Linear Algebra Examples

x - 2 y = 3

$x - 2 y = 3$ ,

- 2 x + 4 y = 6

$- 2 x + 4 y = 6$

Step 1

Find the

A X = B

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

[\begin{matrix} 1 & - 2 - 2 & 4 \end{matrix}] \cdot [\begin{matrix} x y \end{matrix}] = [\begin{matrix} 36 \end{matrix}]

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

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} 1 & - 2 - 2 & 4 \end{matrix}]

$[\begin{array}{c} 1 & - 2 \\ - 2 & 4 \end{array}]$ .

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

determinant [\begin{matrix} 1 & - 2 - 2 & 4 \end{matrix}] = ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 2 & 4 \end{matrix} ∣ ∣ ∣

$determinant [\begin{array}{c} 1 & - 2 \\ - 2 & 4 \end{array}] = | \begin{array}{c} 1 & - 2 \\ - 2 & 4 \end{array} |$

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

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

(1) (4) + 2 \cdot - 2

$(1) (4) + 2 \cdot - 2$

Simplify the determinant.

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Simplify each term.

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Multiply

4

$4$ by

1

$1$ .

4 + 2 \cdot - 2

$4 + 2 \cdot - 2$

Multiply

2

$2$ by

- 2

$- 2$ .

4 - 4

$4 - 4$

4 - 4

$4 - 4$

Subtract

4

$4$ from

4

$4$ .

0

$0$

0

$0$

0

$0$

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

\frac{1}{0} [\begin{matrix} 4 & - (- 2) - (- 2) & 1 \end{matrix}]

$\frac{1}{0} [\begin{array}{c} 4 & - (- 2) \\ - (- 2) & 1 \end{array}]$

Simplify each element in the matrix.

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Rearrange

- (- 2)

$- (- 2)$ .

\frac{1}{0} [\begin{matrix} 4 & 2 - (- 2) & 1 \end{matrix}]

$\frac{1}{0} [\begin{array}{c} 4 & 2 \\ - (- 2) & 1 \end{array}]$

Rearrange

- (- 2)

$- (- 2)$ .

\frac{1}{0} [\begin{matrix} 4 & 2 2 & 1 \end{matrix}]

$\frac{1}{0} [\begin{array}{c} 4 & 2 \\ 2 & 1 \end{array}]$

\frac{1}{0} [\begin{matrix} 4 & 2 2 & 1 \end{matrix}]

$\frac{1}{0} [\begin{array}{c} 4 & 2 \\ 2 & 1 \end{array}]$

Multiply

\frac{1}{0}

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

[\begin{matrix} \frac{1}{0} \cdot 4 & \frac{1}{0} \cdot 2 \frac{1}{0} \cdot 2 & \frac{1}{0} \cdot 1 \end{matrix}]

$[\begin{array}{c} \frac{1}{0} \cdot 4 & \frac{1}{0} \cdot 2 \\ \frac{1}{0} \cdot 2 & \frac{1}{0} \cdot 1 \end{array}]$

Rearrange

\frac{1}{0} \cdot 4

$\frac{1}{0} \cdot 4$ .

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

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

Since the matrix is undefined, it cannot be solved.

Undefined

$Undefined$

Undefined