Linear Algebra Examples

x + y + z = 12

$x + y + z = 12$ ,

2 x - 3 y + 2 z = 4

$2 x - 3 y + 2 z = 4$ ,

x + z = 2 y

$x + z = 2 y$

Step 1

Find the

A X = B

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 2 & - 3 & 2 1 & - 2 & 1 \end{matrix} ⎤ ⎥ ⎦ \cdot ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1240 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 1 \\ 2 & - 3 & 2 \\ 1 & - 2 & 1 \end{array}] \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 12 \\ 4 \\ 0 \end{array}]$

Step 2

Find the inverse of the coefficient matrix.

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Set up a matrix that is broken into two pieces of equal size. On the left side, fill in the elements of the original matrix. On the right side, fill in elements of the identity matrix. In order to find the inverse matrix, use row operations to convert the left side into the identity matrix. After this is complete, the inverse of the original matrix will be on the right side of the double matrix.

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 2 & - 3 & 2 & 0 & 1 & 0 1 & - 2 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

Perform the row operation

R_{2} = - 2 \cdot R_{1} + R_{2}

$R_{2} = - 2 \cdot R_{1} + R_{2}$ on

R_{2}

$R_{2}$ (row

2

$2$ ) in order to convert some elements in the row to

0

$0$ .

Tap for more steps...

Replace

R_{2}

$R_{2}$ (row

2

$2$ ) with the row operation

R_{2} = - 2 \cdot R_{1} + R_{2}

$R_{2} = - 2 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} 1 & - 2 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

R_{2} = - 2 \cdot R_{1} + R_{2}

$R_{2} = - 2 \cdot R_{1} + R_{2}$

Replace

R_{2}

$R_{2}$ (row

2

$2$ ) with the actual values of the elements for the row operation

R_{2} = - 2 \cdot R_{1} + R_{2}

$R_{2} = - 2 \cdot R_{1} + R_{2}$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 (- 2) \cdot (1) + 2 & (- 2) \cdot (1) - 3 & (- 2) \cdot (1) + 2 & (- 2) \cdot (1) + 0 & (- 2) \cdot (0) + 1 & (- 2) \cdot (0) + 0 1 & - 2 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ (- 2) \cdot (1) + 2 & (- 2) \cdot (1) - 3 & (- 2) \cdot (1) + 2 & (- 2) \cdot (1) + 0 & (- 2) \cdot (0) + 1 & (- 2) \cdot (0) + 0 \\ 1 & - 2 & 1 & 0 & 0 & 1 \end{array}]$

R_{2} = - 2 \cdot R_{1} + R_{2}

$R_{2} = - 2 \cdot R_{1} + R_{2}$

Simplify

R_{2}

$R_{2}$ (row

2

$2$ ).

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 0 & - 5 & 0 & - 2 & 1 & 0 1 & - 2 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & - 5 & 0 & - 2 & 1 & 0 \\ 1 & - 2 & 1 & 0 & 0 & 1 \end{array}]$

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 0 & - 5 & 0 & - 2 & 1 & 0 1 & - 2 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & - 5 & 0 & - 2 & 1 & 0 \\ 1 & - 2 & 1 & 0 & 0 & 1 \end{array}]$

Perform the row operation

R_{3} = - 1 \cdot R_{1} + R_{3}

$R_{3} = - 1 \cdot R_{1} + R_{3}$ on

R_{3}

$R_{3}$ (row

3

$3$ ) in order to convert some elements in the row to

0

$0$ .

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Replace

R_{3}

$R_{3}$ (row

3

$3$ ) with the row operation

R_{3} = - 1 \cdot R_{1} + R_{3}

$R_{3} = - 1 \cdot R_{1} + R_{3}$ in order to convert some elements in the row to the desired value

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 0 & - 5 & 0 & - 2 & 1 & 0 - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & - 5 & 0 & - 2 & 1 & 0 \\ - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} \end{array}]$

