Solve the Matrix Equation [[1,3],[1,4]]x=[[6],[8]]

$[\begin{array}{c} 1 & 3 \\ 1 & 4 \end{array}] x = [\begin{array}{c} 6 \\ 8 \end{array}]$

Step 1

Find the inverse of $[\begin{array}{c} 1 & 3 \\ 1 & 4 \end{array}]$ .

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.

Find the determinant.

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

$1 \cdot 4 - 1 \cdot 3$

Simplify the determinant.

Simplify each term.

Multiply $4$ by $1$ .

$4 - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$4 - 3$

Subtract $3$ from $4$ .

$1$

Substitute the known values into the formula for the inverse.

$\frac{1}{1} [\begin{array}{c} 4 & - 3 \\ - 1 & 1 \end{array}]$

Divide $1$ by $1$ .

$1 [\begin{array}{c} 4 & - 3 \\ - 1 & 1 \end{array}]$

Multiply $1$ by each element of the matrix.

$[\begin{array}{c} 1 \cdot 4 & 1 \cdot - 3 \\ 1 \cdot - 1 & 1 \cdot 1 \end{array}]$

Simplify each element in the matrix.

Multiply $4$ by $1$ .

$[\begin{array}{c} 4 & 1 \cdot - 3 \\ 1 \cdot - 1 & 1 \cdot 1 \end{array}]$

Multiply $- 3$ by $1$ .

$[\begin{array}{c} 4 & - 3 \\ 1 \cdot - 1 & 1 \cdot 1 \end{array}]$

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 4 & - 3 \\ - 1 & 1 \cdot 1 \end{array}]$

Multiply $1$ by $1$ .

$[\begin{array}{c} 4 & - 3 \\ - 1 & 1 \end{array}]$

Step 2

Multiply both sides by the inverse of $[\begin{array}{c} 1 & 3 \\ 1 & 4 \end{array}]$ .

$[\begin{array}{c} 4 & - 3 \\ - 1 & 1 \end{array}] [\begin{array}{c} 1 & 3 \\ 1 & 4 \end{array}] x = [\begin{array}{c} 4 & - 3 \\ - 1 & 1 \end{array}] [\begin{array}{c} 6 \\ 8 \end{array}]$

Step 3

Simplify the equation.

Multiply $[\begin{array}{c} 4 & - 3 \\ - 1 & 1 \end{array}] [\begin{array}{c} 1 & 3 \\ 1 & 4 \end{array}]$ .

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

$[\begin{array}{c} 4 \cdot 1 - 3 \cdot 1 & 4 \cdot 3 - 3 \cdot 4 \\ - 1 \cdot 1 + 1 \cdot 1 & - 1 \cdot 3 + 1 \cdot 4 \end{array}] x = [\begin{array}{c} 4 & - 3 \\ - 1 & 1 \end{array}] [\begin{array}{c} 6 \\ 8 \end{array}]$

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

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] x = [\begin{array}{c} 4 & - 3 \\ - 1 & 1 \end{array}] [\begin{array}{c} 6 \\ 8 \end{array}]$

Multiplying the identity matrix by any matrix $A$ is the matrix $A$ itself.

$x = [\begin{array}{c} 4 & - 3 \\ - 1 & 1 \end{array}] [\begin{array}{c} 6 \\ 8 \end{array}]$

Multiply $[\begin{array}{c} 4 & - 3 \\ - 1 & 1 \end{array}] [\begin{array}{c} 6 \\ 8 \end{array}]$ .

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

$x = [\begin{array}{c} 4 \cdot 6 - 3 \cdot 8 \\ - 1 \cdot 6 + 1 \cdot 8 \end{array}]$

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

$x = [\begin{array}{c} 0 \\ 2 \end{array}]$

Solve the Matrix Equation [[1,3],[1,4]]x=[[6],[8]]

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