 1 \cdot - 2 \\ - 3 \cdot 1 + 1 \cdot 3 & - 3 \cdot - 1 + 1 \cdot - 2 \end{array}] x = [\begin{array}{c} - 2 & 1 \\ - 3 & 1 \end{array}] \cdot [\begin{array}{c} 5 \\ 12 \end{array}]$

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

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] x = [\begin{array}{c} - 2 & 1 \\ - 3 & 1 \end{array}] \cdot [\begin{array}{c} 5 \\ 12 \end{array}]$

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

$x = [\begin{array}{c} - 2 & 1 \\ - 3 & 1 \end{array}] \cdot [\begin{array}{c} 5 \\ 12 \end{array}]$

Step 4

Simplify the right side of the equation.

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

$x = [\begin{array}{c} - 2 \cdot 5 + 1 \cdot 12 \\ - 3 \cdot 5 + 1 \cdot 12 \end{array}]$

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

$x = [\begin{array}{c} 2 \\ - 3 \end{array}]$

The matrix is in the most simplified form.

$x = [\begin{array}{c} 2 \\ - 3 \end{array}]$

Step 5

Solving for the variables in $[\begin{array}{c} x \\ y \end{array}]$ , the answer is $x = 2, y = - 3$ .

$x = 2, y = - 3$

Solve the Matrix Equation [[1,-1],[3,-2]][[x],[y]]=[[5],[12]]

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