Precalculus Examples
[1-123][1−123]
Step 1
Write the matrix as a product of a lower triangular matrix and an upper triangular matrix.
[10l211][u11u120u22]=[1-123][10l211][u11u120u22]=[1−123]
Step 2
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 2×22×2 and the second matrix is 2×22×2.
Step 2.2
Multiply each row in the first matrix by each column in the second matrix.
[1u11+0⋅01u12+0u22l21u11+1⋅0l21u12+1u22]=[1-123][1u11+0⋅01u12+0u22l21u11+1⋅0l21u12+1u22]=[1−123]
Step 2.3
Simplify each element of the matrix by multiplying out all the expressions.
[u11u12l21u11l21u12+u22]=[1-123][u11u12l21u11l21u12+u22]=[1−123]
[u11u12l21u11l21u12+u22]=[1-123][u11u12l21u11l21u12+u22]=[1−123]
Step 3
Step 3.1
Write as a linear system of equations.
u11=1u11=1
u12=-1u12=−1
l21u11=2l21u11=2
l21u12+u22=3l21u12+u22=3
Step 3.2
Solve the system of equations.
Step 3.2.1
Replace all occurrences of u11u11 with 11 in each equation.
Step 3.2.1.1
Replace all occurrences of u11u11 in l21u11=2l21u11=2 with 11.
l21⋅1=2l21⋅1=2
u11=1u11=1
u12=-1u12=−1
l21u12+u22=3l21u12+u22=3
Step 3.2.1.2
Simplify the left side.
Step 3.2.1.2.1
Multiply l21l21 by 11.
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
l21u12+u22=3l21u12+u22=3
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
l21u12+u22=3l21u12+u22=3
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
l21u12+u22=3l21u12+u22=3
Step 3.2.2
Replace all occurrences of l21l21 with 22 in each equation.
Step 3.2.2.1
Replace all occurrences of l21l21 in l21u12+u22=3l21u12+u22=3 with 22.
2⋅u12+u22=32⋅u12+u22=3
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
Step 3.2.2.2
Simplify the left side.
Step 3.2.2.2.1
Multiply 22 by u12u12.
2u12+u22=32u12+u22=3
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
2u12+u22=32u12+u22=3
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
2u12+u22=32u12+u22=3
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
Step 3.2.3
Replace all occurrences of u12u12 with -1−1 in each equation.
Step 3.2.3.1
Replace all occurrences of u12u12 in 2u12+u22=32u12+u22=3 with -1−1.
2(-1)+u22=32(−1)+u22=3
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
Step 3.2.3.2
Simplify the left side.
Step 3.2.3.2.1
Multiply 22 by -1−1.
-2+u22=3−2+u22=3
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
-2+u22=3−2+u22=3
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
-2+u22=3−2+u22=3
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
Step 3.2.4
Move all terms not containing u22u22 to the right side of the equation.
Step 3.2.4.1
Add 22 to both sides of the equation.
u22=3+2u22=3+2
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
Step 3.2.4.2
Add 33 and 22.
u22=5u22=5
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
u22=5u22=5
l21=2l21=2
u11=1u11=1
u12=-1u12=−1
Step 3.2.5
Solve the system of equations.
u22=5l21=2u11=1u12=-1
Step 3.2.6
List all of the solutions.
u22=5,l21=2,u11=1,u12=-1
u22=5,l21=2,u11=1,u12=-1
u22=5,l21=2,u11=1,u12=-1
Step 4
Substitute in the solved values.
[1-123]=[1021][1-105]