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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.

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  • $\begingroup$ By the way, the correct determinant is $28$. $\endgroup$ – Von Neumann Mar 11 '18 at 17:02
  • $\begingroup$ Isn't the det of a triangular matrix given by the product of terms in the principal diagonal? $\endgroup$ – Netravat Pendsey Mar 11 '18 at 17:02
  • $\begingroup$ I don't understand the downvote. The OP has shown their own work. $\endgroup$ – user1551 Mar 12 '18 at 11:49
  • $\begingroup$ The correct ans is 28. I think I forget to add a minus sign for the REF form. (Since exchange any two row need to add a minus sign) So I keep getting the incorrect answer. $\endgroup$ – kdhug886 Mar 13 '18 at 7:18
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$$\begin{align}\begin{vmatrix} 3 &0 &0 &2 &0\cr 0 &2 &0 &0 &2\cr 0 &0 &-3 &4/3 &0\cr 0 &0 &0 &-1 &-1\cr 0 &0 &0 &0 &-14/9 \end{vmatrix}&=3\begin{vmatrix}2 &0 &0 &2\cr 0 &-3 &4/3 &0\cr 0 &0 &-1 &-1\cr 0 &0 &0 &-14/9 \end{vmatrix}\\&=3\cdot2\begin{vmatrix}-3 &4/3 &0\cr 0 &-1 &-1\cr 0 &0 &-14/9 \end{vmatrix}\\&=3\cdot2\cdot(-3)\begin{vmatrix}-1 &-1\cr 0 &-14/9 \end{vmatrix}\\&=3\cdot2\cdot(-3)\cdot\left(\frac{14}9\right)\\&=\color{red}{-28}\end{align}$$

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    $\begingroup$ Can you explain why $$\begin{pmatrix}\color{red}1&-1\\ 0&-\frac{14}{9}\end{pmatrix}$$ is 1 instead of -1 $\endgroup$ – kdhug886 Mar 11 '18 at 17:18
  • $\begingroup$ Whoops! Edited. $\endgroup$ – TheSimpliFire Mar 11 '18 at 19:56
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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.

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

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  • $\begingroup$ Note that you have three minus signs so $D$ must be negative. $\endgroup$ – TheSimpliFire Mar 11 '18 at 19:57
  • $\begingroup$ @TheSimpliFire thanks $\endgroup$ – Netravat Pendsey Mar 12 '18 at 4:24
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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.

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