Matrix solution solution and linear combo

charlies1902 · Sep 21, 2012

I've attached the problem.

for the 1st part:
I solve the system for det(A)=0 and find the values of a that the system DOES NOT have an unique solution.

This ended up being a=0, -2, 2

Thus the answer would be a≠0, -2 , 2

For the 2nd part:
I must rearrange the matrix A such that the last column [0 0 1]^T is the rightmost column of an augmented matrix.
I then find RREF of this matrix. In order for the last column to be a linear combo of the 1st 2, there must be no free variables in the 1st 2 columns. I'm not too sure how to row reduce this problem though since we have a variable a in there. I can tell that a=0 is one of the answers by inspection. a can't equal another other scalar because that will yield an inconsistent system. Thus, a=0 should be the only answer.

How do I approach this problem by row reducing? because not all problems can be solved by inspection.

This problem was a practice problem from last year's exam. It was 1 of 6 problems on the 50 minute test. The problem is it took me like 20minutes just to do this problem (However, I have not done any reviewing yet.) The second part took under a minute, but the second part just took me a while to figure out what is going on.

Also, on a side note, the vector b is not used to solve these 2 parts right?

HallsofIvy · Sep 21, 2012

Yes, that is correct. I presume you expanded on the last column so it was relatively easy to find the determinant.

If you just use the definition of "linear independence" there must exist numbers x and y such that [itex]2ax+ a^2y= 0[/itex], [itex]ax+ 2ay= 0[/itex], and [itex]x+ y= 1[/itex]. From ax+ 2ay= 0, as long as a is not 0, x= -2y. Putting that into x+y= 1, -2y+ y= -y= 1 so y= -1 and x= -2y= 2. But then the first equation becomes 4a- 2a= 2a= 0 which is only true if a= 0. If a= 0 then x and be anything and y= 1- x. That shouldn't take 20 minutes.

pyroknife · Sep 22, 2012

HallsofIvy said: ↑

Yes, that is correct. I presume you expanded on the last column so it was relatively easy to find the determinant.

If you just use the definition of "linear independence" there must exist numbers x and y such that [itex]2ax+ a^2y= 0[/itex], [itex]ax+ 2ay= 0[/itex], and [itex]x+ y= 1[/itex]. From ax+ 2ay= 0, as long as a is not 0, x= -2y. Putting that into x+y= 1, -2y+ y= -y= 1 so y= -1 and x= -2y= 2. But then the first equation becomes 4a- 2a= 2a= 0 which is only true if a= 0. If a= 0 then x and be anything and y= 1- x. That shouldn't take 20 minutes.

I think you typoed, it's x+2y=1. Thys -2y+2y=1 wihch gives 0=1, which doesn't make sense, thus a=0/

HallsofIvy · Sep 22, 2012

HallsofIvy said: ↑

Yes, that is correct. I presume you expanded on the last column so it was relatively easy to find the determinant.

If you just use the definition of "linear independence" there must exist numbers x and y such that [itex]2ax+ a^2y= 0[/itex], [itex]ax+ 2ay= 0[/itex], and [itex]x+ y= 1[/itex].

Correction: this last equation should be x+ 2y= 1 (Thanks, pyroknife)
Fron ax+ 2ay= 0, as long as a is not 0, x= -2y. Putting that unto x+ 2y= 1, -2y+ 2y= 0= 1 for all y which is impossible. Therefore a= 0 (easier than I thought!). If a= 0, and the only condition we have is x+y= 1 so x can be anything and then y= 1- x.

From ax+ 2ay= 0, as long as a is not 0, x= -2y. Putting that into x+y= 1, -2y+ y= -y= 1 so y= -1 and x= -2y= 2. But then the first equation becomes 4a- 2a= 2a= 0 which is only true if a= 0. If a= 0 then x and be anything and y= 1- x. That shouldn't take 20 minutes.

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Matrix solution solution and linear combo

charlies1902

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HallsofIvy

pyroknife

HallsofIvy

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