$A =\left(\begin{array}{ccc} 0 & 1 & b \\ -1 & 0 & c \\ -b & -c & 0 \end{array}\right)$
I tried to come up with the RREF($A$), but in the final step I have the matrix:
$A =\left(\begin{array}{ccc} 1 & 0 & -c \\ 0 & 1 & b \\ 0 & 0 & 0 \end{array}\right)$
Which I then conclude that no values of $b,c$ can make the matrix invertible (since the RREF($A$) doesn't have a pivot in the final row- a unique solution does not exist). Is there a way that I can verify this result?
$\endgroup$ 02 Answers
$\begingroup$$A$ is invertible iff $\det(A) \neq 0$.
$\det(A) = 0(0 + c^2 ) - 1(0 + bc)+b(c-0) = 0 - bc + bc = 0$.
Therefore, $\det(A)$ doesn't depend on values of $b, c$. Therefore, $A$ is not invertible for any values of $b,c$.
$\endgroup$ 1 $\begingroup$Hint : Find the determinant and find where it is zero.
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