Solve the Matrix Equation [[3,-2],[-5,4]]x-[[3,-1],[6,2]]=[[0,-4],[-2,3]]

$[\begin{array}{c} 3 & - 2 \\ - 5 & 4 \end{array}] x - [\begin{array}{c} 3 & - 1 \\ 6 & 2 \end{array}] = [\begin{array}{c} 0 & - 4 \\ - 2 & 3 \end{array}]$

Step 1

Add the matrix to both sides of the equation.

$X \cdot X - [\begin{array}{c} 3 & - 1 \\ 6 & 2 \end{array}] + [\begin{array}{c} 3 & - 1 \\ 6 & 2 \end{array}] = [\begin{array}{c} 0 & - 4 \\ - 2 & 3 \end{array}] + [\begin{array}{c} 3 & - 1 \\ 6 & 2 \end{array}]$

Step 2

Simplify both sides of the equation.

Simplify the left side of the equation.

Multiply $X$ by $X$ .

$- [\begin{array}{c} 3 & - 1 \\ 6 & 2 \end{array}] + [\begin{array}{c} 3 & - 1 \\ 6 & 2 \end{array}] + X^{2} = [\begin{array}{c} 0 & - 4 \\ - 2 & 3 \end{array}] + [\begin{array}{c} 3 & - 1 \\ 6 & 2 \end{array}]$

Add the corresponding elements.

$[\begin{array}{c} - 3 + 3 & 1 - 1 \\ - 6 + 6 & - 2 + 2 \end{array}] + X^{2} = [\begin{array}{c} 0 & - 4 \\ - 2 & 3 \end{array}] + [\begin{array}{c} 3 & - 1 \\ 6 & 2 \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} - 3 + 3 & 1 - 1 \\ - 6 + 6 & - 2 + 2 \end{array}]$ .

Simplify $- 3 + 3$ .

$[\begin{array}{c} 0 & 1 - 1 \\ - 6 + 6 & - 2 + 2 \end{array}] + X^{2} = [\begin{array}{c} 0 & - 4 \\ - 2 & 3 \end{array}] + [\begin{array}{c} 3 & - 1 \\ 6 & 2 \end{array}]$

Simplify $1 - 1$ .

$[\begin{array}{c} 0 & 0 \\ - 6 + 6 & - 2 + 2 \end{array}] + X^{2} = [\begin{array}{c} 0 & - 4 \\ - 2 & 3 \end{array}] + [\begin{array}{c} 3 & - 1 \\ 6 & 2 \end{array}]$

Simplify $- 6 + 6$ .

$[\begin{array}{c} 0 & 0 \\ 0 & - 2 + 2 \end{array}] + X^{2} = [\begin{array}{c} 0 & - 4 \\ - 2 & 3 \end{array}] + [\begin{array}{c} 3 & - 1 \\ 6 & 2 \end{array}]$

Simplify $- 2 + 2$ .

$[\begin{array}{c} 0 & 0 \\ 0 & 0 \end{array}] + X^{2} = [\begin{array}{c} 0 & - 4 \\ - 2 & 3 \end{array}] + [\begin{array}{c} 3 & - 1 \\ 6 & 2 \end{array}]$

Simplify the right side of the equation.

Add the corresponding elements.

$[\begin{array}{c} 0 & 0 \\ 0 & 0 \end{array}] + X^{2} = [\begin{array}{c} 0 + 3 & - 4 - 1 \\ - 2 + 6 & 3 + 2 \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} 0 + 3 & - 4 - 1 \\ - 2 + 6 & 3 + 2 \end{array}]$ .

Simplify $0 + 3$ .

$[\begin{array}{c} 0 & 0 \\ 0 & 0 \end{array}] + X^{2} = [\begin{array}{c} 3 & - 4 - 1 \\ - 2 + 6 & 3 + 2 \end{array}]$

Simplify $- 4 - 1$ .

$[\begin{array}{c} 0 & 0 \\ 0 & 0 \end{array}] + X^{2} = [\begin{array}{c} 3 & - 5 \\ - 2 + 6 & 3 + 2 \end{array}]$

Simplify $- 2 + 6$ .

$[\begin{array}{c} 0 & 0 \\ 0 & 0 \end{array}] + X^{2} = [\begin{array}{c} 3 & - 5 \\ 4 & 3 + 2 \end{array}]$

Simplify $3 + 2$ .

$[\begin{array}{c} 0 & 0 \\ 0 & 0 \end{array}] + X^{2} = [\begin{array}{c} 3 & - 5 \\ 4 & 5 \end{array}]$

Adding the null matrix to any matrix is a matrix itself.

$X^{2} = [\begin{array}{c} 3 & - 5 \\ 4 & 5 \end{array}]$

Step 3

The matrix is in the most simplified form.

$X^{2} = [\begin{array}{c} 3 & - 5 \\ 4 & 5 \end{array}]$

Solve the Matrix Equation [[3,-2],[-5,4]]x-[[3,-1],[6,2]]=[[0,-4],[-2,3]]

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