Linear Algebra Examples

x + 3 y = 10

$x + 3 y = 10$ ,

3 x + 9 y = 16

$3 x + 9 y = 16$

Step 1

Find the

A X = B

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

[\begin{matrix} 1 & 3 3 & 9 \end{matrix}] \cdot [\begin{matrix} x y \end{matrix}] = [\begin{matrix} 1016 \end{matrix}]

$[\begin{array}{c} 1 & 3 \\ 3 & 9 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 10 \\ 16 \end{array}]$

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 & 3 3 & 9 \end{matrix}]

$[\begin{array}{c} 1 & 3 \\ 3 & 9 \end{array}]$ .

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

determinant [\begin{matrix} 1 & 3 3 & 9 \end{matrix}] = ∣ ∣ ∣ \begin{matrix} 1 & 3 3 & 9 \end{matrix} ∣ ∣ ∣

$determinant [\begin{array}{c} 1 & 3 \\ 3 & 9 \end{array}] = | \begin{array}{c} 1 & 3 \\ 3 & 9 \end{array} |$

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) (9) - 3 \cdot 3

$(1) (9) - 3 \cdot 3$

Simplify the determinant.

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Simplify each term.

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Multiply

9

$9$ by

1

$1$ .

9 - 3 \cdot 3

$9 - 3 \cdot 3$

Multiply

- 3

$- 3$ by

3

$3$ .

9 - 9

$9 - 9$

9 - 9

$9 - 9$

Subtract

9

$9$ from

9

$9$ .

0

$0$

0

$0$

0

$0$

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

\frac{1}{0} [\begin{matrix} 9 & - (3) - (3) & 1 \end{matrix}]

$\frac{1}{0} [\begin{array}{c} 9 & - (3) \\ - (3) & 1 \end{array}]$

Simplify each element in the matrix.

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Rearrange

- (3)

$- (3)$ .

\frac{1}{0} [\begin{matrix} 9 & - 3 - (3) & 1 \end{matrix}]

$\frac{1}{0} [\begin{array}{c} 9 & - 3 \\ - (3) & 1 \end{array}]$

Rearrange

- (3)

$- (3)$ .

\frac{1}{0} [\begin{matrix} 9 & - 3 - 3 & 1 \end{matrix}]

$\frac{1}{0} [\begin{array}{c} 9 & - 3 \\ - 3 & 1 \end{array}]$

Multiply

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

$[\begin{array}{c} \frac{1}{0} \cdot 9 & \frac{1}{0} \cdot - 3 \\ \frac{1}{0} \cdot - 3 & \frac{1}{0} \cdot 1 \end{array}]$

Rearrange

$\frac{1}{0} \cdot 9$ .

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

Since the matrix is undefined, it cannot be solved.

$Undefined$

Undefined