Linear Algebra Examples

x - y = 7

$x - y = 7$ ,

x - y = 5

$x - y = 5$

Step 1

Find the

A X = B

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

[\begin{array}{c} 1 & - 1 \\ 1 & - 1 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 7 \\ 5 \end{array}]

$[\begin{array}{c} 1 & - 1 \\ 1 & - 1 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 7 \\ 5 \end{array}]$

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{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} 1 & - 1 \\ 1 & - 1 \end{array}]

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

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

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

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

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

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

(1) (- 1) - 1 \cdot - 1

$(1) (- 1) - 1 \cdot - 1$

Simplify the determinant.

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Simplify each term.

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Multiply

- 1

$- 1$ by

1

$1$ .

- 1 - 1 \cdot - 1

$- 1 - 1 \cdot - 1$

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

- 1 + 1

$- 1 + 1$

- 1 + 1

$- 1 + 1$

Add

- 1

$- 1$ and

1

$1$ .

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} - 1 & - (- 1) \\ - (1) & 1 \end{array}]

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

Simplify each element in the matrix.

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Rearrange

- (- 1)

$- (- 1)$ .

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

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

Rearrange

- (1)

$- (1)$ .

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

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

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

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

Multiply

\frac{1}{0}

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

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

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

Rearrange

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

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

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

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

Since the matrix is undefined, it cannot be solved.

Undefined

$Undefined$

Undefined