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.
Lesson : Solving System of Equations using matrices
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3Are you familiar with determinants? Specifically, the fact that a matrix is singular if and only if its determinant is zero? – 2012-05-24
3 Answers
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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0Thanks. I had been trying to solve the system, using row echelon, and kept getting some weird looking results. – 2012-05-24
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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0@RacMain It's a \times {\rm det}\pmatrix{b & 5 \\ 1 & b}. Can you compute that? – 2012-05-24
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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0I tried to do just that and had a hard time figuring it out, so I decided to ask for help with it instead. – 2012-05-24