MHB Find determinant by row reduction in echelon

[karush]

$\textsf{a. Find the determinants by row reduction in echelon form.}$
$$\left|
\begin{array}{rrr}
1&5&-6\\ -1&-4&4 \\ -2&-7 & 9
\end{array}
\right|$$
ok i multiplied $r_1$ by 1 and added it to $r_2$ to get
$$\left|
\begin{array}{rrr}
1&5&-6\\ 0&1&-2 \\ -2&-7 & 9
\end{array}
\right|$$
but how do you get $0 \,0 \, r_3 c_3$
so it will be in echelon form?
the book answer is $3$

 

[skeeter]

$\textsf{a. Find the determinants by row reduction in echelon form.}$
$$\left|
\begin{array}{rrr}
1&5&-6\\ -1&-4&4 \\ -2&-7 & 9
\end{array}
\right|$$
ok i multiplied $r_1$ by 1 and added it to $r_2$ to get
$$\left|
\begin{array}{rrr}
1&5&-6\\ 0&1&-2 \\ -2&-7 & 9
\end{array}
\right|$$
but how do you get $0 \,0 \, r_3 c_3$
so it will be in echelon form?
the book answer is $3$

multiply $r_1$ by 2 and add to $r_3$ ...

$\begin{vmatrix}
1 & 5 & -6\\
0 & 1 & -2 \\
0 & 3 & -3
\end{vmatrix}$

multiply $r_2$ by -3 and add to $r_3$ ...

$\begin{vmatrix}
1 & 5 & -6\\
0 & 1 & -2 \\
0 & 0 & 3
\end{vmatrix}$

 

[karush]

one more

$\textsf{a. Find the determinants by row reduction in echelon form.}$
$$\left|
\begin{array}{rrr}
1&5&-3\\ 3&-3&3 \\ 2&13 &-7
\end{array}
\right|$$
$\textsf{ multiplied $r_1$ by $-2$ and added to $r_3$} $
$$\left|
\begin{array}{rrr}
1&5&-3\\ 3&-3&3 \\ 0&3 &-1
\end{array}
\right|$$
$\textsf{ multiplied $r_1$ by $6$ and added to $r_2$} $
$$\left|
\begin{array}{rrr}
1&5&-3\\ 0&0&-6 \\ 0&3 &-1
\end{array}
\right|$$
$\textsf{ exchange $r_3$ and $r_2$ change sign} $
$$-\left|
\begin{array}{rrr}
1&5&-3\\ 0&3&-1 \\ 0&0 &-6
\end{array}
\right|=18$$

 

[skeeter]

one more

$\textsf{a. Find the determinants by row reduction in echelon form.}$
$$\left|
\begin{array}{rrr}
1&5&-3\\ 3&-3&3 \\ 2&13 &-7
\end{array}
\right|$$
$\textsf{ multiplied $r_1$ by $-2$ and added to $r_3$} $
$$\left|
\begin{array}{rrr}
1&5&-3\\ 3&-3&3 \\ 0&3 &-1
\end{array}
\right|$$
$\textsf{ multiplied $r_1$ by $\color{red}{6}$ and added to $r_2$} $ ?
$$\left|
\begin{array}{rrr}
1&5&-3\\ 0&0&-6 \\ 0&3 &-1
\end{array}
\right|$$
$\textsf{ exchange $r_3$ and $r_2$ change sign} $
$$-\left|
\begin{array}{rrr}
1&5&-3\\ 0&3&-1 \\ 0&0 &-6
\end{array}
\right|=18$$

...

$\begin{bmatrix}
1 & 5 &-3 \\
3 & -3 &3 \\
2 & 13 & -7
\end{bmatrix}$

$-2r_1 + r_3 \to r_3$

$\begin{bmatrix}
1 & 5 &-3 \\
3 & -3 &3 \\
0 & 3 & -1
\end{bmatrix}$

$-3r_1 + r_2 \to r_2$

$\begin{bmatrix}
1 & 5 &-3 \\
0 & -18 &12 \\
0 & 3 & -1
\end{bmatrix}$

$\dfrac{1}{6}r_2 + r_3 \to r_3$

$\begin{bmatrix}
1 & 5 &-3 \\
0 & -18 &12 \\
0 & 0 & 1
\end{bmatrix}$

determinant of an upper triangular matrix is the product of the terms in the diagonal ...

$(1)(-18)(1) = -18$