R_{3} = - 1 \cdot R_{1} + R_{3}

$R_{3} = - 1 \cdot R_{1} + R_{3}$

Replace

R_{3}

$R_{3}$ (row

3

$3$ ) with the actual values of the elements for the row operation

R_{3} = - 1 \cdot R_{1} + R_{3}

$R_{3} = - 1 \cdot R_{1} + R_{3}$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 0 & - 5 & 0 & - 2 & 1 & 0 (- 1) \cdot (1) + 1 & (- 1) \cdot (1) - 2 & (- 1) \cdot (1) + 1 & (- 1) \cdot (1) + 0 & (- 1) \cdot (0) + 0 & (- 1) \cdot (0) + 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & - 5 & 0 & - 2 & 1 & 0 \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (1) - 2 & (- 1) \cdot (1) + 1 & (- 1) \cdot (1) + 0 & (- 1) \cdot (0) + 0 & (- 1) \cdot (0) + 1 \end{array}]$

R_{3} = - 1 \cdot R_{1} + R_{3}

$R_{3} = - 1 \cdot R_{1} + R_{3}$

Simplify

R_{3}

$R_{3}$ (row

3

$3$ ).

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 0 & - 5 & 0 & - 2 & 1 & 0 0 & - 3 & 0 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & - 5 & 0 & - 2 & 1 & 0 \\ 0 & - 3 & 0 & - 1 & 0 & 1 \end{array}]$

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 0 & - 5 & 0 & - 2 & 1 & 0 0 & - 3 & 0 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & - 5 & 0 & - 2 & 1 & 0 \\ 0 & - 3 & 0 & - 1 & 0 & 1 \end{array}]$

Perform the row operation

R_{2} = - \frac{1}{5} R_{2}

$R_{2} = - \frac{1}{5} R_{2}$ on

R_{2}

$R_{2}$ (row

2

$2$ ) in order to convert some elements in the row to

1

$1$ .

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Replace

R_{2}

$R_{2}$ (row

2

$2$ ) with the row operation

R_{2} = - \frac{1}{5} R_{2}

$R_{2} = - \frac{1}{5} R_{2}$ in order to convert some elements in the row to the desired value

1

$1$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} 0 & - 3 & 0 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} \\ 0 & - 3 & 0 & - 1 & 0 & 1 \end{array}]$

R_{2} = - \frac{1}{5} R_{2}

$R_{2} = - \frac{1}{5} R_{2}$

Replace

R_{2}

$R_{2}$ (row

2

$2$ ) with the actual values of the elements for the row operation

R_{2} = - \frac{1}{5} R_{2}

$R_{2} = - \frac{1}{5} R_{2}$ .

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 (- \frac{1}{5}) \cdot (0) & (- \frac{1}{5}) \cdot (- 5) & (- \frac{1}{5}) \cdot (0) & (- \frac{1}{5}) \cdot (- 2) & (- \frac{1}{5}) \cdot (1) & (- \frac{1}{5}) \cdot (0) 0 & - 3 & 0 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ (- \frac{1}{5}) \cdot (0) & (- \frac{1}{5}) \cdot (- 5) & (- \frac{1}{5}) \cdot (0) & (- \frac{1}{5}) \cdot (- 2) & (- \frac{1}{5}) \cdot (1) & (- \frac{1}{5}) \cdot (0) \\ 0 & - 3 & 0 & - 1 & 0 & 1 \end{array}]$

R_{2} = - \frac{1}{5} R_{2}

$R_{2} = - \frac{1}{5} R_{2}$

Simplify

R_{2}

$R_{2}$ (row

2

$2$ ).

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & - 3 & 0 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & - 3 & 0 & - 1 & 0 & 1 \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & - 3 & 0 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & - 3 & 0 & - 1 & 0 & 1 \end{array}]$

Perform the row operation

R_{1} = - 1 \cdot R_{2} + R_{1}

$R_{1} = - 1 \cdot R_{2} + R_{1}$ on

R_{1}

$R_{1}$ (row

1

$1$ ) in order to convert some elements in the row to

0

$0$ .

Tap for more steps...

Replace

R_{1}

$R_{1}$ (row

1

$1$ ) with the row operation

R_{1} = - 1 \cdot R_{2} + R_{1}

$R_{1} = - 1 \cdot R_{2} + R_{1}$ in order to convert some elements in the row to the desired value

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & - 3 & 0 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & - 3 & 0 & - 1 & 0 & 1 \end{array}]$

R_{1} = - 1 \cdot R_{2} + R_{1}

$R_{1} = - 1 \cdot R_{2} + R_{1}$

Replace

R_{1}

$R_{1}$ (row

1

$1$ ) with the actual values of the elements for the row operation

R_{1} = - 1 \cdot R_{2} + R_{1}

$R_{1} = - 1 \cdot R_{2} + R_{1}$ .

