Solve the Matrix Equation [[2,5],[1,3]]x=[[2],[0]]

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

Step 1

Find the inverse of $[\begin{array}{c} 2 & 5 \\ 1 & 3 \end{array}]$ .

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 $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$2 \cdot 3 - 1 \cdot 5$

Simplify the determinant.

Simplify each term.

Multiply $2$ by $3$ .

$6 - 1 \cdot 5$

Multiply $- 1$ by $5$ .

$6 - 5$

Subtract $5$ from $6$ .

$1$

Substitute the known values into the formula for the inverse.

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

Divide $1$ by $1$ .

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

Multiply $1$ by each element of the matrix.

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

Simplify each element in the matrix.

Multiply $3$ by $1$ .

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

Multiply $- 5$ by $1$ .

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

Multiply $- 1$ by $1$ .

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

Multiply $2$ by $1$ .

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

Step 2

Multiply both sides by the inverse of $[\begin{array}{c} 2 & 5 \\ 1 & 3 \end{array}]$ .

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

Step 3

Simplify the equation.

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

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

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

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

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

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

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

Multiply $[\begin{array}{c} 3 & - 5 \\ - 1 & 2 \end{array}] [\begin{array}{c} 2 \\ 0 \end{array}]$ .

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

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

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

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

Solve the Matrix Equation [[2,5],[1,3]]x=[[2],[0]]

Related Questions

Support LCMGCF.com by disabling your adblocker.