Examples

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

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

Step 1

Write the matrix as a product of a lower triangular matrix and an upper triangular matrix.

[\begin{array}{c} 1 & 0 \\ l_{21} & 1 \end{array}] [\begin{array}{c} u_{11} & u_{12} \\ 0 & u_{22} \end{array}] = [\begin{array}{c} 1 & - 1 \\ 2 & 3 \end{array}]

$[\begin{array}{c} 1 & 0 \\ l_{21} & 1 \end{array}] [\begin{array}{c} u_{11} & u_{12} \\ 0 & u_{22} \end{array}] = [\begin{array}{c} 1 & - 1 \\ 2 & 3 \end{array}]$

Step 2

Multiply

[\begin{array}{c} 1 & 0 \\ l_{21} & 1 \end{array}] [\begin{array}{c} u_{11} & u_{12} \\ 0 & u_{22} \end{array}]

$[\begin{array}{c} 1 & 0 \\ l_{21} & 1 \end{array}] [\begin{array}{c} u_{11} & u_{12} \\ 0 & u_{22} \end{array}]$ .

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Step 2.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is

2 \times 2

$2 \times 2$ and the second matrix is

2 \times 2

$2 \times 2$ .

Step 2.2

Multiply each row in the first matrix by each column in the second matrix.

[\begin{array}{c} 1 u_{11} + 0 \cdot 0 & 1 u_{12} + 0 u_{22} \\ l_{21} u_{11} + 1 \cdot 0 & l_{21} u_{12} + 1 u_{22} \end{array}] = [\begin{array}{c} 1 & - 1 \\ 2 & 3 \end{array}]

$[\begin{array}{c} 1 u_{11} + 0 \cdot 0 & 1 u_{12} + 0 u_{22} \\ l_{21} u_{11} + 1 \cdot 0 & l_{21} u_{12} + 1 u_{22} \end{array}] = [\begin{array}{c} 1 & - 1 \\ 2 & 3 \end{array}]$

Step 2.3

Simplify each element of the matrix by multiplying out all the expressions.

[\begin{array}{c} u_{11} & u_{12} \\ l_{21} u_{11} & l_{21} u_{12} + u_{22} \end{array}] = [\begin{array}{c} 1 & - 1 \\ 2 & 3 \end{array}]

$[\begin{array}{c} u_{11} & u_{12} \\ l_{21} u_{11} & l_{21} u_{12} + u_{22} \end{array}] = [\begin{array}{c} 1 & - 1 \\ 2 & 3 \end{array}]$

[\begin{array}{c} u_{11} & u_{12} \\ l_{21} u_{11} & l_{21} u_{12} + u_{22} \end{array}] = [\begin{array}{c} 1 & - 1 \\ 2 & 3 \end{array}]

$[\begin{array}{c} u_{11} & u_{12} \\ l_{21} u_{11} & l_{21} u_{12} + u_{22} \end{array}] = [\begin{array}{c} 1 & - 1 \\ 2 & 3 \end{array}]$

Step 3

Solve.

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Step 3.1

Write as a linear system of equations.

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

l_{21} u_{11} = 2

$l_{21} u_{11} = 2$

l_{21} u_{12} + u_{22} = 3

$l_{21} u_{12} + u_{22} = 3$

Step 3.2

Solve the system of equations.

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Step 3.2.1

Replace all occurrences of

u_{11}

$u_{11}$ with

1

$1$ in each equation.

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Step 3.2.1.1

Replace all occurrences of

u_{11}

$u_{11}$ in

l_{21} u_{11} = 2

$l_{21} u_{11} = 2$ with

1

$1$ .

l_{21} \cdot 1 = 2

$l_{21} \cdot 1 = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

l_{21} u_{12} + u_{22} = 3

$l_{21} u_{12} + u_{22} = 3$

Step 3.2.1.2

Simplify the left side.

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Step 3.2.1.2.1

Multiply

l_{21}

$l_{21}$ by

1

$1$ .

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

l_{21} u_{12} + u_{22} = 3

$l_{21} u_{12} + u_{22} = 3$

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

l_{21} u_{12} + u_{22} = 3

$l_{21} u_{12} + u_{22} = 3$

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

l_{21} u_{12} + u_{22} = 3

$l_{21} u_{12} + u_{22} = 3$

Step 3.2.2

Replace all occurrences of

l_{21}

$l_{21}$ with

2

$2$ in each equation.

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Step 3.2.2.1

Replace all occurrences of

l_{21}

$l_{21}$ in

l_{21} u_{12} + u_{22} = 3

$l_{21} u_{12} + u_{22} = 3$ with

2

$2$ .

2 \cdot u_{12} + u_{22} = 3

$2 \cdot u_{12} + u_{22} = 3$

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

Step 3.2.2.2

Simplify the left side.

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Step 3.2.2.2.1

Multiply

2

$2$ by

u_{12}

$u_{12}$ .

2 u_{12} + u_{22} = 3

$2 u_{12} + u_{22} = 3$

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

2 u_{12} + u_{22} = 3

$2 u_{12} + u_{22} = 3$

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

2 u_{12} + u_{22} = 3

$2 u_{12} + u_{22} = 3$

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

Step 3.2.3

Replace all occurrences of

u_{12}

$u_{12}$ with

- 1

$- 1$ in each equation.

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Step 3.2.3.1

Replace all occurrences of

u_{12}

$u_{12}$ in

2 u_{12} + u_{22} = 3

$2 u_{12} + u_{22} = 3$ with

- 1

$- 1$ .

2 (- 1) + u_{22} = 3

$2 (- 1) + u_{22} = 3$

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

Step 3.2.3.2

Simplify the left side.

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Step 3.2.3.2.1

Multiply

2

$2$ by

- 1

$- 1$ .

- 2 + u_{22} = 3

$- 2 + u_{22} = 3$

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

- 2 + u_{22} = 3

$- 2 + u_{22} = 3$

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

- 2 + u_{22} = 3

$- 2 + u_{22} = 3$

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

Step 3.2.4

Move all terms not containing

u_{22}

$u_{22}$ to the right side of the equation.

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Step 3.2.4.1

Add

2

$2$ to both sides of the equation.

u_{22} = 3 + 2

$u_{22} = 3 + 2$

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

Step 3.2.4.2

Add

3

$3$ and

2

$2$ .

u_{22} = 5

$u_{22} = 5$

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

u_{22} = 5

$u_{22} = 5$

l_{21} = 2

$l_{21} = 2$

u_{11} = 1

$u_{11} = 1$

u_{12} = - 1

$u_{12} = - 1$

Step 3.2.5

Solve the system of equations.

u_{22} = 5 l_{21} = 2 u_{11} = 1 u_{12} = - 1

$u_{22} = 5 l_{21} = 2 u_{11} = 1 u_{12} = - 1$

Step 3.2.6

List all of the solutions.

u_{22} = 5, l_{21} = 2, u_{11} = 1, u_{12} = - 1

$u_{22} = 5, l_{21} = 2, u_{11} = 1, u_{12} = - 1$

u_{22} = 5, l_{21} = 2, u_{11} = 1, u_{12} = - 1

$u_{22} = 5, l_{21} = 2, u_{11} = 1, u_{12} = - 1$

u_{22} = 5, l_{21} = 2, u_{11} = 1, u_{12} = - 1

$u_{22} = 5, l_{21} = 2, u_{11} = 1, u_{12} = - 1$

Step 4

Substitute in the solved values.

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

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

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