Linear Algebra Examples

2 x + y = 8

$2 x + y = 8$ ,

4 x + 2 y = 24

$4 x + 2 y = 24$

Step 1

Find the

A X = B

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

[\begin{array}{c} 2 & 1 \\ 4 & 2 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 8 \\ 24 \end{array}]

$[\begin{array}{c} 2 & 1 \\ 4 & 2 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 8 \\ 24 \end{array}]$

Step 2

Find the inverse of the coefficient matrix.

Tap for more steps...

The inverse of a

2 \times 2

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

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

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

| A |

$| A |$ is the determinant of

A

$A$ .

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

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

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

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

Find the determinant of

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

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

Tap for more steps...

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

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

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

The determinant of a

2 \times 2

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

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

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

(2) (2) - 4 \cdot 1

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

Simplify the determinant.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply

2

$2$ by

2

$2$ .

4 - 4 \cdot 1

$4 - 4 \cdot 1$

Multiply

- 4

$- 4$ by

1

$1$ .

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{array}{c} 2 & - (1) \\ - (4) & 2 \end{array}]

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

Simplify each element in the matrix.

Tap for more steps...

Rearrange

- (1)

$- (1)$ .

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

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

Rearrange

- (4)

$- (4)$ .

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

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

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

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

Multiply

\frac{1}{0}

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

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

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

Rearrange

\frac{1}{0} \cdot 2

$\frac{1}{0} \cdot 2$ .

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

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

Since the matrix is undefined, it cannot be solved.

Undefined

$Undefined$

Undefined