2

Finding a 3x3 matrix is easy, but how can I find the determinant of this 5x5 matrix?? I just need an example of the first couple steps to mimic

$A =$ $\begin{bmatrix} 7&1&9&-4&3\\0&-3&4&9&-6\\0&0&-6&-6&-9\\0&0&0&7&6\\0&0&0&0&2\end{bmatrix}$

then the $\det(A) = ?$

By the way, I did put the matrix in REF form and tried to multiply the diagonal and it didn't work at all.

I ended up getting

$\begin{bmatrix} 464,679,936&0&0&0&0\\0&-14,112&0&0&0\\0&0&-168&0&0\\0&0&0&14&0\\0&0&0&0&2\end{bmatrix}$

Shammy
  • 1,153
  • 2
    The determinant of an upper triangular matrix is just the product of the diagonal entries. (Expand along the first column.) – symplectomorphic Jul 04 '16 at 16:31
  • 1
    The matrix is upper triangular. Hence, its determinant is the product of the entries on the main diagonal. – Rodrigo de Azevedo Jul 04 '16 at 16:32
  • Thank you, I thought for some reason it had to be in REF – Shammy Jul 04 '16 at 16:33
  • Elementary row operations change the determinant. They do so in a predictable way so you can keep track, but you definitely can't just put it in REF first and then find the determinant and expect it to be the same as the determinant of the original matrix. –  Jul 04 '16 at 16:40
  • It is just $7^2\cdot 6^2$ because of a quite known property of triangular matrix. – Piquito Jul 04 '16 at 17:00

1 Answers1

1

Developing with respect the last row, you get that the determinant equals

$$2\cdot\begin{vmatrix}7&1&9&\!-4\\0&\!-3&4&9\\0&0&\!-6&\!-6\\0&0&0&7\end{vmatrix}$$

Again deloping the last row we get

$$2\cdot7\cdot\begin{vmatrix}7&1&9\\0&\!-3&4\\0&0&\!-6\end{vmatrix}=2\cdot7\cdot(-6)\cdot\begin{vmatrix}7&1\\0&\!-3\end{vmatrix}=-84(-21)=1,764$$