Algebra Examples

$[\begin{array}{c} - 3 & 5 \\ 25 & - 2 \end{array}] - 3 [\begin{array}{c} 0 & - 2 \\ x & 4 \end{array}] = [\begin{array}{c} - 3 & 11 \\ 15 & - 14 \end{array}]$

Step 1

Simplify each term.

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Step 1.1

Multiply $- 3$ by each element of the matrix.

$[\begin{array}{c} - 3 & 5 \\ 25 & - 2 \end{array}] + [\begin{array}{c} - 3 \cdot 0 & - 3 \cdot - 2 \\ - 3 x & - 3 \cdot 4 \end{array}] = [\begin{array}{c} - 3 & 11 \\ 15 & - 14 \end{array}]$

Step 1.2

Simplify each element in the matrix.

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Step 1.2.1

Multiply $- 3$ by $0$ .

$[\begin{array}{c} - 3 & 5 \\ 25 & - 2 \end{array}] + [\begin{array}{c} 0 & - 3 \cdot - 2 \\ - 3 x & - 3 \cdot 4 \end{array}] = [\begin{array}{c} - 3 & 11 \\ 15 & - 14 \end{array}]$

Step 1.2.2

Multiply $- 3$ by $- 2$ .

$[\begin{array}{c} - 3 & 5 \\ 25 & - 2 \end{array}] + [\begin{array}{c} 0 & 6 \\ - 3 x & - 3 \cdot 4 \end{array}] = [\begin{array}{c} - 3 & 11 \\ 15 & - 14 \end{array}]$

Step 1.2.3

Multiply $- 3$ by $4$ .

$[\begin{array}{c} - 3 & 5 \\ 25 & - 2 \end{array}] + [\begin{array}{c} 0 & 6 \\ - 3 x & - 12 \end{array}] = [\begin{array}{c} - 3 & 11 \\ 15 & - 14 \end{array}]$

Step 2

Add the corresponding elements.

$[\begin{array}{c} - 3 + 0 & 5 + 6 \\ 25 - 3 x & - 2 - 12 \end{array}] = [\begin{array}{c} - 3 & 11 \\ 15 & - 14 \end{array}]$

Step 3

Simplify each element.

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Step 3.1

Add $- 3$ and $0$ .

$[\begin{array}{c} - 3 & 5 + 6 \\ 25 - 3 x & - 2 - 12 \end{array}] = [\begin{array}{c} - 3 & 11 \\ 15 & - 14 \end{array}]$

Step 3.2

Add $5$ and $6$ .

$[\begin{array}{c} - 3 & 11 \\ 25 - 3 x & - 2 - 12 \end{array}] = [\begin{array}{c} - 3 & 11 \\ 15 & - 14 \end{array}]$

Step 3.3

Subtract $12$ from $- 2$ .

$[\begin{array}{c} - 3 & 11 \\ 25 - 3 x & - 14 \end{array}] = [\begin{array}{c} - 3 & 11 \\ 15 & - 14 \end{array}]$

Step 4

Write as a linear system of equations.

$- 3 = - 3$

$11 = 11$

$25 - 3 x = 15$

$- 14 = - 14$

Step 5

Since $- 14 = - 14$ is always true, the matrix equation has infinite solutions.

Infinite solutions satisfying $25 - 3 x = 15$ .