anonymous
  • anonymous
how to find determinant of Matrices A = {{1,-2,3}{2,1,-1}{1,0,2}} with Elementary Row Operations methode
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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goformit100
  • goformit100
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ganeshie8
  • ganeshie8
\[ \left| \begin{array}{ccc} 1 & -2 & 3 \\ 2 & 1 & -1 \\ 1 & 0 & 2 \end{array} \right| \]
ganeshie8
  • ganeshie8
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

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anonymous
  • anonymous
http://www.wikihow.com/Find-the-Determinant-of-a-3X3-Matrix If u want to learn to do it in only 6 steps
ganeshie8
  • ganeshie8
R1+2R2 \[ \left| \begin{array}{ccc} 5 & 0 & 1 \\ 2 & 1 & -1 \\ 1 & 0 & 2 \end{array} \right| = 10 - 1 = 9 \]
anonymous
  • anonymous
i mean the Elementary Row Operations method :)
ganeshie8
  • ganeshie8
yup ! we did one elementary row operation R1+2R2
ganeshie8
  • ganeshie8
you want to convert the whole matrix to upper triangular matrix, and multiple the diagonal is it ?
anonymous
  • anonymous
yaph that's what i mean :)
ganeshie8
  • ganeshie8
cool, then convert it to upper triangular matrix ? :)
ganeshie8
  • ganeshie8
\( \left| \begin{array}{ccc} 1 & -2 & 3 \\ 2 & 1 & -1 \\ 1 & 0 & 2 \end{array} \right| \)
ganeshie8
  • ganeshie8
R2-2R1 R3-R1 \( \left| \begin{array}{ccc} 1 & -2 & 3 \\ 0 & 5 & -7 \\ 0 & 2 & -1 \end{array} \right| \)
ganeshie8
  • ganeshie8
wat next ?
anonymous
  • anonymous
i don't understand hoe to make it into upper triangular matrice -_-
ganeshie8
  • ganeshie8
so far, u okays ? :)
anonymous
  • anonymous
okay so far i understand )
anonymous
  • anonymous
what next ?
ganeshie8
  • ganeshie8
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
  • ganeshie8
our matrix still missing one 0, right ? at 32 position we have 2, lets make it 0
anonymous
  • anonymous
yaph there is still 1 missing zero okay so far so good :)
ganeshie8
  • ganeshie8
\( \left| \begin{array}{ccc} 1 & -2 & 3 \\ 0 & 5 & -7 \\ 0 & 2 & -1 \end{array} \right| \)
ganeshie8
  • ganeshie8
how to make it 0 ? will below work ? R3 - (2/5)R2
ganeshie8
  • ganeshie8
R3 - (2/5)R2 \( \left| \begin{array}{ccc} 1 & -2 & 3 \\ 0 & 5 & -7 \\ 0 & 0 & 9/5 \end{array} \right| \)
ganeshie8
  • ganeshie8
now its complete! to find det, just multiply the diagonal elements
anonymous
  • anonymous
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
  • ganeshie8
np :) no trial and error, matlab does this in a systematic way
ganeshie8
  • ganeshie8
\( \left| \begin{array}{ccc} 1 & -2 & 3 \\ 2 & 1 & -1 \\ 1 & 0 & 2 \end{array} \right| \)
ganeshie8
  • ganeshie8
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
  • ganeshie8
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
  • ganeshie8
Notice that, I am subtracting multiples of R1 in both cases; 11 element is our pivot when making 21 and 31 elements 0
anonymous
  • anonymous
i see so we find the equation by seeing the 21, 31 and 32 right ? okay thank you :)
ganeshie8
  • ganeshie8
exactly !, and in that order

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