Linear Algebra Examples

2 x - y = 9

$2 x - y = 9$ ,

6 x - 3 y = 9

$6 x - 3 y = 9$

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} 99 \end{matrix}]

$[\begin{array}{c} 2 & - 1 \\ 6 & - 3 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 9 \\ 9 \end{array}]$

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)

$- (6)$ .

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

Multiply

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

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

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

Since the matrix is undefined, it cannot be solved.

$Undefined$

Undefined