# Thread: 3.2.9 find 4x4 matrix det by row replacement

1. ## 3.2.9 find 4x4 matrix det by row replacement

$\textsf{ compute the determinant by cofactor expansions}$
$$\left[\begin{array}{rrrr} 6& \,0& \,0& \,5\\ 1& 7& 2&-5\\ 2& 0& 0& 0\\8& 3& 1& 8 \end{array}\right]\stackrel{a}{\sim} % \left[\begin{array}{rrrr} 0& \,0& \,0& \,5\\ 1& 7& 2&-5\\ 2& 0& 0& 0\\8& 3& 1& 8 \end{array}\right]\stackrel{b}{\sim} % \left[\begin{array}{rrrr} 1& \,7& \,2& \,0\\ 1& 7& 2&-5\\ 2& 0& 0& 0\\8& 3& 1& 8 \end{array}\right]\stackrel{c}{\sim} % \left[\begin{array}{rrrr} 2& \,0& \,0& \,0\\ -15& 1& 0& -21\\ 1& 7& 2& 0\\8& 3& 1& 8 \end{array}\right]$$

$$\stackrel{d}{\sim}\left[\begin{array}{rrrr} 2& \,0& \,0& \,0\\ 9& 10& 3&3\\ 1& 7& 2& 0\\8& 3& 1& 8 \end{array}\right]\stackrel{e}{\sim} % \left[\begin{array}{rrrr} 2& \,0& \,0& \,0\\ 9& 10& 3&3\\ 1& 7& 2& 0\\8& 3& 1& 8 \end{array}\right]\stackrel{f}{\sim} % \left[\begin{array}{rrrr} 2& \,0& \,0& \,0\\ 9& 10& 3&3\\ 1& 7& 2& 0\\ \frac{23}{2}& \frac{55}{2}& 8& 8 \end{array}\right]\stackrel{g}{\sim} \left[\begin{array}{rrrr} 2& \,0& \,0& \,0\\ \frac{75}{16}& \frac{-5}{16}& 0&0\\ 1& 7& 2& 0\\ \frac{23}{2}& \frac{55}{2}& 8& 8 \end{array}\right]$$
$\textbf{Solution:}\\$
$\textit{a.$-3\times R_3$add to$R_1$}$
$\textit{b.$1 \times R_2$add to$R_1$}$
$\textit{c. exchange$R_3$and$R_1$}$
$\textit{d.$-2\times R_4$add to$R_2$}$
$\textit{e.$3\times R_4$add to$R_2$}$
$\textit{f.$7/2\times R_3$add to$R_4$}$
$\textit{g.$-3/8\times R_4$add to$R_2$}$
$(2)(-5/16)(2)(8)=-10$
ok the bk answer is 10 so not sure where the ??? is

2. ## Re: 3.2.9 find 4x4 matrix det by row replacement

I thought this was to be done by co-factor expansion?

ie.

$\displaystyle \left | \begin{matrix} 6 & 0 & 0 & 5 \\ 1 & 7 & 2 & -5 \\ 2 & 0 & 0 & 0 \\ 8 & 3 & 1 & 8 \end{matrix} \right | = 6 \left | \begin{matrix} 7 & 2 & -5 \\ 0 & 0 & 0 \\ 3 & 1 & 8 \end{matrix} \right | - 5 \left | \begin{matrix} 1 & 7 & 2 \\ 2 & 0 & 0 \\ 8 & 3 & 1 \end{matrix} \right |$

etc.

And yes, the determinant is 10.

By the way... Step c) multiplies your determinant by -1. Just switching rows does this.

