Examples

Find the Basis and Dimension for the Column Space of the Matrix
[3-1021-1]310211
Step 1
Find the reduced row echelon form.
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Step 1.1
Multiply each element of R1R1 by 1313 to make the entry at 1,11,1 a 11.
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Step 1.1.1
Multiply each element of R1R1 by 1313 to make the entry at 1,11,1 a 11.
[33-13021-1]⎢ ⎢33130211⎥ ⎥
Step 1.1.2
Simplify R1R1.
[1-13021-1]⎢ ⎢1130211⎥ ⎥
[1-13021-1]⎢ ⎢1130211⎥ ⎥
Step 1.2
Perform the row operation R3=R3-R1R3=R3R1 to make the entry at 3,13,1 a 00.
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Step 1.2.1
Perform the row operation R3=R3-R1R3=R3R1 to make the entry at 3,13,1 a 00.
[1-13021-1-1+13]⎢ ⎢11302111+13⎥ ⎥
Step 1.2.2
Simplify R3R3.
[1-13020-23]⎢ ⎢11302023⎥ ⎥
[1-13020-23]⎢ ⎢11302023⎥ ⎥
Step 1.3
Multiply each element of R2R2 by 1212 to make the entry at 2,22,2 a 11.
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Step 1.3.1
Multiply each element of R2R2 by 1212 to make the entry at 2,22,2 a 11.
[1-1302220-23]⎢ ⎢ ⎢1130222023⎥ ⎥ ⎥
Step 1.3.2
Simplify R2R2.
[1-13010-23]⎢ ⎢11301023⎥ ⎥
[1-13010-23]⎢ ⎢11301023⎥ ⎥
Step 1.4
Perform the row operation R3=R3+23R2R3=R3+23R2 to make the entry at 3,23,2 a 00.
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Step 1.4.1
Perform the row operation R3=R3+23R2R3=R3+23R2 to make the entry at 3,23,2 a 00.
[1-13010+230-23+231]⎢ ⎢113010+23023+231⎥ ⎥
Step 1.4.2
Simplify R3R3.
[1-130100]⎢ ⎢1130100⎥ ⎥
[1-130100]⎢ ⎢1130100⎥ ⎥
Step 1.5
Perform the row operation R1=R1+13R2R1=R1+13R2 to make the entry at 1,21,2 a 00.
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Step 1.5.1
Perform the row operation R1=R1+13R2R1=R1+13R2 to make the entry at 1,21,2 a 00.
[1+130-13+1310100]⎢ ⎢1+13013+1310100⎥ ⎥
Step 1.5.2
Simplify R1R1.
[100100]100100
[100100]100100
[100100]100100
Step 2
The pivot positions are the locations with the leading 11 in each row. The pivot columns are the columns that have a pivot position.
Pivot Positions: a11a11 and a22a22
Pivot Columns: 11 and 22
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)Col(A) is the number of vectors in a basis for Col(A)Col(A).
Basis of Col(A)Col(A): {[301],[-12-1]}301,121
Dimension of Col(A)Col(A): 22
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