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what would the value of determinant of a matrix be if a specific entry changed?
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up vote
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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
linear-algebra matrices
edited Sep 7 at 15:23
Bernard
112k635102
asked Sep 7 at 14:01
B.B.
1467
add a comment |Â
up vote
10
down vote
favorite
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
linear-algebra matrices
edited Sep 7 at 15:23
Bernard
112k635102
asked Sep 7 at 14:01
B.B.
1467
Hint: Expand by cofactor.
– Yuta
Sep 7 at 14:07
@Yuta: Make me an illustrating answer. Thanks
– B.B.
Sep 7 at 14:18
add a comment |Â
up vote
10
down vote
favorite
up vote
10
down vote
favorite
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
linear-algebra matrices
edited Sep 7 at 15:23
Bernard
112k635102
asked Sep 7 at 14:01
B.B.
1467
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
linear-algebra matrices
edited Sep 7 at 15:23
Bernard
112k635102
asked Sep 7 at 14:01
B.B.
1467
edited Sep 7 at 15:23
Bernard
112k635102
edited Sep 7 at 15:23
Bernard
112k635102
edited Sep 7 at 15:23
Bernard
112k635102
112k635102
asked Sep 7 at 14:01
B.B.
1467
asked Sep 7 at 14:01
B.B.
1467
asked Sep 7 at 14:01
B.B.
1467
1467
Hint: Expand by cofactor.
– Yuta
Sep 7 at 14:07
@Yuta: Make me an illustrating answer. Thanks
– B.B.
Sep 7 at 14:18
add a comment |Â
Hint: Expand by cofactor.
– Yuta
Sep 7 at 14:07
@Yuta: Make me an illustrating answer. Thanks
– B.B.
Sep 7 at 14:18
Hint: Expand by cofactor.
– Yuta
Sep 7 at 14:07
Hint: Expand by cofactor.
– Yuta
Sep 7 at 14:07
@Yuta: Make me an illustrating answer. Thanks
– B.B.
Sep 7 at 14:18
@Yuta: Make me an illustrating answer. Thanks
– B.B.
Sep 7 at 14:18
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
18
down vote
accepted
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?
answered Sep 7 at 14:07
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
1
I did not notice this version.
– B.B.
Sep 7 at 14:13
1
So we should put that $0$s instead of$3$ and $4$? Am I right?
– B.B.
Sep 7 at 14:17
2
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
– Babelfish
Sep 7 at 14:20
Yes, both of you are right.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 7 at 14:44
2
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?
– Todd Wilcox
Sep 7 at 21:19
 |Â
show 1 more comment
up vote
6
down vote
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
answered Sep 7 at 14:28
Yuta
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
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 $x$.
– jpmc26
Sep 7 at 17:55
add a comment |Â
up vote
5
down vote
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).
answered Sep 7 at 15:18
farruhota
15.1k2734
Thanks for the extended version!+
– B.B.
Sep 8 at 10:34
You are welcome! Thanks for the good question! Good luck.
– farruhota
Sep 8 at 10:35
add a comment |Â
up vote
3
down vote
+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.
answered Sep 7 at 14:21
Anand Joshi
313
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Anand Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
18
down vote
accepted
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?
answered Sep 7 at 14:07
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
1
I did not notice this version.
– B.B.
Sep 7 at 14:13
1
So we should put that $0$s instead of$3$ and $4$? Am I right?
– B.B.
Sep 7 at 14:17
2
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
– Babelfish
Sep 7 at 14:20
Yes, both of you are right.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 7 at 14:44
2
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?
– Todd Wilcox
Sep 7 at 21:19
 |Â
show 1 more comment
up vote
18
down vote
accepted
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?
answered Sep 7 at 14:07
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
1
I did not notice this version.
– B.B.
Sep 7 at 14:13
1
So we should put that $0$s instead of$3$ and $4$? Am I right?
– B.B.
Sep 7 at 14:17
2
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
– Babelfish
Sep 7 at 14:20
Yes, both of you are right.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 7 at 14:44
2
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?
– Todd Wilcox
Sep 7 at 21:19
 |Â
show 1 more comment
up vote
18
down vote
accepted
up vote
18
down vote
accepted
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?
answered Sep 7 at 14:07
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
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?
answered Sep 7 at 14:07
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
edited Sep 7 at 15:53
answered Sep 7 at 14:07
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
answered Sep 7 at 14:07
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
answered Sep 7 at 14:07
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
33.5k32870
1
I did not notice this version.
– B.B.
Sep 7 at 14:13
1
So we should put that $0$s instead of$3$ and $4$? Am I right?
– B.B.
Sep 7 at 14:17
2
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
– Babelfish
Sep 7 at 14:20
Yes, both of you are right.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 7 at 14:44
2
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?