-Dan

3. ## Re: 3.2.9 find 4x4 matrix det by row replacement

Originally Posted by bigwave
$\textsf{ compute the determinant by cofactor expansions}$
$$\left[\begin{array}{rrrr} 6& \,0& \,0& \,5\\ 1& 7& 2&-5\\ 2& 0& 0& 0\\8& 3& 1& 8 \end{array}\right]\stackrel{a}{\sim} % \left[\begin{array}{rrrr} 0& \,0& \,0& \,5\\ 1& 7& 2&-5\\ 2& 0& 0& 0\\8& 3& 1& 8 \end{array}\right]\stackrel{b}{\sim} % \left[\begin{array}{rrrr} 1& \,7& \,2& \,0\\ 1& 7& 2&-5\\ 2& 0& 0& 0\\8& 3& 1& 8 \end{array}\right]\stackrel{c}{\sim} % \left[\begin{array}{rrrr} 2& \,0& \,0& \,0\\ -15& 1& 0& -21\\ 1& 7& 2& 0\\8& 3& 1& 8 \end{array}\right]$$

$$\stackrel{d}{\sim}\left[\begin{array}{rrrr} 2& \,0& \,0& \,0\\ 9& 10& 3&3\\ 1& 7& 2& 0\\8& 3& 1& 8 \end{array}\right]\stackrel{e}{\sim} % \left[\begin{array}{rrrr} 2& \,0& \,0& \,0\\ 9& 10& 3&3\\ 1& 7& 2& 0\\8& 3& 1& 8 \end{array}\right]\stackrel{f}{\sim} % \left[\begin{array}{rrrr} 2& \,0& \,0& \,0\\ 9& 10& 3&3\\ 1& 7& 2& 0\\ \frac{23}{2}& \frac{55}{2}& 8& 8 \end{array}\right]\stackrel{g}{\sim} \left[\begin{array}{rrrr} 2& \,0& \,0& \,0\\ \frac{75}{16}& \frac{-5}{16}& 0&0\\ 1& 7& 2& 0\\ \frac{23}{2}& \frac{55}{2}& 8& 8 \end{array}\right]$$
$\textbf{Solution:}\\$
$\textit{a.$-3\times R_3$add to$R_1$}$
$\textit{b.$1 \times R_2$add to$R_1$}$
$\textit{c. exchange$R_3$and$R_1$}$
$\textit{d.$-2\times R_4$add to$R_2$}$
$\textit{e.$3\times R_4$add to$R_2$}$
$\textit{f.$7/2\times R_3$add to$R_4$}$
$\textit{g.$-3/8\times R_4$add to$R_2$}$
$(2)(-5/16)(2)(8)=-10$
ok the bk answer is 10 so not sure where the ??? is
Interchanging two rows causes a sign change in the determinant.

4. ## Re: 3.2.9 find 4x4 matrix det by row replacement

Originally Posted by JaguarXJS
Interchanging two rows causes a sign change in the determinant.
Oh come one! Just evaluate the minor factors

$\displaystyle 6 \left | \begin{matrix} 7 & 2 & -5 \\ 0 & 0 & 0 \\ 3 & 1 & 8 \end{matrix}\right| =0$. This results from the all zero row.

Then
$\displaystyle - 5 \left | \begin{matrix} 1 & 7 & 2 \\ 2 & 0 & 0 \\ 8 & 3 & 1 \end{matrix} \right|=-5(-2)[(7)(1)-(2)(3)]=10$

5. ## Re: 3.2.9 find 4x4 matrix det by row replacement

Originally Posted by Plato
Oh come one! Just evaluate the minor factors

$\displaystyle 6 \left | \begin{matrix} 7 & 2 & -5 \\ 0 & 0 & 0 \\ 3 & 1 & 8 \end{matrix}\right| =0$. This results from the all zero row.

Then
$\displaystyle - 5 \left | \begin{matrix} 1 & 7 & 2 \\ 2 & 0 & 0 \\ 8 & 3 & 1 \end{matrix} \right|=-5(-2)[(7)(1)-(2)(3)]=10$
Agreed. I simply pointed that out to show him how to correct his work with the row reduction. And, of course, it's also as easy to expand along row 3...

-Dan