Lesson : Solving System of Equations using matrices 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.
 A: 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. $b^2=5$.
A: 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 $a(b^2-5)$.)
A: 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.
