Trouble with $4\times4$ matrix determinant

Question

$$ \begin{vmatrix} 1 & -6 & 7 & 5 \\ 0 & 0 & 3 & 0 \\ 3 & -2 & -8 & 6\\ 2 & 0 & 5 & 4\\ \end{vmatrix} $$

Clearly I want to expand along the second row yielding:

$((-1)^5)3$ times the following matrix

$$ \begin{vmatrix} 1 & -6 & 5 \\ 3 & -2 & 6 \\ 2 & 0 & 4 \\ \end{vmatrix} $$

and then breaks down into several smaller matrices:

2 times $$ \begin{vmatrix} -6 & 5 \\ -2 & 6 \\ \end{vmatrix} $$

and 4 times

$$ \begin{vmatrix} 1 & -6 \\ 3 & -2 \\ \end{vmatrix} $$

which should come out to be $-3[(2(-36+10))+(4(-2+18))]$

$-3[(2(-16))+(4(16))]$

$-3(-32+64)=32 \times -3$

but the answer is -36 I don't know what went wrong?

Answer 1

Your computations are almost correct. But it turns out that $-36+10=-26$, not $-16$.

Answer 2

Note that, in your $3\times 3$ matrix, you could have subtracted twice the first column to the last, obtaining instantly the determinant: $$\begin{vmatrix} 1 & -6 & 5 \\ 3 & -2 & 6 \\ 2 & 0 & 4 \\ \end{vmatrix}=\begin{vmatrix} 1 & -6 & 3 \\ 3 & -2 & 0 \\ 2 & 0 & 0 \\ \end{vmatrix}=3\cdot 4\quad_\text{(expanding along the last column),}$$ whence the final determinant $-3\cdot 12=-36$.

Answer 3

Another way to approach this 4x4 matrix is to row reduce to a triangular matrix with zeros underneath the diagonal.

\begin{bmatrix}1&-6&7&5\\0&0&3&0\\3&-2&-8&6\\2&0&5&4\end{bmatrix}

Row reducing to the triangular matrix yields:

\begin{bmatrix}1&-6&7&5\\0&16&-29&-9\\0&0&3&0\\0&0&0&\frac{3} {4}\end{bmatrix}

From here, we can multiply the numbers in the diagonal:

-(1 $\cdot$ 16 $\cdot$ 3 $\cdot$ $\frac{3} {4}$) = -36