# Help using Cramer's rule to show properties of determinants

#### GimB0id

New here, have an assignment concerning Cramer's rule which I think I have a decent understanding of - I can use it to find determinants - but am a little lost on a few questions.

Given the set of linear equations:
a11 x1 + a12 x2 + a13 x3 = 0
a22 x2 + a23 x3 = 0
a33 x3 = 0

assume D = 0 (determinant), this implies that at least 1 of a22 or a33 is 0. show if x !=0 then D != 0

My try:
when I do D I get -> (a11 a22 a33), but if a22 or a33 is 0, then D=0. So I'm not sure what is going on here as if a33 = 0 then x3 can be anything, implying x!=0?

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#### aPhilosopher

D = 0 could imply that a11 = 0 as well.

On the whole, the problem doesn't make sense to me as stated. The determinant of a linear transformation is independent of the value of the vector that it operates on so x being zero would have nothing to do with it.

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#### GimB0id

D = 0 could imply that a11 = 0 as well.

On the whole, the problem doesn't make sense to me as stated. The determinate of a linear transformation is independent of the value of the vector that it operates on so x being zero would have nothing to do with it.
right, a11 could be 0 also. I also don't quite understand the question, I was hoping it was a lack of knowledge on my part as to why I didn't understand it.

#### HallsofIvy

Science Advisor
Homework Helper
New here, have an assignment concerning Cramer's rule which I think I have a decent understanding of - I can use it to find determinants - but am a little lost on a few questions.

Given the set of linear equations:
a11 x1 + a12 x2 + a13 x3 = 0
a22 x2 + a23 x3 = 0
a33 x3 = 0
So this can be expressed as the matrix equation
$$\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ 0 & a_{22} & a_{23} \\ 0 & 0 & a_{33}\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x3\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

It easy to show that the determinant of the matrix is $a_{11}a_{22}a_{33}$

assume D = 0 (determinant), this implies that at least 1 of a22 or a33 is 0. show if x !=0 then D != 0

My try:
when I do D I get -> (a11 a22 a33), but if a22 or a33 is 0, then D=0. So I'm not sure what is going on here as if a33 = 0 then x3 can be anything, implying x!=0?
Yes, the determinant is $a_{11}a_{22}a_{33}$ as you say and that equals 0 if and only if any one of $a_{11}$, $a_{22}$, $a_{33}$ is 0. The part about "if x != 0" I don't understand. Are you sure you have copied the problem correctly? It is true that if there exist a non-zero x satifying that equation, the the determinant must be 0. If the determinant is 0, there is only the single solution x= 0.

#### GimB0id

I think this is a poorly written question, my prof. gave us some minimal notes and there's typos all over the place so I'm going to talk to him to find out what he wants as I suspect there's typos in the question.

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