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (- 1) \cdot (0) + 1 & (- 1) \cdot (1) + 1 & (- 1) \cdot (0) + 1 & (- 1) \cdot (\frac{2}{5}) + 1 & (- 1) \cdot (- \frac{1}{5}) + 0 & (- 1) \cdot (0) + 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & - 3 & 0 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} (- 1) \cdot (0) + 1 & (- 1) \cdot (1) + 1 & (- 1) \cdot (0) + 1 & (- 1) \cdot (\frac{2}{5}) + 1 & (- 1) \cdot (- \frac{1}{5}) + 0 & (- 1) \cdot (0) + 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & - 3 & 0 & - 1 & 0 & 1 \end{array}]$

R_{1} = - 1 \cdot R_{2} + R_{1}

$R_{1} = - 1 \cdot R_{2} + R_{1}$

Simplify

R_{1}

$R_{1}$ (row

1

$1$ ).

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & - 3 & 0 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & - 3 & 0 & - 1 & 0 & 1 \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & - 3 & 0 & - 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & - 3 & 0 & - 1 & 0 & 1 \end{array}]$

Perform the row operation

R_{3} = 3 \cdot R_{2} + R_{3}

$R_{3} = 3 \cdot R_{2} + R_{3}$ on

R_{3}

$R_{3}$ (row

3

$3$ ) in order to convert some elements in the row to

0

$0$ .

Tap for more steps...

Replace

R_{3}

$R_{3}$ (row

3

$3$ ) with the row operation

R_{3} = 3 \cdot R_{2} + R_{3}

$R_{3} = 3 \cdot R_{2} + R_{3}$ in order to convert some elements in the row to the desired value

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 3 \cdot R_{2} + R_{3} & 3 \cdot R_{2} + R_{3} & 3 \cdot R_{2} + R_{3} & 3 \cdot R_{2} + R_{3} & 3 \cdot R_{2} + R_{3} & 3 \cdot R_{2} + R_{3} \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 3 \cdot R_{2} + R_{3} & 3 \cdot R_{2} + R_{3} & 3 \cdot R_{2} + R_{3} & 3 \cdot R_{2} + R_{3} & 3 \cdot R_{2} + R_{3} & 3 \cdot R_{2} + R_{3} \end{array}]$

R_{3} = 3 \cdot R_{2} + R_{3}

$R_{3} = 3 \cdot R_{2} + R_{3}$

Replace

R_{3}

$R_{3}$ (row

3

$3$ ) with the actual values of the elements for the row operation

R_{3} = 3 \cdot R_{2} + R_{3}

$R_{3} = 3 \cdot R_{2} + R_{3}$ .

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 (3) \cdot (0) + 0 & (3) \cdot (1) - 3 & (3) \cdot (0) + 0 & (3) \cdot (\frac{2}{5}) - 1 & (3) \cdot (- \frac{1}{5}) + 0 & (3) \cdot (0) + 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ (3) \cdot (0) + 0 & (3) \cdot (1) - 3 & (3) \cdot (0) + 0 & (3) \cdot (\frac{2}{5}) - 1 & (3) \cdot (- \frac{1}{5}) + 0 & (3) \cdot (0) + 1 \end{array}]$

R_{3} = 3 \cdot R_{2} + R_{3}

$R_{3} = 3 \cdot R_{2} + R_{3}$

Simplify

R_{3}

$R_{3}$ (row

3

$3$ ).

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & 0 & 0 & \frac{1}{5} & - \frac{3}{5} & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & \frac{1}{5} & - \frac{3}{5} & 1 \end{array}]$

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & 0 & 0 & \frac{1}{5} & - \frac{3}{5} & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & \frac{1}{5} & - \frac{3}{5} & 1 \end{array}]$

Perform the row operation

R_{3} = 5 \cdot R_{3}

$R_{3} = 5 \cdot R_{3}$ on

R_{3}

$R_{3}$ (row

3

$3$ ) in order to convert some elements in the row to

1

$1$ .

Tap for more steps...

