compute determinant and is A invertible?

Question

Compute the determinant of

A = \begin{bmatrix} 0 & 1+i & 2 \\ -2i & 0 & 1-i \\ 3 & 4i & 0 \end{bmatrix}

along the third row.

Is A invertible?

Answer 1

Since you're stating "along the 3rd row", I'll assume you want to use the method using the "submatrices" :

So we end up with :

$$(-1)^0\cdot3\cdot \det\left(\begin{bmatrix}1+i&2\\0&1-i\end{bmatrix}\right) + (-1)^1\cdot 4i \cdot \det\left(\begin{bmatrix}0&2\\-2i&1-i\end{bmatrix}\right) + (-1)^2\cdot 0 \cdot \det\left(\begin{bmatrix}0&i+1\\-2i&0\end{bmatrix}\right)$$ Which you should be able to process, since those are 2x2 matrices, and we get so : $$(-1)^0\cdot3\cdot (1-i^2) + (-1)^1\cdot 4i \cdot 4i + (-1)^2\cdot 0 \cdot (2i)(i+1)=3\cdot 2+16 = 22$$ Hence since it has a non-zero determinant, it is invertible.