Linear Algebra Examples

2 x + y = 4

$2 x + y = 4$ ,

- 6 x - 3 y = - 12

$- 6 x - 3 y = - 12$

Step 1

Find the

A X = B

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

[\begin{matrix} 2 & 1 - 6 & - 3 \end{matrix}] \cdot [\begin{matrix} x y \end{matrix}] = [\begin{matrix} 4 - 12 \end{matrix}]

$[\begin{array}{c} 2 & 1 \\ - 6 & - 3 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 4 \\ - 12 \end{array}]$

Step 2

Find the inverse of the coefficient matrix.

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The inverse of a

2 \times 2

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

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

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

| A |

$| A |$ is the determinant of

A

$A$ .

A = [\begin{matrix} a & b c & d \end{matrix}]

$A = [\begin{array}{c} a & b \\ c & d \end{array}]$ then

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

$A^{- 1} = \frac{1}{| A |} [\begin{array}{c} d & - b \\ - c & a \end{array}]$

Find the determinant of

[\begin{matrix} 2 & 1 - 6 & - 3 \end{matrix}]

$[\begin{array}{c} 2 & 1 \\ - 6 & - 3 \end{array}]$ .

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These are both valid notations for the determinant of a matrix.

determinant [\begin{matrix} 2 & 1 - 6 & - 3 \end{matrix}] = ∣ ∣ ∣ \begin{matrix} 2 & 1 - 6 & - 3 \end{matrix} ∣ ∣ ∣

$determinant [\begin{array}{c} 2 & 1 \\ - 6 & - 3 \end{array}] = | \begin{array}{c} 2 & 1 \\ - 6 & - 3 \end{array} |$

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

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

(2) (- 3) + 6 \cdot 1

$(2) (- 3) + 6 \cdot 1$

Simplify the determinant.

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Simplify each term.

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Multiply

2

$2$ by

- 3

$- 3$ .

- 6 + 6 \cdot 1

$- 6 + 6 \cdot 1$

Multiply

6

$6$ by

1

$1$ .

- 6 + 6

$- 6 + 6$

- 6 + 6

$- 6 + 6$

Add

- 6

$- 6$ and

6

$6$ .

0

$0$

0

$0$

0

$0$

Substitute the known values into the formula for the inverse of a matrix.

\frac{1}{0} [\begin{matrix} - 3 & - (1) - (- 6) & 2 \end{matrix}]

$\frac{1}{0} [\begin{array}{c} - 3 & - (1) \\ - (- 6) & 2 \end{array}]$

Simplify each element in the matrix.

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Rearrange

- (1)

$- (1)$ .

\frac{1}{0} [\begin{matrix} - 3 & - 1 - (- 6) & 2 \end{matrix}]

$\frac{1}{0} [\begin{array}{c} - 3 & - 1 \\ - (- 6) & 2 \end{array}]$

Rearrange

- (- 6)

$- (- 6)$ .

\frac{1}{0} [\begin{matrix} - 3 & - 1 6 & 2 \end{matrix}]

$\frac{1}{0} [\begin{array}{c} - 3 & - 1 \\ 6 & 2 \end{array}]$

\frac{1}{0} [\begin{matrix} - 3 & - 1 6 & 2 \end{matrix}]

$\frac{1}{0} [\begin{array}{c} - 3 & - 1 \\ 6 & 2 \end{array}]$

Multiply

\frac{1}{0}

$\frac{1}{0}$ by each element of the matrix.

[\begin{matrix} \frac{1}{0} \cdot - 3 & \frac{1}{0} \cdot - 1 \frac{1}{0} \cdot 6 & \frac{1}{0} \cdot 2 \end{matrix}]

$[\begin{array}{c} \frac{1}{0} \cdot - 3 & \frac{1}{0} \cdot - 1 \\ \frac{1}{0} \cdot 6 & \frac{1}{0} \cdot 2 \end{array}]$

Rearrange

\frac{1}{0} \cdot - 3

$\frac{1}{0} \cdot - 3$ .

[\begin{matrix} Undefined & \frac{1}{0} \cdot - 1 \frac{1}{0} \cdot 6 & \frac{1}{0} \cdot 2 \end{matrix}]

$[\begin{array}{c} Undefined & \frac{1}{0} \cdot - 1 \\ \frac{1}{0} \cdot 6 & \frac{1}{0} \cdot 2 \end{array}]$

Since the matrix is undefined, it cannot be solved.

Undefined

$Undefined$

Undefined