Compute the determinant using the Laplace expansion (using the first row) and the Sarrus rule for $$A=\begin{bmatrix} 1 & 3 & 5\\ 2 & 4 & 6\\ 0 & 2 & 4\end{bmatrix}.$$

Solution: By Laplace expansion using the first row, we have \begin{align*}\det(A)=&\ 1\cdot \begin{vmatrix}4 & 6\\ 2 & 4\end{vmatrix}-3\cdot \begin{vmatrix}2 & 6\\ 0 & 4\end{vmatrix} + 5\cdot \begin{vmatrix}2 & 4\\ 0 & 2\end{vmatrix}\\=&\ 1(4\cdot 4-6\cdot 2)-3\cdot(2\cdot 4-6\cdot 0)+5\cdot(2\cdot 2-4\cdot 0)\\ =& \ 4-3\cdot 8+5\cdot 4=0.\end{align*}

Or by Sarrus rule, we have \begin{align*}\det(A)=&\ 1\cdot 4\cdot 4+2\cdot 2\cdot 5+0\cdot 3\cdot 6-0\cdot 4\cdot 5-1\cdot 2\cdot 6-2\cdot 3\cdot 4\\=&\ 16+20-12-24=0.\end{align*}

## Biying

2 Oct 2020Hi Can you help on the Exercise 4.12? Thanks so much!

## Linearity

2 Oct 2020See Solution to Mathematics for Machine Learning Exercise 4.12.