LINEAR ALGEBRA
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For details see post on
26-Feb-2015 at 6:00 am.
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Determine whether b is in the column space of A. If it is, then write b as a linear combination of the column vectors of A. (Use v1, v2, and v3, respectively, for the three columns. If not possible, enter IMPOSSIBLE.)
A = [(1,3,0);(-1,1,0);(2,0,1)]
b= [(2,3,-4)]
| A= | ||||||
| U | V | W | B= | X= | ||
| 1 | -1 | 2 | 2 | P | ||
| 3 | 1 | 0 | 3 | Q | ||
| 0 | 0 | 1 | -4 | R | ||
| AX=B………X=A INVERSE *B | ||||||
| A INVERSE = | ||||||
| 0.25 | 0.25 | -0.5 | ||||
| -0.75 | 0.25 | 1.5 | ||||
| 0 | 0 | 1 | ||||
| X= A INVERSE * B = | ||||||
| 3.25 | ||||||
| -6.75 | ||||||
| -4 | ||||||
| HENCE B IS IN COLUMN SPACE OF A | ||||||
| B = 3.25*U-6.75*V-4*W……………….ANSWER | ||||||


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