Equation with one variable unknow how to solve easily
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I have
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| = 0$$
How can I solve this with a fast way? I thought I can do
$$x-2+3x-6=0 implies 4x=-8 implies x=-2$$
and I will continue with the other two, but I don't know if I am right.
linear-algebra differential-equations
edited Sep 17 at 5:48
Blue
44.8k868142
asked Sep 17 at 5:41
m.s
475
add a comment |Â
up vote
2
down vote
favorite
I have
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| = 0$$
How can I solve this with a fast way? I thought I can do
$$x-2+3x-6=0 implies 4x=-8 implies x=-2$$
and I will continue with the other two, but I don't know if I am right.
linear-algebra differential-equations
edited Sep 17 at 5:48
Blue
44.8k868142
asked Sep 17 at 5:41
m.s
475
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| = 0$$
How can I solve this with a fast way? I thought I can do
$$x-2+3x-6=0 implies 4x=-8 implies x=-2$$
and I will continue with the other two, but I don't know if I am right.
linear-algebra differential-equations
edited Sep 17 at 5:48
Blue
44.8k868142
asked Sep 17 at 5:41
m.s
475
I have
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| = 0$$
How can I solve this with a fast way? I thought I can do
$$x-2+3x-6=0 implies 4x=-8 implies x=-2$$
and I will continue with the other two, but I don't know if I am right.
linear-algebra differential-equations
linear-algebra differential-equations
edited Sep 17 at 5:48
Blue
44.8k868142
asked Sep 17 at 5:41
m.s
475
edited Sep 17 at 5:48
Blue
44.8k868142
asked Sep 17 at 5:41
m.s
475
edited Sep 17 at 5:48
Blue
44.8k868142
edited Sep 17 at 5:48
Blue
44.8k868142
edited Sep 17 at 5:48
Blue
44.8k868142
44.8k868142
asked Sep 17 at 5:41
m.s
475
asked Sep 17 at 5:41
m.s
475
asked Sep 17 at 5:41
m.s
475
475
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
accepted
Elements of the first column are all multiples of $x$; elements in the third column are all multiples of $x-2$. You can factor the determinant thusly:
$$0 = left|beginarrayccc
phantom-x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| =
x(x-2) left|beginarrayccc
phantom-1 & -2 & phantom-3 \
phantom-2& phantom-0 & -1 \
-1& phantom-5 & phantom-1
endarrayright| =
37x(x-2) quadimpliesquad x=0,2$$
To add a bit of detail ...
Determinant rules allow you to factor-out common multipliers from an individual colunmn (or row):
$$left|beginarrayccc
pa & b & c \
pd & e & f \
ph & i & j
endarrayright| ;=; pleft|beginarrayccc
a & b & c \
d & e & f \
h & i & j
endarrayright|$$
The justification is immediately clear if you're familiar with the relationship between a determinant and the volume of the parallelepiped determined by the column vectors. But it also makes sense if you pay attention to how determinants are expanded. Here's a quick example with a $2times 2$:
$$left|beginarraycc pa & b \ pc & d endarrayright| = pacdot d -pccdot b = p(ad-cb) = pleft|beginarraycc a & b \ c & d endarrayright|$$
Every term in the expansion includes the multiplier, which can be factored-out.
Anyway ... For the problem in question, I factored-out $x$ and $x-2$ from the first and third columns. Then I calculated the value of the remaining determinant (just to make sure it wasn't zero, in which case $x$ could be anything).
answered Sep 17 at 5:57
Blue
44.8k868142
can you explain please how you find 37x(x-2)?
– m.s
Sep 17 at 6:25
3
The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
– Mathematician 42
Sep 17 at 6:27
coolmath.com/algebra/14-determinants-cramers-rule/… does this way is useful? For solve?
– m.s
Sep 17 at 8:34
@blue I find 3x3 but not with x inside. If I work as on the link I said will it work?
– m.s
Sep 17 at 8:38
@m.s: The link shows how to expand a $3times 3$ determinant, so it's helpful. If you use it on the original determinant in the problem, you could find yourself multiplying-out a bunch of terms that you'll then have to factor. My solution saves a lot of that effort by factoring-out common multipliers first. (This isn't always possible. It just so happens that the problem allows for it.) To get that the remaining numbers-only determinant has value $37$, use the technique described in that CoolMath link.
– Blue
Sep 17 at 8:47
 |Â
show 3 more comments
up vote
1
down vote
This is wrong:
$$x-2+3x-6=0 implies 4x=-8 implies x=-2 $$
This is the right way to do it
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| = 0$$
Apply $R3 rightarrow R3+R1$
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
0 & phantom-3 & 4x-8
endarrayright| = 0$$
Apply $R2rightarrow R2-2R1$
$$left|beginarrayccc
x & -2 & 3x-6 \
0 & phantom-4 & 14-7x \
0 & phantom-3 & 4x-8
endarrayright| = 0$$
$$xleft( 4(4x-8)-3(14-7x)right)=0$$
$$37x(x-2)=0 implies x=0,2$$
answered Sep 17 at 5:45
Deepesh Meena
4,46421025
yes sorry.But is the right way this?
