62 lines
1.2 KiB
TeX
62 lines
1.2 KiB
TeX
\parindent=0pt
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Find the determinant of the following $4\times4$ matrix.
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$$D=\pmatrix{
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1&0&4&-1\cr
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-4&0&2&0\cr
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3&2&2&1\cr
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2&5&3&2\cr
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}$$
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\bigskip
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Solution: Clear out the 4th column.
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\bigskip
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$R_1+R_3\rightarrow R_1$
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$R_4-2R_3\rightarrow R_4$
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$$\pmatrix{
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4&2&6&0\cr
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-4&0&2&0\cr
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3&2&2&1\cr
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-4&1&-1&0\cr
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}$$
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\bigskip
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Now this is very important and I missed it the first time.
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We have to multiply the final result by $(-1)^{3+4}=-1$ due to
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the pivot of the cofactor expansion being on row 3, column 4.
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\bigskip
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Continuing with the cofactor expansion we now need to compute the
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determinant of the following minor.
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$$\pmatrix{
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4&2&6\cr
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-4&0&2\cr
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-4&1&-1\cr
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}$$
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\bigskip
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The cofactor expansion of the first row is
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$$
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4\bigg[(0)(-1)-(1)(2)\bigg]
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-2\bigg[(-4)(-1)-(-4)(2)\bigg]
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+6\bigg[(-4)(1)-(-4)(0)\bigg]
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$$
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which is
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$$4(-2)-2(12)+6(-4)=-56$$
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\bigskip
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Multiply this by $-1$ due to the $(-1)^{3+4}$ in the original pivot to obtain
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$$\det D=56$$
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\bigskip
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For some reason, when computing determinants I have a tendency to foul up
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when products involve zero.
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For example, I'll see $(-4)(0)$ and get the result $-4$.
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Don't forget, something times zero is zero!
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\bigskip
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{\it 38.tex}
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\end
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