Linear Algebra Examples

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

Step 1

Multiply

$A$ by each element of the matrix.

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

Step 2

Simplify each element in the matrix.

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Step 2.1

Multiply

$A$ by

$1$ .

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

Step 2.2

Move

$2$ to the left of

$A$ .

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

Step 2.3

Move

$3$ to the left of

$A$ .

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

Step 2.4

Move

$- 1$ to the left of

$A$ .

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

Step 2.5

Rewrite

$- 1 A$ as

$- A$ .

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

Step 3

Write as a linear system of equations.

$A = 2$

$2 A = 1$

$3 A = 3$

$- A = - 2$

Step 4

Solve the system of equations.

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Step 4.1

Replace all occurrences of

$A$ with

$2$ in each equation.

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Step 4.1.1

Replace all occurrences of

$A$ in

$2 A = 1$ with

$2$ .

$2 (2) = 1$

$A = 2$

$3 A = 3$

$- A = - 2$

Step 4.1.2

Simplify the left side.

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Step 4.1.2.1

Multiply

$2$ by

$2$ .

$4 = 1$

$A = 2$

$3 A = 3$

$- A = - 2$

$4 = 1$

$A = 2$

$3 A = 3$

$- A = - 2$

Step 4.1.3

Replace all occurrences of

$A$ in

$3 A = 3$ with

$2$ .

$3 (2) = 3$

$4 = 1$

$A = 2$

$- A = - 2$

Step 4.1.4

Simplify the left side.

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Step 4.1.4.1

Multiply

$3$ by

$2$ .

$6 = 3$

$4 = 1$

$A = 2$

$- A = - 2$

$6 = 3$

$4 = 1$

$A = 2$

$- A = - 2$

Step 4.1.5

Replace all occurrences of

$A$ in

$- A = - 2$ with

$2$ .

$- (2) = - 2$

$6 = 3$

$4 = 1$

$A = 2$

Step 4.1.6

Simplify the left side.

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Step 4.1.6.1

Multiply

$- 1$ by

$2$ .

$- 2 = - 2$

$6 = 3$

$4 = 1$

$A = 2$

$- 2 = - 2$

$6 = 3$

$4 = 1$

$A = 2$

$- 2 = - 2$

$6 = 3$

$4 = 1$

$A = 2$

Step 4.2

Since

$6 = 3$ is not true, there is no solution.

$No solution$