Linear Algebra Examples

12 x - 3 y = - 6

$12 x - 3 y = - 6$ ,

2 x + 15 y = 30

$2 x + 15 y = 30$

Step 1

Find the

A X = B

$A X = B$ from the system of equations.

[\begin{matrix} 12 & - 3 2 & 15 \end{matrix}] \cdot [\begin{matrix} x y \end{matrix}] = [\begin{matrix} - 6 30 \end{matrix}]

$[\begin{array}{c} 12 & - 3 \\ 2 & 15 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 6 \\ 30 \end{array}]$

Step 2

Find the inverse of the coefficient matrix.

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Step 2.1

The inverse of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

\frac{1}{a d - b c} [\begin{matrix} d & - b - c & a \end{matrix}]

$\frac{1}{a d - b c} [\begin{array}{c} d & - b \\ - c & a \end{array}]$ where

a d - b c

$a d - b c$ is the determinant.

Step 2.2

Find the determinant.

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Step 2.2.1

The determinant of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

12 \cdot 15 - 2 \cdot - 3

$12 \cdot 15 - 2 \cdot - 3$

Step 2.2.2

Simplify the determinant.

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

Multiply

12

$12$ by

15

$15$ .

180 - 2 \cdot - 3

$180 - 2 \cdot - 3$

Step 2.2.2.1.2

Multiply

- 2

$- 2$ by

- 3

$- 3$ .

180 + 6

$180 + 6$

180 + 6

$180 + 6$

Step 2.2.2.2

Add

180

$180$ and

6

$6$ .

186

$186$

186

$186$

186

$186$

Step 2.3

Since the determinant is non-zero, the inverse exists.

Step 2.4

Substitute the known values into the formula for the inverse.

\frac{1}{186} [\begin{matrix} 15 & 3 - 2 & 12 \end{matrix}]

$\frac{1}{186} [\begin{array}{c} 15 & 3 \\ - 2 & 12 \end{array}]$

Step 2.5

Multiply

\frac{1}{186}

$\frac{1}{186}$ by each element of the matrix.

[\begin{matrix} \frac{1}{186} \cdot 15 & \frac{1}{186} \cdot 3 \frac{1}{186} \cdot - 2 & \frac{1}{186} \cdot 12 \end{matrix}]

$[\begin{array}{c} \frac{1}{186} \cdot 15 & \frac{1}{186} \cdot 3 \\ \frac{1}{186} \cdot - 2 & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6

Simplify each element in the matrix.

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Step 2.6.1

Cancel the common factor of

3

$3$ .

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Step 2.6.1.1

Factor

3

$3$ out of

186

$186$ .

⎡ ⎣ \begin{matrix} \frac{1}{3 (62)} \cdot 15 & \frac{1}{186} \cdot 3 \frac{1}{186} \cdot - 2 & \frac{1}{186} \cdot 12 \end{matrix} ⎤ ⎦

$[\begin{array}{c} \frac{1}{3 (62)} \cdot 15 & \frac{1}{186} \cdot 3 \\ \frac{1}{186} \cdot - 2 & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.1.2

Factor

3

$3$ out of

15

$15$ .

[\begin{matrix} \frac{1}{3 \cdot 62} \cdot (3 \cdot 5) & \frac{1}{186} \cdot 3 \frac{1}{186} \cdot - 2 & \frac{1}{186} \cdot 12 \end{matrix}]

$[\begin{array}{c} \frac{1}{3 \cdot 62} \cdot (3 \cdot 5) & \frac{1}{186} \cdot 3 \\ \frac{1}{186} \cdot - 2 & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.1.3

Cancel the common factor.

$[\begin{array}{c} \frac{1}{3 \cdot 62} \cdot (3 \cdot 5) & \frac{1}{186} \cdot 3 \\ \frac{1}{186} \cdot - 2 & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.1.4

Rewrite the expression.

$[\begin{array}{c} \frac{1}{62} \cdot 5 & \frac{1}{186} \cdot 3 \\ \frac{1}{186} \cdot - 2 & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.2

Combine

$\frac{1}{62}$ and

$5$ .

