# Row operations on a matrix

• Jun 16th 2008, 12:44 PM
michael11
Row operations on a matrix
A = [1 2 5 3 8]
[3 6 11 6 20]
[1 3 6 0 16]
[6 12 23 12 40]
[5 11 22 9 40]

Hi there i'm really struggling to find the determinant of this matrix using row operations
• Jun 16th 2008, 01:27 PM
flyingsquirrel
Hello
Quote:

Originally Posted by michael11
A = [1 2 5 3 8]
[3 6 11 6 20]
[1 3 6 0 16]
[6 12 23 12 40]
[5 11 22 9 40]

Hi there i'm really struggling to find the determinant of this matrix using row operations

Notice that the fourth row is almost the second one multiplied by two : Let's do $\displaystyle \mathrm{row}4=\mathrm{row}4-2\times \mathrm{row} 2$ (it'll make appear "a lot" of 0) :

$\displaystyle D=\left| \begin{tabular}{ccccc} 1&2&5&3&8\\ 3&6&11&6&20\\ 1&3&6&0&16\\ 0&0&1&0&0\\ 5&11&22&9&40\\ \end{tabular} \right|=-1\times\left| \begin{tabular}{ccccc} 1&2&3&8\\ 3&6&6&20\\ 1&3&0&16\\ 5&11&9&40\\ \end{tabular} \right|$

Now let's do $\displaystyle \mathrm{row}4=\mathrm{row}4-\mathrm{row}3-\mathrm{row}2-\mathrm{row}1$ (same idea as before, it makes appear several 0) :

$\displaystyle D=-\left| \begin{tabular}{cccc} 1&2&3&8\\ 3&6&6&20\\ 1&3&0&16\\ 0&0&0&-4\\ \end{tabular} \right|=4\times\left| \begin{tabular}{ccc} 1&2&3\\ 3&6&6\\ 1&3&0\\ \end{tabular} \right|$

Can you go on ?
• Jun 16th 2008, 02:00 PM
michael11
hi,

i tried to finish it and i got detA = -3408

don't think thats right some how
• Jun 16th 2008, 02:07 PM
flyingsquirrel
Wooo, that's a big number. :D

How did you expand $\displaystyle D=4\times\left| \begin{tabular}{ccc} 1&2&3\\ 3&6&6\\ 1&3&0\\ \end{tabular} \right|$ ?
• Jun 16th 2008, 02:10 PM
michael11
I multiplied the 4 through, then found the determinant using the lattice method?

but something tells me that is wrong
• Jun 17th 2008, 12:09 AM
flyingsquirrel
I assume that the lattice method is Sarrus' rule ?

Repeat the first two columns :

$\displaystyle \begin{tabular}{|ccc|cc} {\color{red} 1}&{\color{green}2}&{\color{blue}3}&1&2\\ 3&{\color{red}6}&{\color{green}6}&{\color{blue}3}& 6\\ 1&3&{\color{red}0}&{\color{green}1}&{\color{blue}3 }\\ \end{tabular}$

Now multiply the red numbers together, multiply the green numbers together, multiply the blue numbers together and add these three products. Let's call the result $\displaystyle D_1$.

$\displaystyle \begin{tabular}{|ccc|cc} 1&2&{\color{magenta}3}&{\color{yellow}1}&{\color{c yan}2}\\ 3&{\color{magenta}6}&{\color{yellow}6}&{\color{cya n}3}&6\\ {\color{magenta}1}&{\color{yellow}3}&{\color{cyan} 0}&1&3\\ \end{tabular}$

Same thing : multiply together the numbers of the same color (except black :D) and add the three products. Again, let's call the result $\displaystyle D_2$.

As $\displaystyle D=4\times\left| \begin{tabular}{ccc} 1&2&3\\ 3&6&6\\ 1&3&0\\ \end{tabular} \right|$, the determinant is given by $\displaystyle D=4(D_1-D_2)$.

Hope that helps.
• Jun 17th 2008, 11:57 AM
Soroban
Hello, michael11!

I assumed that they wanted Row Operations all the way . . .

flyingsquirrel had: .$\displaystyle \left|\begin{array}{ccc} 1 & 2 & 3 \\ 3 & 6 & 6 \\ 1 & 3 & 0 \end{array}\right| \times 4$

$\displaystyle \begin{array}{c}\\ R_2 - 3R_1 \\ R_3-R_1\end{array}\left|\begin{array}{ccc}1 & 2 & 3 \\ 0 & 0 & \text{-}3 \\ 0 & 1 & \text{-}3 \end{array}\right| \times 4 \quad = \quad \left|\begin{array}{cc}1 & 2 \\ 0 & 1 \end{array} \right| \times 12$

$\displaystyle \begin{array}{c}R_1-2R_2 \\ \\ \end{array}\left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right| \times 12 \quad=\quad\boxed{ 12}$

Terrific job with the colors, flyingsquirrel!
.