Linear Algebra Examples

$[\begin{array}{c} - 1 & - 3 \\ 3 & 4 \end{array}] \cdot [\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} 12 \\ - 11 \end{array}]$

Step 1

Multiply $[\begin{array}{c} - 1 & - 3 \\ 3 & 4 \end{array}] [\begin{array}{c} a \\ b \end{array}]$ .

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Step 1.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is $2 \times 2$ and the second matrix is $2 \times 1$ .

Step 1.2

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

$[\begin{array}{c} - a - 3 b \\ 3 a + 4 b \end{array}] = [\begin{array}{c} 12 \\ - 11 \end{array}]$

Step 2

Write as a linear system of equations.

$- a - 3 b = 12$

$3 a + 4 b = - 11$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $a$ in $- a - 3 b = 12$ .

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Step 3.1.1

Add $3 b$ to both sides of the equation.

$- a = 12 + 3 b$

$3 a + 4 b = - 11$

Step 3.1.2

Divide each term in $- a = 12 + 3 b$ by $- 1$ and simplify.

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Step 3.1.2.1

Divide each term in $- a = 12 + 3 b$ by $- 1$ .

$\frac{- a}{- 1} = \frac{12}{- 1} + \frac{3 b}{- 1}$

$3 a + 4 b = - 11$

Step 3.1.2.2

Simplify the left side.

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Step 3.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{a}{1} = \frac{12}{- 1} + \frac{3 b}{- 1}$

$3 a + 4 b = - 11$

Step 3.1.2.2.2

Divide $a$ by $1$ .

$a = \frac{12}{- 1} + \frac{3 b}{- 1}$

$3 a + 4 b = - 11$

$a = \frac{12}{- 1} + \frac{3 b}{- 1}$

$3 a + 4 b = - 11$

Step 3.1.2.3

Simplify the right side.

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Step 3.1.2.3.1

Simplify each term.

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Step 3.1.2.3.1.1

Divide $12$ by $- 1$ .

$a = - 12 + \frac{3 b}{- 1}$

$3 a + 4 b = - 11$

Step 3.1.2.3.1.2

Move the negative one from the denominator of $\frac{3 b}{- 1}$ .

$a = - 12 - 1 \cdot (3 b)$

$3 a + 4 b = - 11$

Step 3.1.2.3.1.3

Rewrite $- 1 \cdot (3 b)$ as $- (3 b)$ .

$a = - 12 - (3 b)$

$3 a + 4 b = - 11$

Step 3.1.2.3.1.4

Multiply $3$ by $- 1$ .

$a = - 12 - 3 b$

$3 a + 4 b = - 11$

$a = - 12 - 3 b$

$3 a + 4 b = - 11$

$a = - 12 - 3 b$

$3 a + 4 b = - 11$

$a = - 12 - 3 b$

$3 a + 4 b = - 11$

$a = - 12 - 3 b$

$3 a + 4 b = - 11$

Step 3.2

Replace all occurrences of $a$ with $- 12 - 3 b$ in each equation.

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Step 3.2.1

Replace all occurrences of $a$ in $3 a + 4 b = - 11$ with $- 12 - 3 b$ .

$3 (- 12 - 3 b) + 4 b = - 11$

$a = - 12 - 3 b$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Simplify $3 (- 12 - 3 b) + 4 b$ .

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Step 3.2.2.1.1

Simplify each term.

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Step 3.2.2.1.1.1

Apply the distributive property.

$3 \cdot - 12 + 3 (- 3 b) + 4 b = - 11$

$a = - 12 - 3 b$

Step 3.2.2.1.1.2

Multiply $3$ by $- 12$ .

$- 36 + 3 (- 3 b) + 4 b = - 11$

$a = - 12 - 3 b$

Step 3.2.2.1.1.3

Multiply $- 3$ by $3$ .

$- 36 - 9 b + 4 b = - 11$

$a = - 12 - 3 b$

$- 36 - 9 b + 4 b = - 11$

$a = - 12 - 3 b$

Step 3.2.2.1.2

Add $- 9 b$ and $4 b$ .

$- 36 - 5 b = - 11$

$a = - 12 - 3 b$

$- 36 - 5 b = - 11$

$a = - 12 - 3 b$

$- 36 - 5 b = - 11$

$a = - 12 - 3 b$

$- 36 - 5 b = - 11$

$a = - 12 - 3 b$

Step 3.3

Solve for $b$ in $- 36 - 5 b = - 11$ .

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Step 3.3.1

Move all terms not containing $b$ to the right side of the equation.

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Step 3.3.1.1

Add $36$ to both sides of the equation.

$- 5 b = - 11 + 36$

$a = - 12 - 3 b$

Step 3.3.1.2

Add $- 11$ and $36$ .

$- 5 b = 25$

$a = - 12 - 3 b$

$- 5 b = 25$

$a = - 12 - 3 b$

Step 3.3.2

Divide each term in $- 5 b = 25$ by $- 5$ and simplify.

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Step 3.3.2.1

Divide each term in $- 5 b = 25$ by $- 5$ .

$\frac{- 5 b}{- 5} = \frac{25}{- 5}$

$a = - 12 - 3 b$

Step 3.3.2.2

Simplify the left side.

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Step 3.3.2.2.1

Cancel the common factor of $- 5$ .

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Step 3.3.2.2.1.1

Cancel the common factor.

$\frac{- 5 b}{- 5} = \frac{25}{- 5}$

$a = - 12 - 3 b$

Step 3.3.2.2.1.2

Divide $b$ by $1$ .

$b = \frac{25}{- 5}$

$a = - 12 - 3 b$

$b = \frac{25}{- 5}$

$a = - 12 - 3 b$

$b = \frac{25}{- 5}$

$a = - 12 - 3 b$

Step 3.3.2.3

Simplify the right side.

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Step 3.3.2.3.1

Divide $25$ by $- 5$ .

$b = - 5$

$a = - 12 - 3 b$

$b = - 5$

$a = - 12 - 3 b$

$b = - 5$

$a = - 12 - 3 b$

$b = - 5$

$a = - 12 - 3 b$

Step 3.4

Replace all occurrences of $b$ with $- 5$ in each equation.

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Step 3.4.1

Replace all occurrences of $b$ in $a = - 12 - 3 b$ with $- 5$ .

$a = - 12 - 3 \cdot - 5$

$b = - 5$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $- 12 - 3 \cdot - 5$ .

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Step 3.4.2.1.1

Multiply $- 3$ by $- 5$ .

$a = - 12 + 15$

$b = - 5$

Step 3.4.2.1.2

Add $- 12$ and $15$ .

$a = 3$

$b = - 5$

$a = 3$

$b = - 5$

$a = 3$

$b = - 5$

$a = 3$

$b = - 5$

Step 3.5

List all of the solutions.

$a = 3, b = - 5$