how to find determinant of Matrices A = {{1,-2,3}{2,1,-1}{1,0,2}} with Elementary Row Operations methode

Question

goformit100 · Answer

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ganeshie8 · Answer

$$

 \left| \begin{array}{ccc}
1 & -2  & 3 \
2 & 1  & -1 \
1 & 0 & 2 \end{array} \right|

$$

ganeshie8 · Answer

since 0 is there in 32, lets try to expand the determinant wid that column, 
before that, lets make another element in that column 0

anonymous · Answer

http://www.wikihow.com/Find-the-Determinant-of-a-3X3-Matrix If u want to learn to do it in only 6 steps

ganeshie8 · Answer

R1+2R2

$$

 \left| \begin{array}{ccc}
5 & 0  & 1 \
2 & 1  & -1 \
1 & 0 & 2 \end{array} \right|

= 10 - 1  = 9
$$

anonymous · Answer

i mean the  Elementary Row Operations method :)

ganeshie8 · Answer

yup ! we did one elementary row operation
R1+2R2

ganeshie8 · Answer

you want to convert the whole matrix to upper triangular matrix, and multiple the diagonal is it ?

anonymous · Answer

yaph that's what i mean :)

ganeshie8 · Answer

cool, then convert it to upper triangular matrix ? :)

ganeshie8 · Answer

$
\left| \begin{array}{ccc}
1 & -2  & 3 \
2 & 1  & -1 \
1 & 0 & 2 \end{array} \right|
$

ganeshie8 · Answer

R2-2R1
R3-R1

$
\left| \begin{array}{ccc}
1 & -2  & 3 \
0 & 5  & -7 \
0 & 2 & -1 \end{array} \right|
$

ganeshie8 · Answer

wat next ?

anonymous · Answer

i don't understand hoe to make it into upper triangular matrice -_-

ganeshie8 · Answer

so far, u okays ? :)

anonymous · Answer

okay so far i understand )

anonymous · Answer

what next ?

ganeshie8 · Answer

good :), upper triangular means,  the matrix must have 0s in the lower triangle region

like below

$
\left| \begin{array}{ccc}
a & b  & c \
0 & d  & e \
0 & 0 & f \end{array} \right|
$

ganeshie8 · Answer

our matrix still missing one 0, right ? 
at 32 position we have 2, lets make it 0

anonymous · Answer

yaph there is still 1 missing zero
okay so far so good :)

ganeshie8 · Answer

$
\left| \begin{array}{ccc}
1 & -2  & 3 \
0 & 5  & -7 \
0 & 2 & -1 \end{array} \right|

$

ganeshie8 · Answer

how to make it 0 ?
will below work ?
R3 - (2/5)R2

ganeshie8 · Answer

R3 - (2/5)R2

$
\left| \begin{array}{ccc}
1 & -2  & 3 \
0 & 5  & -7 \
0 & 0 & 9/5 \end{array} \right|

$

ganeshie8 · Answer

now its complete! to find det, just multiply the diagonal elements

anonymous · Answer

wow you're so cool :D thank you very much 
i really confused how to solve this before, and another question
when we make the equation like R1+2R2 are we make it by trying and error or there is special terms ?

ganeshie8 · Answer

np :)

no trial and error, matlab does this in a systematic way

ganeshie8 · Answer

$

\left| \begin{array}{ccc}
1 & -2  & 3 \
2 & 1  & -1 \
1 & 0 & 2 \end{array} \right|
$

ganeshie8 · Answer

thats the given matrix, 
First i want to make 21 element 0, i see that 2 is sitting there. so, to make it 0 using row operations, i have to R2-2R1

ganeshie8 · Answer

Next i want to make 31 element also 0, i see that 1 is sitting there. so, to make it 0 using row operations, i have to do R2-R1

ganeshie8 · Answer

Notice that, I am subtracting multiples of R1 in both cases; 11 element is our pivot when making 21 and 31 elements 0

anonymous · Answer

i see so we find the equation by seeing the 21, 31 and 32 right ?
okay thank you :)

ganeshie8 · Answer

exactly !, and in that order