Precalculus Examples
[-132110110]⎡⎢⎣−132110110⎤⎥⎦
Step 1
Write as an augmented matrix for Ax=0Ax=0.
[-132011001100]⎡⎢
⎢⎣−132011001100⎤⎥
⎥⎦
Step 2
Step 2.1
Multiply each element of R1R1 by -1−1 to make the entry at 1,11,1 a 11.
Step 2.1.1
Multiply each element of R1R1 by -1−1 to make the entry at 1,11,1 a 11.
[--1-1⋅3-1⋅2-011001100]⎡⎢
⎢⎣−−1−1⋅3−1⋅2−011001100⎤⎥
⎥⎦
Step 2.1.2
Simplify R1R1.
[1-3-2011001100]⎡⎢
⎢⎣1−3−2011001100⎤⎥
⎥⎦
[1-3-2011001100]⎡⎢
⎢⎣1−3−2011001100⎤⎥
⎥⎦
Step 2.2
Perform the row operation R2=R2-R1R2=R2−R1 to make the entry at 2,12,1 a 00.
Step 2.2.1
Perform the row operation R2=R2-R1R2=R2−R1 to make the entry at 2,12,1 a 00.
[1-3-201-11+30+20-01100]⎡⎢
⎢⎣1−3−201−11+30+20−01100⎤⎥
⎥⎦
Step 2.2.2
Simplify R2R2.
[1-3-2004201100]⎡⎢
⎢⎣1−3−2004201100⎤⎥
⎥⎦
[1-3-2004201100]⎡⎢
⎢⎣1−3−2004201100⎤⎥
⎥⎦
Step 2.3
Perform the row operation R3=R3-R1R3=R3−R1 to make the entry at 3,13,1 a 00.
Step 2.3.1
Perform the row operation R3=R3-R1R3=R3−R1 to make the entry at 3,13,1 a 00.
[1-3-2004201-11+30+20-0]⎡⎢
⎢⎣1−3−2004201−11+30+20−0⎤⎥
⎥⎦
Step 2.3.2
Simplify R3R3.
[1-3-2004200420]⎡⎢
⎢⎣1−3−2004200420⎤⎥
⎥⎦
[1-3-2004200420]⎡⎢
⎢⎣1−3−2004200420⎤⎥
⎥⎦
Step 2.4
Multiply each element of R2R2 by 1414 to make the entry at 2,22,2 a 11.
Step 2.4.1
Multiply each element of R2R2 by 1414 to make the entry at 2,22,2 a 11.
[1-3-20044424040420]⎡⎢
⎢⎣1−3−20044424040420⎤⎥
⎥⎦
Step 2.4.2
Simplify R2R2.
[1-3-20011200420]⎡⎢
⎢⎣1−3−20011200420⎤⎥
⎥⎦
[1-3-20011200420]⎡⎢
⎢⎣1−3−20011200420⎤⎥
⎥⎦
Step 2.5
Perform the row operation R3=R3-4R2R3=R3−4R2 to make the entry at 3,23,2 a 0.
Step 2.5.1
Perform the row operation R3=R3-4R2 to make the entry at 3,2 a 0.
[1-3-20011200-4⋅04-4⋅12-4(12)0-4⋅0]
Step 2.5.2
Simplify R3.
[1-3-20011200000]
[1-3-20011200000]
Step 2.6
Perform the row operation R1=R1+3R2 to make the entry at 1,2 a 0.
Step 2.6.1
Perform the row operation R1=R1+3R2 to make the entry at 1,2 a 0.
[1+3⋅0-3+3⋅1-2+3(12)0+3⋅0011200000]
Step 2.6.2
Simplify R1.
[10-120011200000]
[10-120011200000]
[10-120011200000]
Step 3
Use the result matrix to declare the final solution to the system of equations.
x-12z=0
y+12z=0
0=0
Step 4
Write a solution vector by solving in terms of the free variables in each row.
[xyz]=[z2-z2z]
Step 5
Write the solution as a linear combination of vectors.
[xyz]=z[12-121]
Step 6
Write as a solution set.
{z[12-121]|z∈R}
Step 7
The solution is the set of vectors created from the free variables of the system.
Basis of Nul(A): {[12-121]}
Dimension of Nul(A): 1