Wrong answer on Linear Algebra and Its Applications 4th Ed.

[mafagafo]

Homework Statement

The Attempt at a Solution

<br /> \left[<br /> \begin{array}{cccc}<br /> 1 &amp; 0 &amp; 5 &amp; 2 \\<br /> -2 &amp; 1 &amp; -6 &amp; -1 \\<br /> 0 &amp; 2 &amp; 8 &amp; 6<br /> \end{array}<br /> \right]<br /> \sim<br /> \left[<br /> \begin{array}{cccc}<br /> 1 &amp; 0 &amp; 5 &amp; 2 \\<br /> 0 &amp; 1 &amp; 4 &amp; 3 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0<br /> \end{array}<br /> \right]<br />

From the RREF it is clear to me that (2, 3, 0) are valid scalars for a linear combination and there are infinitely many such solutions. However, the answer in the back of the book suggests otherwise.

More than anything, I am asking for confirmation that my understanding is correct and that the provided answer is wrong.

 

[Charles Link]

I'm no expert in linear algebra, (I received a B in a college course in it many years ago), but your assessment looks correct. I tried to do a Kramer's rule solution for this one, but the denominator determinant along with all 3 numerator determinants are zero==>> Kramer's rule didn't supply any solutions, but you correctly identified a solution. For this problem a3=5(a1)+4(a2) so you don't have 3 vectors that are linearly independent. B can be written as all kinds of (infinitely many) linear combinations of a1, a2, and a3 so the question is really somewhat ambiguous in my opinion for the problem that they gave you. Perhaps a better worded question is, Can "b" be written as one and only one specific linear combination of a1, a2, and a3? In any case I think you analyzed it correctly. editing...Another way to look at this one is the vectors a1, a2, and a3 all lie in the same plane and b also lies in this plane. additional editing...And if "b" didn't lie in this plane as well, then it could not have been written as a linear combination of a1, a2, and a3. Then there simply would have been no solution instead of infinitely many solutions.

 

[mafagafo]

In any case I think you analyzed it correctly.

Thanks for taking the time to reply.

 

[Charles Link]

@mafagafo For a little additional info, I edited post #2=I added a couple of things.

 

[SammyS]

Homework Statement

The Attempt at a Solution

[/B]<br /> \left[<br /> \begin{array}{cccc}<br /> 1 &amp; 0 &amp; 5 &amp; 2 \\<br /> -2 &amp; 1 &amp; -6 &amp; -1 \\<br /> 0 &amp; 2 &amp; 8 &amp; 6<br /> \end{array}<br /> \right]<br /> \sim<br /> \left[<br /> \begin{array}{cccc}<br /> 1 &amp; 0 &amp; 5 &amp; 2 \\<br /> 0 &amp; 1 &amp; 4 &amp; 3 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0<br /> \end{array}<br /> \right]<br />From the RREF it is clear to me that (2, 3, 0) are valid scalars for a linear combination and there are infinitely many such solutions. However, the answer in the back of the book suggests otherwise.

More than anything, I am asking for confirmation that my understanding is correct and that the provided answer is wrong.

You are correct. There are infinite linear combinations of a₁, a₂, and a₃ which gibe b .

Another is a₃ - a₂ - 3 a₁ = b .

As Charles points out, a₁, a₂, and a₃ lie in the same plane. Thus, they are not linearly independent, but the question does not ask about that.

If the third row of the Reduced Row Echelon Form had a non-zero entry only in the 4th column, then no linear combination would be possible.

 

[mafagafo]

You are correct.

(...)

If the third row of the Reduced Row Echelon Form had a non-zero entry only in the 4th column, then no linear combination would be possible.

Thanks for confirming.