Linear Algebra Examples

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

Step 1

Multiply $X$ by each element of the matrix.

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

Step 2

Simplify each element in the matrix.

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Step 2.1

Multiply $X$ by $1$ .

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

Step 2.2

Move $- 1$ to the left of $X$ .

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

Step 2.3

Rewrite $- 1 X$ as $- X$ .

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

Step 2.4

Move $2$ to the left of $X$ .

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

Step 2.5

Move $2$ to the left of $X$ .

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

Step 2.6

Multiply $X$ by $0$ .

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

Step 2.7

Multiply $X$ by $1$ .

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

Step 3

Write as a linear system of equations.

$X = - 6$

$- X = - 2$

$2 X = 0$

$2 X = 4$

$0 = - 2$

$X = 5$

Step 4

Since $0 \neq - 2$ , the matrix equation has no solution.

No solution