SOLUTION: Consider the accompanying matrix. Use the test for linear independence to find a basis for the space spanned by the rows of the matrix. Suppose that this matrix is augmented matrix

Question 26364: Consider the accompanying matrix. Use the test for linear independence to find a basis for the space spanned by the rows of the matrix. Suppose that this matrix is augmented matrix for a system of equations. What is the rank of this systen? Which equations can be discarded?
{[1 0 1 1
2 1 3 0
3 3 6 -3
4 1 5 2]}
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Consider the accompanying matrix. Use the test for linear independence to find a basis for the space spanned by the rows of the matrix. Suppose that this matrix is augmented matrix for a system of equations. What is the rank of this systen? Which equations can be discarded?
{[1 0 1 1
2 1 3 0
3 3 6 -3
4 1 5 2]}
THE SECOND QUESTION FIRST ...AS PER THE SECOND QUESTION ,THIS IS AN AUGMENTED MATRIX.HENCE LAST COLUMN IS CONSTANTS COLUMN.
AND THE FIRST 3 COLUMNS ARE COEFFICIENT MATRIX.BUT THERE ARE 4 ROWS.THAT IS THERE ARE 3 UNKNOWNS AND 4 EQNS.LET US FIND RANK OF ASUGMENTED MATRIX...
1 0 1 1
2 1 3 0
3 3 6 -3
4 1 5 2
R1=R1�� 1 0 1 1
R2=R2-2R1�.. 0 1 1 -2
R3=R3-3R1�.. 0 3 3 -6
R4=R4-4R1�� 0 1 1 -2
R1=R1�� 1 0 1 1
R2=R2��.�.. 0 1 1 -2
R3=R3-3R2�.. 0 0 0 0
R4=R4-4R1�� 0 0 0 0

HENCE RANK = 2
EQNS.2 AND 4 ARE LEADING TO SAME RESULT.
AND EQN 3 AND 4 ARE ALL ZEROES . SO WE CAN DISCARD EQNS.3 AND 4 IN THIS SYSTEM.
SO WE REALLY HAVE 2 INDEPENDENT EQNS. IN 3 UNKNOWNS LEADING TO INFINITE SOLUTIONS.
NOW COMING TO YOUR FIRST QUESTION ,THE DIMENSIONS OF THE BASIS IS NOT GIVEN.
TAKING 4 DIMENSIONAL BASIS FOR 4 EQNS.,WE GET

EQN.1 �..R1= A+C+D
EQN2��.R2= 2A+B+3C
EQN.3��R3= 3A+3B+6C-3D
EQN.4��R4= 4A+B+5C+2D

BUT WE GOT R4-4R1=B+C-2D=R2-2R1��..OR��.R4=R2+2R1
AND��..R3-3R2+3R1=0��.OR��.R3=3R2-3R1
THAT IS TAKING R1 AND R2 AS 2 INDEPENDENT EQNS. WE SHOWED R3 AND R4 AS A LINEAR COMBINATION OF
R1 AND R2.
HENCE THE BASIS FOR THESE SET OF 4 EQNS.CAN BE TAKEN AS
R1=A+C+D....AND.....R2=2A+B+3C

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