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I have looked over this problem five times and I am not getting the right answer so I was wondering if anybody can perhaps spot the error.

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

$$\det(A) = 5?$$

I am getting it's supposed to be $15$ buy I'm not sure where I erred.

So I choose to expand across row 3 so I get

$$B=\begin{bmatrix} & 0 & 0 & 5 \\ & 7 & 2 & -5 \\ & 3 & 1 & 7 \end{bmatrix}$$

$$\det(A) = 3 \det B$$

Then finally if you expand across the first row, only the last column should matter because all the other terms go to 0, right? In which case you'd get

$$C=\begin{bmatrix} 7 & 2 \\ 3 & 1 \end{bmatrix}$$

$\det(A) = 5 \det C = 5(7-6) = 5$?

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  • $\begingroup$ You forgot to multiply by the original $3$ when you expanded about the third row. $\endgroup$ – Jack Pfaffinger Oct 9 at 18:31
  • $\begingroup$ Yeah but wouldn't that be 3 * 0 from matrix b? $\endgroup$ – Shinji-san Oct 9 at 18:32
  • $\begingroup$ Well, we know that $det(A) = 3 \cdot det(B)$ and $det(B) = 5 \cdot det(C)$. Since you calculated that $det(C) = 1$ you should have your answer. $\endgroup$ – Jack Pfaffinger Oct 9 at 18:37
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You wrote that the $\det(A) = 3 \det(B)$, which you would be correct on.

Though, the $\det(B) = 5\det(C)$ where

$C = \begin{bmatrix} 7&2\\ 3&1 \end{bmatrix} $

In this case $\det(C) = 7(1)-2(3) = 1$.

Therefore, $\det(A) = 3(\det(B)) = 3(5\det(C)) = 3(5)(1) = 15$.

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