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I have a matrix
$$ A = \begin{pmatrix} a & 0 & 0 \\ 2 & b & 5 \\ -3 & 1 & b \end{pmatrix} $$ in my try, I came up with $$ bx1 = 0,\quad x2 + 5/a x3 = 0,\quad x2 + a x3 = 0 $$ The question is to find all possible choices of $a$, and $b$, that would make the matrix singular.

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Are you familiar with determinants? Specifically, the fact that a matrix is singular if and only if its determinant is zero? – anon May 24 '12 at 18:56
up vote 2 down vote accepted

The determinant is $a(b^2-5)$, so the matrix is singular if $a=0$ or if $b=\pm\sqrt 5$.

You can also see this without referring to the determinant. If $a=0$, then the top row is all zeros, so the matrix is singular. If $a\neq 0$, then the first row is clearly not in the space spanned by columns 2 and 3. Therefore, the only way you'll get a singular matrix is when those two columns are linearly dependent (scalar multiples of one another). This happens when the ratios of coordinates are the same: $b/1=5/b$, i.e. $b^2=5$.

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Thanks. I had been trying to solve the system, using row echelon, and kept getting some weird looking results. – Rac Main May 24 '12 at 19:21

The matrix $A$ is singular if and only if its determinant is $0$. The determinant of $A$ is isn’t hard to calculate; it turns out to be a very simple function of $a$ and $b$. If you set that expression to $0$ and factor it, you should be able to determine quite easily what values of $a$ and $b$ make it $0$.

Added: $$\det A = \left| \begin{array}{c} a & 0 & 0 \\ 2 & b & 5 \\ -3 & 1 & b \end{array}\right|= a\left|\begin{array}{c} b&5\\1&b \end{array}\right|=a(b^2-5)\;.$$

(There is also a shortcut for calculating the determinant of a $3\times 3$ matrix that you can find here; it gives $ab^2-5a$, which is then readily factored to $a(b^2-5)$.)

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@RacMain It's $a \times {\rm det}\pmatrix{b & 5 \\ 1 & b}.$ Can you compute that? – user2468 May 24 '12 at 19:09

You can reduce by rows (or columns: whatever) your matrix. It will be singular iff at least one of the rows (columns) becomes all zeroes at some point.

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I tried to do just that and had a hard time figuring it out, so I decided to ask for help with it instead. – Rac Main May 24 '12 at 19:21

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