Examples

Find the Basis and Dimension for the Null Space of the Matrix
[-36-11-71-223-12-458-4]
Step 1
Write as an augmented matrix for Ax=0.
[-36-11-701-223-102-458-40]
Step 2
Find the reduced row echelon form.
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Step 2.1
Multiply each element of R1 by -13 to make the entry at 1,1 a 1.
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Step 2.1.1
Multiply each element of R1 by -13 to make the entry at 1,1 a 1.
[-13-3-136-13-1-131-13-7-1301-223-102-458-40]
Step 2.1.2
Simplify R1.
[1-213-137301-223-102-458-40]
[1-213-137301-223-102-458-40]
Step 2.2
Perform the row operation R2=R2-R1 to make the entry at 2,1 a 0.
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Step 2.2.1
Perform the row operation R2=R2-R1 to make the entry at 2,1 a 0.
[1-213-137301-1-2+22-133+13-1-730-02-458-40]
Step 2.2.2
Simplify R2.
[1-213-137300053103-10302-458-40]
[1-213-137300053103-10302-458-40]
Step 2.3
Perform the row operation R3=R3-2R1 to make the entry at 3,1 a 0.
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Step 2.3.1
Perform the row operation R3=R3-2R1 to make the entry at 3,1 a 0.
[1-213-137300053103-10302-21-4-2-25-2(13)8-2(-13)-4-2(73)0-20]
Step 2.3.2
Simplify R3.
[1-213-137300053103-103000133263-2630]
[1-213-137300053103-103000133263-2630]
Step 2.4
Multiply each element of R2 by 35 to make the entry at 2,3 a 1.
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Step 2.4.1
Multiply each element of R2 by 35 to make the entry at 2,3 a 1.
[1-213-1373035035035533510335(-103)35000133263-2630]
Step 2.4.2
Simplify R2.
[1-213-137300012-2000133263-2630]
[1-213-137300012-2000133263-2630]
Step 2.5
Perform the row operation R3=R3-133R2 to make the entry at 3,3 a 0.
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Step 2.5.1
Perform the row operation R3=R3-133R2 to make the entry at 3,3 a 0.
[1-213-137300012-200-13300-1330133-1331263-1332-263-133-20-1330]
Step 2.5.2
Simplify R3.
[1-213-137300012-20000000]
[1-213-137300012-20000000]
Step 2.6
Perform the row operation R1=R1-13R2 to make the entry at 1,3 a 0.
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Step 2.6.1
Perform the row operation R1=R1-13R2 to make the entry at 1,3 a 0.
[1-130-2-13013-131-13-13273-13-20-1300012-20000000]
Step 2.6.2
Simplify R1.
[1-20-1300012-20000000]
[1-20-1300012-20000000]
[1-20-1300012-20000000]
Step 3
Use the result matrix to declare the final solution to the system of equations.
x1-2x2-x4+3x5=0
x3+2x4-2x5=0
0=0
Step 4
Write a solution vector by solving in terms of the free variables in each row.
[x1x2x3x4x5]=[2x2+x4-3x5x2-2x4+2x5x4x5]
Step 5
Write the solution as a linear combination of vectors.
[x1x2x3x4x5]=x2[21000]+x4[10-210]+x5[-30201]
Step 6
Write as a solution set.
{x2[21000]+x4[10-210]+x5[-30201]|x2,x4,x5R}
Step 7
The solution is the set of vectors created from the free variables of the system.
{[21000],[10-210],[-30201]}
Step 8
Check if the vectors are linearly independent.
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Step 8.1
List the vectors.
[21000],[10-210],[-30201]
Step 8.2
Write the vectors as a matrix.
[21-31000-22010001]
Step 8.3
To determine if the columns in the matrix are linearly dependent, determine if the equation Ax=0 has any non-trivial solutions.
Step 8.4
Write as an augmented matrix for Ax=0.
[21-3010000-22001000010]
Step 8.5
Find the reduced row echelon form.
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Step 8.5.1
Multiply each element of R1 by 12 to make the entry at 1,1 a 1.
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Step 8.5.1.1
Multiply each element of R1 by 12 to make the entry at 1,1 a 1.
[2212-320210000-22001000010]
Step 8.5.1.2
Simplify R1.
