Find the determinant of the matrix

Question

Find the determinant of the matrix $$M = \left[\begin{array}{ccccc} 3 &0 &0 &2 &0\cr -2 &0 &-3 &0 &0\cr 0 &2 &0 &0 &2\cr 0 &0 &0 &-1 &-1\cr 0 &2 &-1 &0 &0 \end{array}\right].$$

I got the REF and tried to find the solution: $$M = \left[\begin{array}{ccccc} 3 &0 &0 &2 &0\cr 0 &2 &0 &0 &2\cr 0 &0 &-3 &4/3 &0\cr 0 &0 &0 &-1 &-1\cr 0 &0 &0 &0 &-14/9 \end{array}\right].$$

And I think $\text{det}(M)$ is
$$ \Bigg[ 3\begin{pmatrix}2&0\\ \:0&-3\end{pmatrix}-0\begin{pmatrix}0&0\\ \:0&-3\end{pmatrix}+0\begin{pmatrix}0&2\\ \:0&0\end{pmatrix}\Bigg].\begin{pmatrix}-1&-1\\ 0&-\frac{14}{9}\end{pmatrix}=-18\cdot\frac{14}{9}=-28$$

So I want to know which part I am wrong.

Answer 1

$$\begin{align}\begin{vmatrix} 3 &0 &0 &2 &0\cr 0 &2 &0 &0 &2\cr 0 &0 &-3 &4/3 &0\cr 0 &0 &0 &-1 &-1\cr 0 &0 &0 &0 &-14/9 \end{vmatrix}&=3\begin{vmatrix}2 &0 &0 &2\cr 0 &-3 &4/3 &0\cr 0 &0 &-1 &-1\cr 0 &0 &0 &-14/9 \end{vmatrix}\\&=3\cdot2\begin{vmatrix}-3 &4/3 &0\cr 0 &-1 &-1\cr 0 &0 &-14/9 \end{vmatrix}\\&=3\cdot2\cdot(-3)\begin{vmatrix}-1 &-1\cr 0 &-14/9 \end{vmatrix}\\&=3\cdot2\cdot(-3)\cdot\left(\frac{14}9\right)\\&=\color{red}{-28}\end{align}$$

Answer 2

You have already found an expression of $M$ as an upper triangular matrix! So the determinant is just the product of the elements in the diagonal.

Answer 3

Since the matrix in the question is a upper triangular matrix, the detrrminant is directly given by the product of terms in the principal diagonal.

$$D=3 \cdot 2 \cdot -3 \cdot -1 \cdot \frac{-14}{9} $$ $$D = -28$$