$[\begin{array}{c} \frac{5}{62} & \frac{1}{186} \cdot 3 \\ \frac{1}{186} \cdot - 2 & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.3

Cancel the common factor of

$3$ .

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Step 2.6.3.1

Factor

$3$ out of

$186$ .

$[\begin{array}{c} \frac{5}{62} & \frac{1}{3 (62)} \cdot 3 \\ \frac{1}{186} \cdot - 2 & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.3.2

Cancel the common factor.

$[\begin{array}{c} \frac{5}{62} & \frac{1}{3 \cdot 62} \cdot 3 \\ \frac{1}{186} \cdot - 2 & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.3.3

Rewrite the expression.

$[\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ \frac{1}{186} \cdot - 2 & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.4

Cancel the common factor of

$2$ .

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Step 2.6.4.1

Factor

$2$ out of

$186$ .

$[\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ \frac{1}{2 (93)} \cdot - 2 & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.4.2

Factor

$2$ out of

$- 2$ .

$[\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ \frac{1}{2 \cdot 93} \cdot (2 \cdot - 1) & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.4.3

Cancel the common factor.

$[\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ \frac{1}{2 \cdot 93} \cdot (2 \cdot - 1) & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.4.4

Rewrite the expression.

$[\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ \frac{1}{93} \cdot - 1 & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.5

Combine

$\frac{1}{93}$ and

$- 1$ .

$[\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ \frac{- 1}{93} & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.6

Move the negative in front of the fraction.

$[\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ - \frac{1}{93} & \frac{1}{186} \cdot 12 \end{array}]$

Step 2.6.7

Cancel the common factor of

$6$ .

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Step 2.6.7.1

Factor

$6$ out of

$186$ .

$[\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ - \frac{1}{93} & \frac{1}{6 (31)} \cdot 12 \end{array}]$

Step 2.6.7.2

Factor

$6$ out of

$12$ .

$[\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ - \frac{1}{93} & \frac{1}{6 \cdot 31} \cdot (6 \cdot 2) \end{array}]$

Step 2.6.7.3

Cancel the common factor.

$[\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ - \frac{1}{93} & \frac{1}{6 \cdot 31} \cdot (6 \cdot 2) \end{array}]$

Step 2.6.7.4

Rewrite the expression.

$[\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ - \frac{1}{93} & \frac{1}{31} \cdot 2 \end{array}]$

Step 2.6.8

Combine

$\frac{1}{31}$ and

$2$ .

$[\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ - \frac{1}{93} & \frac{2}{31} \end{array}]$

Step 3

Left multiply both sides of the matrix equation by the inverse matrix.

$([\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ - \frac{1}{93} & \frac{2}{31} \end{array}] \cdot [\begin{array}{c} 12 & - 3 \\ 2 & 15 \end{array}]) \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ - \frac{1}{93} & \frac{2}{31} \end{array}] \cdot [\begin{array}{c} - 6 \\ 30 \end{array}]$

Step 4

Any matrix multiplied by its inverse is equal to

$1$ all the time.

$A \cdot A^{- 1} = 1$ .

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ - \frac{1}{93} & \frac{2}{31} \end{array}] \cdot [\begin{array}{c} - 6 \\ 30 \end{array}]$

Step 5

Multiply

$[\begin{array}{c} \frac{5}{62} & \frac{1}{62} \\ - \frac{1}{93} & \frac{2}{31} \end{array}] [\begin{array}{c} - 6 \\ 30 \end{array}]$ .

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Step 5.1

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 5.2

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

$[\begin{array}{c} \frac{5}{62} \cdot - 6 + \frac{1}{62} \cdot 30 \\ - \frac{1}{93} \cdot - 6 + \frac{2}{31} \cdot 30 \end{array}]$

Step 5.3

Simplify each element of the matrix by multiplying out all the expressions.

$[\begin{array}{c} 0 \\ 2 \end{array}]$

Step 6

Simplify the left and right side.

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 0 \\ 2 \end{array}]$

Step 7

Find the solution.

$x = 0$

$y = 2$