Replace

R_{3}

$R_{3}$ (row

3

$3$ ) with the row operation

R_{3} = 5 \cdot R_{3}

$R_{3} = 5 \cdot R_{3}$ in order to convert some elements in the row to the desired value

1

$1$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 5 \cdot R_{3} & 5 \cdot R_{3} & 5 \cdot R_{3} & 5 \cdot R_{3} & 5 \cdot R_{3} & 5 \cdot R_{3} \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 5 \cdot R_{3} & 5 \cdot R_{3} & 5 \cdot R_{3} & 5 \cdot R_{3} & 5 \cdot R_{3} & 5 \cdot R_{3} \end{array}]$

R_{3} = 5 \cdot R_{3}

$R_{3} = 5 \cdot R_{3}$

Replace

R_{3}

$R_{3}$ (row

3

$3$ ) with the actual values of the elements for the row operation

R_{3} = 5 \cdot R_{3}

$R_{3} = 5 \cdot R_{3}$ .

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 (5) \cdot (0) & (5) \cdot (0) & (5) \cdot (0) & (5) \cdot (\frac{1}{5}) & (5) \cdot (- \frac{3}{5}) & (5) \cdot (1) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ (5) \cdot (0) & (5) \cdot (0) & (5) \cdot (0) & (5) \cdot (\frac{1}{5}) & (5) \cdot (- \frac{3}{5}) & (5) \cdot (1) \end{array}]$

R_{3} = 5 \cdot R_{3}

$R_{3} = 5 \cdot R_{3}$

Simplify

R_{3}

$R_{3}$ (row

3

$3$ ).

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & 0 & 0 & 1 & - 3 & 5 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & - 3 & 5 \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & 0 & 0 & 1 & - 3 & 5 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & \frac{3}{5} & \frac{1}{5} & 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & - 3 & 5 \end{array}]$

Perform the row operation

R_{1} = - \frac{3}{5} R_{3} + R_{1}

$R_{1} = - \frac{3}{5} R_{3} + R_{1}$ on

R_{1}

$R_{1}$ (row

1

$1$ ) in order to convert some elements in the row to

0

$0$ .

Tap for more steps...

Replace

R_{1}

$R_{1}$ (row

1

$1$ ) with the row operation

R_{1} = - \frac{3}{5} R_{3} + R_{1}

$R_{1} = - \frac{3}{5} R_{3} + R_{1}$ in order to convert some elements in the row to the desired value

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} - \frac{3}{5} R_{3} + R_{1} & - \frac{3}{5} R_{3} + R_{1} & - \frac{3}{5} R_{3} + R_{1} & - \frac{3}{5} R_{3} + R_{1} & - \frac{3}{5} R_{3} + R_{1} & - \frac{3}{5} R_{3} + R_{1} 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & 0 & 0 & 1 & - 3 & 5 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} - \frac{3}{5} R_{3} + R_{1} & - \frac{3}{5} R_{3} + R_{1} & - \frac{3}{5} R_{3} + R_{1} & - \frac{3}{5} R_{3} + R_{1} & - \frac{3}{5} R_{3} + R_{1} & - \frac{3}{5} R_{3} + R_{1} \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & - 3 & 5 \end{array}]$

R_{1} = - \frac{3}{5} R_{3} + R_{1}

$R_{1} = - \frac{3}{5} R_{3} + R_{1}$

Replace

R_{1}

$R_{1}$ (row

1

$1$ ) with the actual values of the elements for the row operation

R_{1} = - \frac{3}{5} R_{3} + R_{1}

$R_{1} = - \frac{3}{5} R_{3} + R_{1}$ .

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (- \frac{3}{5}) \cdot (0) + 1 & (- \frac{3}{5}) \cdot (0) + 0 & (- \frac{3}{5}) \cdot (0) + 1 & (- \frac{3}{5}) \cdot (1) + \frac{3}{5} & (- \frac{3}{5}) \cdot (- 3) + \frac{1}{5} & (- \frac{3}{5}) \cdot (5) + 0 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & 0 & 0 & 1 & - 3 & 5 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} (- \frac{3}{5}) \cdot (0) + 1 & (- \frac{3}{5}) \cdot (0) + 0 & (- \frac{3}{5}) \cdot (0) + 1 & (- \frac{3}{5}) \cdot (1) + \frac{3}{5} & (- \frac{3}{5}) \cdot (- 3) + \frac{1}{5} & (- \frac{3}{5}) \cdot (5) + 0 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & - 3 & 5 \end{array}]$

R_{1} = - \frac{3}{5} R_{3} + R_{1}

$R_{1} = - \frac{3}{5} R_{3} + R_{1}$

Simplify

R_{1}

$R_{1}$ (row

1

$1$ ).

