Determine whether a matrix C has an inverse without lengthy calculations

Question

How do I determine whether the following matrix has an inverse without lengthy calculations? $$C=\begin{bmatrix} 14 & -2i & 14 & -2i\\ 3 & 41 & 1-i & 41\\ 0 & 5 & \frac{7}{2} & 5\\ 9 & -1 & 77 & -1 \end{bmatrix}$$

I think the introduction of complex entries might be confusing me slightly, I notice also that in each row, column 2 and 4 have the same entries, but I don't know what to do with this information.

Answer 1

A matrix is invertible iff its determinant is non-zero.

In this case, the determinant is zero, which is clear since it has two identical columns. To see this, consider the elementary transformation $C_2 \mapsto C_2 - C_4$.

The second column is now zero. Expanding along that column shows that the determinant is zero.

Answer 2

Since two columns are the same, the determinant is zero, and the matrix is not invertible.

Answer 3

Here's a slightly different way from the methods of determinant that works even for noncommutative fields.

Recall a matrix over a field has an inverse iff it defines an injective linear transformation.

Notice that $C\begin{bmatrix}0\\ 1 \\ 0 \\ -1\end{bmatrix}=C\begin{bmatrix}0\\ 0\\ 0\\ 0\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\\ 0\end{bmatrix}$, so it is not, in fact, injective.