what would the value of determinant of a matrix be if a specific entry changed?
what would the value of determinant of a matrix be if a specific entry changed?
What will the value of determinant of matrix $A=pmatrix1&3&4\5&2&a\6&-2&3$ be change if we change $a$ to $a+2$.
This is an easy problem because $|A|=-127+20a$ and if we did that changing, we would get $-87+20a$. My question here is:
Can we do this request by doing another way? Where does that $+40$ come from regarding the whole entries of the matrix? Thanks
$begingroup$
@Yuta: Make me an illustrating answer. Thanks
$endgroup$
– user582150
Sep 7 '18 at 14:18
4 Answers
4
The determinant, is after all a multilinear map on the column / row vectors.
Therefore,
$$
detbeginpmatrix
1&3&4 \ 5&2&(a+2) \6&-2&3
endpmatrix
=
detbeginpmatrix
1&3&4 \ 5&2&a \6&-2&3
endpmatrix
+detbeginpmatrix
1&3&0 \ 5&2&2 \6&-2&0
endpmatrix
$$
The determinant of the second matrix is?
If I changed $2$ to $3$ or $4$ or $b$, the answer would be?
$begingroup$
I did not notice this version.
$endgroup$
– user582150
Sep 7 '18 at 14:13
$begingroup$
So we should put that $0$s instead of$3$ and $4$? Am I right?
$endgroup$
– user582150
Sep 7 '18 at 14:17
$begingroup$
Alternatively, $ detbeginpmatrix 1&3&4 \ 5&2&(a+2) \6&-2&3 endpmatrix = detbeginpmatrix 1&3&0 \ 5&2&a \6&-2&0 endpmatrix +detbeginpmatrix 1&3&4 \ 5&2&2 \6&-2&3 endpmatrix $, then you can expand for the last column
$endgroup$
– Babelfish
Sep 7 '18 at 14:20
$begingroup$
Yes, both of you are right.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 7 '18 at 14:44
$begingroup$
What about $$ detbeginpmatrix 1&3&4 \ 5&2&(a+2) \6&-2&3 endpmatrix stackrel?= detbeginpmatrix 1&3&4 \ 5&2&a \6&-2&3 endpmatrix +detbeginpmatrix 1&3&4 \ 0&0&2 \6&-2&3 endpmatrix $$ ? Is that also equivalent?
$endgroup$
– Todd Wilcox
Sep 7 '18 at 21:19
Expand the $|A|$ along column 3.
beginalign
|A|&=
beginvmatrix
1&3&4\
5&2&a\
6&-2&3
endvmatrix \
&=
4
beginvmatrix
5&2\
6&-2
endvmatrix-a
beginvmatrix
1&3\
6&-2
endvmatrix+3
beginvmatrix
1&3\
5&2
endvmatrix \
&=4(-22)-a(-20)+3(-13) \
&=-127+20a
endalign
$begingroup$
The final step would be to substitute $a$ with $a+2$, of course. We could potentially make the logic more clear if we instead computed the determinant using some third variable, such as $x$.
$endgroup$
– jpmc26
Sep 7 '18 at 17:55
The property of the matrix (in general, for any size $ntimes n$):
$$beginvmatrix
a+b&c+d \ e&f
endvmatrix
=
beginvmatrix
a&c \ e&f
endvmatrix+beginvmatrix
b&d \ e&f
endvmatrix;\
beginvmatrix
a+b&c \ d+e&f
endvmatrix
=
beginvmatrix
a&c \ d&f
endvmatrix+beginvmatrix
b&c \ e&f
endvmatrix;$$
Hence:
$$beginvmatrix
1&3&4 \ 5&2&(a+2) \6&-2&3
endvmatrix
=beginvmatrix
1&3&c+(4-c) \ 5&2&a+2 \6&-2&d+(3-d)
endvmatrix=\
beginvmatrix
1&3&c \ 5&2&a \6&-2&d
endvmatrix+beginvmatrix
1&3&4-c \ 5&2&2 \6&-2&3-d
endvmatrix
$$
For easy calculation of determinant you can create as many zeros as possible (especially in rows/columns).
$begingroup$
Thanks for the extended version!+
$endgroup$
– user582150
Sep 8 '18 at 10:34
$begingroup$
You are welcome! Thanks for the good question! Good luck.
$endgroup$
– farruhota
Sep 8 '18 at 10:35
+40 comes from the cofactor of the element A(2,3). It is calculated as:
$$C_2,3 = (-1)^2+3 beginvmatrix
1 & 3\
6 & -2
endvmatrix = 20$$
When multiplied with $2$, it becomes 40.
Thanks for contributing an answer to Mathematics Stack Exchange!
But avoid …
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Required, but never shown
Required, but never shown
By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.
$begingroup$
Hint: Expand by cofactor.
$endgroup$
– Yuta
Sep 7 '18 at 14:07