[112-32010000-22001000010]
[112-32010000-22001000010]
Step 8.5.2
Perform the row operation R2=R2-R1 to make the entry at 2,1 a 0.
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Step 8.5.2.1
Perform the row operation R2=R2-R1 to make the entry at 2,1 a 0.
[112-3201-10-120+320-00-22001000010]
Step 8.5.2.2
Simplify R2.
[112-3200-123200-22001000010]
[112-3200-123200-22001000010]
Step 8.5.3
Multiply each element of R2 by -2 to make the entry at 2,2 a 1.
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Step 8.5.3.1
Multiply each element of R2 by -2 to make the entry at 2,2 a 1.
[112-320-20-2(-12)-2(32)-200-22001000010]
Step 8.5.3.2
Simplify R2.
[112-32001-300-22001000010]
[112-32001-300-22001000010]
Step 8.5.4
Perform the row operation R3=R3+2R2 to make the entry at 3,2 a 0.
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Step 8.5.4.1
Perform the row operation R3=R3+2R2 to make the entry at 3,2 a 0.
[112-32001-300+20-2+212+2-30+2001000010]
Step 8.5.4.2
Simplify R3.
[112-32001-3000-4001000010]
[112-32001-3000-4001000010]
Step 8.5.5
Perform the row operation R4=R4-R2 to make the entry at 4,2 a 0.
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Step 8.5.5.1
Perform the row operation R4=R4-R2 to make the entry at 4,2 a 0.
[112-32001-3000-400-01-10+30-00010]
Step 8.5.5.2
Simplify R4.
[112-32001-3000-4000300010]
[112-32001-3000-4000300010]
Step 8.5.6
Multiply each element of R3 by -14 to make the entry at 3,3 a 1.
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Step 8.5.6.1
Multiply each element of R3 by -14 to make the entry at 3,3 a 1.
[112-32001-30-140-140-14-4-14000300010]
Step 8.5.6.2
Simplify R3.
[112-32001-30001000300010]
[112-32001-30001000300010]
Step 8.5.7
Perform the row operation R4=R4-3R3 to make the entry at 4,3 a 0.
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Step 8.5.7.1
Perform the row operation R4=R4-3R3 to make the entry at 4,3 a 0.
[112-32001-3000100-300-303-310-300010]
Step 8.5.7.2
Simplify R4.
[112-32001-30001000000010]
[112-32001-30001000000010]
Step 8.5.8
Perform the row operation R5=R5-R3 to make the entry at 5,3 a 0.
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Step 8.5.8.1
Perform the row operation R5=R5-R3 to make the entry at 5,3 a 0.
[112-32001-30001000000-00-01-10-0]
Step 8.5.8.2
Simplify R5.
[112-32001-30001000000000]
[112-32001-30001000000000]
Step 8.5.9
Perform the row operation R2=R2+3R3 to make the entry at 2,3 a 0.
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Step 8.5.9.1
Perform the row operation R2=R2+3R3 to make the entry at 2,3 a 0.
[112-3200+301+30-3+310+30001000000000]
Step 8.5.9.2
Simplify R2.
[112-3200100001000000000]
[112-3200100001000000000]
Step 8.5.10
Perform the row operation R1=R1+32R3 to make the entry at 1,3 a 0.
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Step 8.5.10.1
Perform the row operation R1=R1+32R3 to make the entry at 1,3 a 0.
[1+32012+320-32+3210+3200100001000000000]
Step 8.5.10.2
Simplify R1.
[112000100001000000000]
[112000100001000000000]
Step 8.5.11
Perform the row operation R1=R1-12R2 to make the entry at 1,2 a 0.
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Step 8.5.11.1
Perform the row operation R1=R1-12R2 to make the entry at 1,2 a 0.
[1-12012-1210-1200-1200100001000000000]
Step 8.5.11.2
Simplify R1.
[10000100001000000000]
[10000100001000000000]
[10000100001000000000]
Step 8.6
Remove rows that are all zeros.
[100001000010]
Step 8.7
Write the matrix as a system of linear equations.
x=0
y=0
z=0
Step 8.8
Since the only solution to Ax=0 is the trivial solution, the vectors are linearly independent.
Linearly Independent
Linearly Independent
Step 9
Since the vectors are linearly independent, they form a basis for the null space of the matrix.
Basis of Nul(A): {[21000],[10-210],[-30201]}
Dimension of Nul(A): 3
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