# Calculating Determinant of Matrix using Cofactors

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

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

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

• You forgot to multiply by the original $3$ when you expanded about the third row. – Jack Pfaffinger Oct 9 at 18:31
• Yeah but wouldn't that be 3 * 0 from matrix b? – Shinji-san Oct 9 at 18:32
• 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. – Jack Pfaffinger Oct 9 at 18:37

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