Precalculus Examples

Find the Basis and Dimension for the Column Space of the Matrix
[3-1021-1]
Step 1
Find the reduced row echelon form.
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Step 1.1
Multiply each element of R1 by 13 to make the entry at 1,1 a 1.
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Step 1.1.1
Multiply each element of R1 by 13 to make the entry at 1,1 a 1.
[33-13021-1]
Step 1.1.2
Simplify R1.
[1-13021-1]
[1-13021-1]
Step 1.2
Perform the row operation R3=R3-R1 to make the entry at 3,1 a 0.
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Step 1.2.1
Perform the row operation R3=R3-R1 to make the entry at 3,1 a 0.
[1-13021-1-1+13]
Step 1.2.2
Simplify R3.
[1-13020-23]
[1-13020-23]
Step 1.3
Multiply each element of R2 by 12 to make the entry at 2,2 a 1.
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Step 1.3.1
Multiply each element of R2 by 12 to make the entry at 2,2 a 1.
[1-1302220-23]
Step 1.3.2
Simplify R2.
[1-13010-23]
[1-13010-23]
Step 1.4
Perform the row operation R3=R3+23R2 to make the entry at 3,2 a 0.
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Step 1.4.1
Perform the row operation R3=R3+23R2 to make the entry at 3,2 a 0.
[1-13010+230-23+231]
Step 1.4.2
Simplify R3.
[1-130100]
[1-130100]
Step 1.5
Perform the row operation R1=R1+13R2 to make the entry at 1,2 a 0.
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Step 1.5.1
Perform the row operation R1=R1+13R2 to make the entry at 1,2 a 0.
[1+130-13+1310100]
Step 1.5.2
Simplify R1.
[100100]
[100100]
[100100]
Step 2
The pivot positions are the locations with the leading 1 in each row. The pivot columns are the columns that have a pivot position.
Pivot Positions: a11 and a22
Pivot Columns: 1 and 2
Step 3
The basis for the column space of a matrix is formed by considering corresponding pivot columns in the original matrix. The dimension of Col(A) is the number of vectors in a basis for Col(A).
Basis of Col(A): {[301],[-12-1]}
Dimension of Col(A): 2
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 [x2  12  π  xdx ] 
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