– Todd Wilcox
Sep 7 at 21:19
 |Â
show 1 more comment
1
I did not notice this version.
– B.B.
Sep 7 at 14:13
1
So we should put that $0$s instead of$3$ and $4$? Am I right?
– B.B.
Sep 7 at 14:17
2
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
– Babelfish
Sep 7 at 14:20
Yes, both of you are right.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 7 at 14:44
2
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?
– Todd Wilcox
Sep 7 at 21:19
1
1
I did not notice this version.
– B.B.
Sep 7 at 14:13
I did not notice this version.
– B.B.
Sep 7 at 14:13
1
1
So we should put that $0$s instead of$3$ and $4$? Am I right?
– B.B.
Sep 7 at 14:17
So we should put that $0$s instead of$3$ and $4$? Am I right?
– B.B.
Sep 7 at 14:17
2
2
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
– Babelfish
Sep 7 at 14:20
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
– Babelfish
Sep 7 at 14:20
Yes, both of you are right.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 7 at 14:44
Yes, both of you are right.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 7 at 14:44
2
2
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?
– Todd Wilcox
Sep 7 at 21:19
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?
– Todd Wilcox
Sep 7 at 21:19
 |Â
show 1 more comment
up vote
6
down vote
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
answered Sep 7 at 14:28
Yuta
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
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 $x$.
– jpmc26
Sep 7 at 17:55
add a comment |Â
up vote
6
down vote
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
answered Sep 7 at 14:28
Yuta
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
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 $x$.
– jpmc26
Sep 7 at 17:55
add a comment |Â
up vote
6
down vote
up vote
6
down vote
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
answered Sep 7 at 14:28
Yuta
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
answered Sep 7 at 14:28
Yuta
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Sep 7 at 14:28
Yuta
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Sep 7 at 14:28
Yuta
62929
answered Sep 7 at 14:28
Yuta
62929
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
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 $x$.
– jpmc26
Sep 7 at 17:55
add a comment |Â
1
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 $x$.
– jpmc26
Sep 7 at 17:55
1
1
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 $x$.
– jpmc26
Sep 7 at 17:55
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 $x$.
– jpmc26
Sep 7 at 17:55
add a comment |Â
up vote
5
down vote
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).
answered Sep 7 at 15:18
farruhota
15.1k2734
Thanks for the extended version!+
– B.B.
Sep 8 at 10:34
You are welcome! Thanks for the good question! Good luck.
– farruhota
Sep 8 at 10:35
add a comment |Â
up vote
5
down vote
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).
answered Sep 7 at 15:18
farruhota
15.1k2734
Thanks for the extended version!+
– B.B.
Sep 8 at 10:34
You are welcome! Thanks for the good question! Good luck.
– farruhota
Sep 8 at 10:35
add a comment |Â
up vote
5
down vote
up vote
5
down vote
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).
answered Sep 7 at 15:18
farruhota
15.1k2734
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).
answered Sep 7 at 15:18
farruhota
15.1k2734
answered Sep 7 at 15:18
farruhota
15.1k2734
answered Sep 7 at 15:18
farruhota
15.1k2734
answered Sep 7 at 15:18
farruhota
15.1k2734
15.1k2734
Thanks for the extended version!+
– B.B.
Sep 8 at 10:34
You are welcome! Thanks for the good question! Good luck.
– farruhota
Sep 8 at 10:35
add a comment |Â
Thanks for the extended version!+
– B.B.
Sep 8 at 10:34
You are welcome! Thanks for the good question! Good luck.
– farruhota
Sep 8 at 10:35
Thanks for the extended version!+
– B.B.
Sep 8 at 10:34
Thanks for the extended version!+
– B.B.
Sep 8 at 10:34
You are welcome! Thanks for the good question! Good luck.
– farruhota
Sep 8 at 10:35
You are welcome! Thanks for the good question! Good luck.
– farruhota
Sep 8 at 10:35
add a comment |Â
up vote
3
down vote
+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.
answered Sep 7 at 14:21
Anand Joshi
313
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Anand Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
3
down vote
+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.
answered Sep 7 at 14:21
Anand Joshi
313
New contributor
Anand Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
up vote
3
down vote
+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.
answered Sep 7 at 14:21
Anand Joshi
313
New contributor
Anand Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
+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.
answered Sep 7 at 14:21
Anand Joshi
313
New contributor
Anand Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Sep 7 at 14:26
answered Sep 7 at 14:21
Anand Joshi
313
New contributor
Anand Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Sep 7 at 14:21
Anand Joshi
313
answered Sep 7 at 14:21
Anand Joshi
313
313
New contributor
Anand Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Anand Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Anand Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
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Hint: Expand by cofactor.
– Yuta
Sep 7 at 14:07
@Yuta: Make me an illustrating answer. Thanks
– B.B.
Sep 7 at 14:18