Solve the Matrix Equation [[4,8],[2,5]]*[[x],[y]]=[[0],[6]]

$[\begin{array}{c} 4 & 8 \\ 2 & 5 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 0 \\ 6 \end{array}]$

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 $[\begin{array}{c} 4 & 8 \\ 2 & 5 \end{array}]$ .

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

$determinant [\begin{array}{c} 4 & 8 \\ 2 & 5 \end{array}] = | \begin{array}{c} 4 & 8 \\ 2 & 5 \end{array} |$

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

$(4) (5) - 2 \cdot 8$

Simplify the determinant.

Simplify each term.

Multiply $4$ by $5$ .

$20 - 2 \cdot 8$

Multiply $- 2$ by $8$ .

$20 - 16$

Subtract $16$ from $20$ .

$4$

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

$\frac{1}{4} [\begin{array}{c} 5 & - (8) \\ - (2) & 4 \end{array}]$

Simplify each element in the matrix.

Rearrange $- (8)$ .

$\frac{1}{4} [\begin{array}{c} 5 & - 8 \\ - (2) & 4 \end{array}]$

Rearrange $- (2)$ .

$\frac{1}{4} [\begin{array}{c} 5 & - 8 \\ - 2 & 4 \end{array}]$

Multiply $\frac{1}{4}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{4} \cdot 5 & \frac{1}{4} \cdot - 8 \\ \frac{1}{4} \cdot - 2 & \frac{1}{4} \cdot 4 \end{array}]$

Simplify each element in the matrix.

Rearrange $\frac{1}{4} \cdot 5$ .

$[\begin{array}{c} \frac{5}{4} & \frac{1}{4} \cdot - 8 \\ \frac{1}{4} \cdot - 2 & \frac{1}{4} \cdot 4 \end{array}]$

Rearrange $\frac{1}{4} \cdot - 8$ .

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

Rearrange $\frac{1}{4} \cdot - 2$ .

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

Rearrange $\frac{1}{4} \cdot 4$ .

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

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

$[\begin{array}{c} \frac{5}{4} & - 2 \\ - \frac{1}{2} & 1 \end{array}] \cdot [\begin{array}{c} 4 & 8 \\ 2 & 5 \end{array}] \cdot x = [\begin{array}{c} \frac{5}{4} & - 2 \\ - \frac{1}{2} & 1 \end{array}] \cdot [\begin{array}{c} 0 \\ 6 \end{array}]$

Simplify the left side of the equation.

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

$[\begin{array}{c} \frac{5}{4} \cdot 4 - 2 \cdot 2 & \frac{5}{4} \cdot 8 - 2 \cdot 5 \\ - \frac{1}{2} \cdot 4 + 1 \cdot 2 & - \frac{1}{2} \cdot 8 + 1 \cdot 5 \end{array}] x = [\begin{array}{c} \frac{5}{4} & - 2 \\ - \frac{1}{2} & 1 \end{array}] \cdot [\begin{array}{c} 0 \\ 6 \end{array}]$

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

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] x = [\begin{array}{c} \frac{5}{4} & - 2 \\ - \frac{1}{2} & 1 \end{array}] \cdot [\begin{array}{c} 0 \\ 6 \end{array}]$

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

$x = [\begin{array}{c} \frac{5}{4} & - 2 \\ - \frac{1}{2} & 1 \end{array}] \cdot [\begin{array}{c} 0 \\ 6 \end{array}]$

Simplify the right side of the equation.

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

$x = [\begin{array}{c} \frac{5}{4} \cdot 0 - 2 \cdot 6 \\ - \frac{1}{2} \cdot 0 + 1 \cdot 6 \end{array}]$

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

$x = [\begin{array}{c} - 12 \\ 6 \end{array}]$

The matrix is in the most simplified form.

$x = [\begin{array}{c} - 12 \\ 6 \end{array}]$

Solving for the variables in $[\begin{array}{c} x \\ y \end{array}]$ , the answer is $x = - 12, y = 6$ .

$x = - 12, y = 6$

Solve the Matrix Equation [[4,8],[2,5]]*[[x],[y]]=[[0],[6]]

Related Questions

Support LCMGCF.com by disabling your adblocker.