Use determinants to find which real values of c make each of the following matrices invertible

Question

So i was given this question

Use determinants to find which real values of c make each of the following matrices invertible

$ \left[ {\begin{array}{cc} 0 & c & -c \\ -1 & 2& 1 \\ c & -c & -c \end{array} } \right] $

When i look at the solutions for this question they usually adding or subtracting a column or row to one another, until there is two consecutive $0s$ in the row or column then make a 2x2 matrix. I understand the logic behind finding a determinant, but for this question and similar ones is there a rule or method to go about solving these kinds of problems, and how to do it?

Answer 1

Note that if you multiply a row or column by a constant, then since the determinant is multi linear, the determinant of the resulting matrix is a constant times the original matrix.

Hence $\det \begin{bmatrix} 0 & c & -c \\ -1 & 2& 1 \\ c & -c & -c \end{bmatrix} = c \det \begin{bmatrix} 0 & 1 & -1 \\ -1 & 2& 1 \\ c & -c & -c \end{bmatrix} = c^2 \det \begin{bmatrix} 0 & 1 & -1 \\ -1 & 2& 1 \\ 1 & -1 & -1 \end{bmatrix}$.

We also have $\det \begin{bmatrix} 0 & 1 & -1 \\ -1 & 2& 1 \\ 1 & -1 & -1 \end{bmatrix} = 1$.

Answer 2

If a given square matrix is invertible, then determinant of it should be non zero. The determinant of this matrix is $-c(c-c)-c(c-2c)=c^2$. So, any non zero $c$ would make this matrix invertible.

Answer 3

First, by multilinearity, you have: $$\begin{vmatrix} 0 & c & -c \\ -1 & 2& 1 \\ c & -c & -c \end{vmatrix} =c^2\begin{vmatrix} 0 & 1 & -1 \\ -1 & 2& 1 \\ 1 & -1 & -1 \end{vmatrix} $$ Then, with row or column operations: $$\begin{vmatrix} 0 & 1 & -1 \\ -1 & 2& 1 \\ 1 & -1 & -1 \end{vmatrix}=\begin{vmatrix} 0 & 1 & -1 \\ 0 & 1& 0 \\ 1 & -1 & -1 \end{vmatrix}=1$$ (develop along the first column). Thus the determinant is equal to $c^2$, and it is non-zero if and only if $c\neq0$.