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 0 & 2 & - 3 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & 0 & 0 & 1 & - 3 & 5 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & 0 & 2 & - 3 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & - 3 & 5 \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 0 & 2 & - 3 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 0 & 0 & 0 & 1 & - 3 & 5 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & 0 & 2 & - 3 \\ 0 & 1 & 0 & \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & - 3 & 5 \end{array}]$

Perform the row operation

R_{2} = - \frac{2}{5} R_{3} + R_{2}

$R_{2} = - \frac{2}{5} R_{3} + R_{2}$ on

R_{2}

$R_{2}$ (row

2

$2$ ) in order to convert some elements in the row to

0

$0$ .

Tap for more steps...

Replace

R_{2}

$R_{2}$ (row

2

$2$ ) with the row operation

R_{2} = - \frac{2}{5} R_{3} + R_{2}

$R_{2} = - \frac{2}{5} R_{3} + R_{2}$ in order to convert some elements in the row to the desired value

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 0 & 2 & - 3 - \frac{2}{5} R_{3} + R_{2} & - \frac{2}{5} R_{3} + R_{2} & - \frac{2}{5} R_{3} + R_{2} & - \frac{2}{5} R_{3} + R_{2} & - \frac{2}{5} R_{3} + R_{2} & - \frac{2}{5} R_{3} + R_{2} 0 & 0 & 0 & 1 & - 3 & 5 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & 0 & 2 & - 3 \\ - \frac{2}{5} R_{3} + R_{2} & - \frac{2}{5} R_{3} + R_{2} & - \frac{2}{5} R_{3} + R_{2} & - \frac{2}{5} R_{3} + R_{2} & - \frac{2}{5} R_{3} + R_{2} & - \frac{2}{5} R_{3} + R_{2} \\ 0 & 0 & 0 & 1 & - 3 & 5 \end{array}]$

R_{2} = - \frac{2}{5} R_{3} + R_{2}

$R_{2} = - \frac{2}{5} R_{3} + R_{2}$

Replace

R_{2}

$R_{2}$ (row

2

$2$ ) with the actual values of the elements for the row operation

R_{2} = - \frac{2}{5} R_{3} + R_{2}

$R_{2} = - \frac{2}{5} R_{3} + R_{2}$ .

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 0 & 2 & - 3 (- \frac{2}{5}) \cdot (0) + 0 & (- \frac{2}{5}) \cdot (0) + 1 & (- \frac{2}{5}) \cdot (0) + 0 & (- \frac{2}{5}) \cdot (1) + \frac{2}{5} & (- \frac{2}{5}) \cdot (- 3) - \frac{1}{5} & (- \frac{2}{5}) \cdot (5) + 0 0 & 0 & 0 & 1 & - 3 & 5 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & 0 & 2 & - 3 \\ (- \frac{2}{5}) \cdot (0) + 0 & (- \frac{2}{5}) \cdot (0) + 1 & (- \frac{2}{5}) \cdot (0) + 0 & (- \frac{2}{5}) \cdot (1) + \frac{2}{5} & (- \frac{2}{5}) \cdot (- 3) - \frac{1}{5} & (- \frac{2}{5}) \cdot (5) + 0 \\ 0 & 0 & 0 & 1 & - 3 & 5 \end{array}]$

R_{2} = - \frac{2}{5} R_{3} + R_{2}

$R_{2} = - \frac{2}{5} R_{3} + R_{2}$

Simplify

R_{2}

$R_{2}$ (row

2

$2$ ).

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 0 & 2 & - 3 0 & 1 & 0 & 0 & 1 & - 2 0 & 0 & 0 & 1 & - 3 & 5 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & 0 & 2 & - 3 \\ 0 & 1 & 0 & 0 & 1 & - 2 \\ 0 & 0 & 0 & 1 & - 3 & 5 \end{array}]$

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 0 & 2 & - 3 0 & 1 & 0 & 0 & 1 & - 2 0 & 0 & 0 & 1 & - 3 & 5 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 1 & 0 & 2 & - 3 \\ 0 & 1 & 0 & 0 & 1 & - 2 \\ 0 & 0 & 0 & 1 & - 3 & 5 \end{array}]$

Since the determinant of the matrix is zero, there is no inverse.

No inverse

Step 3

Since the matrix has no inverse, it cannot be solved using the inverse matrix.

No solution