– m.s
Sep 17 at 5:46
I can't understand what you are asking can you edit your question using mathjax
– Deepesh Meena
Sep 17 at 5:48
1
imgur.com/a/qCgZxLj . this is what i ask
– m.s
Sep 17 at 5:48
no , i cant get it ,where you going it.If you could continue i would be greatful
– m.s
Sep 17 at 5:55
see the updated naswer
– Deepesh Meena
Sep 17 at 5:58
 |Â
show 4 more comments
up vote
1
down vote
You can also use the Laplace formula on the second column and get
$$beginvmatrix
x & -2 & 3x-6 \
2x & 0 & 2-x \
-x & 5 & x-2
endvmatrix = 2 beginvmatrix 2x & 2-x \ -x & x-2 endvmatrix - 5 beginvmatrix x & 3x-6 \ 2x & 2-x endvmatrix=37x(x-2) $$
Hence $x in 0,2$.
answered Sep 17 at 9:26
Marvin
2,3783920
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Elements of the first column are all multiples of $x$; elements in the third column are all multiples of $x-2$. You can factor the determinant thusly:
$$0 = left|beginarrayccc
phantom-x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| =
x(x-2) left|beginarrayccc
phantom-1 & -2 & phantom-3 \
phantom-2& phantom-0 & -1 \
-1& phantom-5 & phantom-1
endarrayright| =
37x(x-2) quadimpliesquad x=0,2$$
To add a bit of detail ...
Determinant rules allow you to factor-out common multipliers from an individual colunmn (or row):
$$left|beginarrayccc
pa & b & c \
pd & e & f \
ph & i & j
endarrayright| ;=; pleft|beginarrayccc
a & b & c \
d & e & f \
h & i & j
endarrayright|$$
The justification is immediately clear if you're familiar with the relationship between a determinant and the volume of the parallelepiped determined by the column vectors. But it also makes sense if you pay attention to how determinants are expanded. Here's a quick example with a $2times 2$:
$$left|beginarraycc pa & b \ pc & d endarrayright| = pacdot d -pccdot b = p(ad-cb) = pleft|beginarraycc a & b \ c & d endarrayright|$$
Every term in the expansion includes the multiplier, which can be factored-out.
Anyway ... For the problem in question, I factored-out $x$ and $x-2$ from the first and third columns. Then I calculated the value of the remaining determinant (just to make sure it wasn't zero, in which case $x$ could be anything).
answered Sep 17 at 5:57
Blue
44.8k868142
can you explain please how you find 37x(x-2)?
– m.s
Sep 17 at 6:25
3
The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
– Mathematician 42
Sep 17 at 6:27
coolmath.com/algebra/14-determinants-cramers-rule/… does this way is useful? For solve?
– m.s
Sep 17 at 8:34
@blue I find 3x3 but not with x inside. If I work as on the link I said will it work?
– m.s
Sep 17 at 8:38
@m.s: The link shows how to expand a $3times 3$ determinant, so it's helpful. If you use it on the original determinant in the problem, you could find yourself multiplying-out a bunch of terms that you'll then have to factor. My solution saves a lot of that effort by factoring-out common multipliers first. (This isn't always possible. It just so happens that the problem allows for it.) To get that the remaining numbers-only determinant has value $37$, use the technique described in that CoolMath link.
– Blue
Sep 17 at 8:47
 |Â
show 3 more comments
up vote
5
down vote
accepted
Elements of the first column are all multiples of $x$; elements in the third column are all multiples of $x-2$. You can factor the determinant thusly:
$$0 = left|beginarrayccc
phantom-x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| =
x(x-2) left|beginarrayccc
phantom-1 & -2 & phantom-3 \
phantom-2& phantom-0 & -1 \
-1& phantom-5 & phantom-1
endarrayright| =
37x(x-2) quadimpliesquad x=0,2$$
To add a bit of detail ...
Determinant rules allow you to factor-out common multipliers from an individual colunmn (or row):
$$left|beginarrayccc
pa & b & c \
pd & e & f \
ph & i & j
endarrayright| ;=; pleft|beginarrayccc
a & b & c \
d & e & f \
h & i & j
endarrayright|$$
The justification is immediately clear if you're familiar with the relationship between a determinant and the volume of the parallelepiped determined by the column vectors. But it also makes sense if you pay attention to how determinants are expanded. Here's a quick example with a $2times 2$:
$$left|beginarraycc pa & b \ pc & d endarrayright| = pacdot d -pccdot b = p(ad-cb) = pleft|beginarraycc a & b \ c & d endarrayright|$$
Every term in the expansion includes the multiplier, which can be factored-out.
Anyway ... For the problem in question, I factored-out $x$ and $x-2$ from the first and third columns. Then I calculated the value of the remaining determinant (just to make sure it wasn't zero, in which case $x$ could be anything).
answered Sep 17 at 5:57
Blue
44.8k868142
can you explain please how you find 37x(x-2)?
– m.s
Sep 17 at 6:25
3
The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
– Mathematician 42
Sep 17 at 6:27
coolmath.com/algebra/14-determinants-cramers-rule/… does this way is useful? For solve?
– m.s
Sep 17 at 8:34
@blue I find 3x3 but not with x inside. If I work as on the link I said will it work?
– m.s
Sep 17 at 8:38
@m.s: The link shows how to expand a $3times 3$ determinant, so it's helpful. If you use it on the original determinant in the problem, you could find yourself multiplying-out a bunch of terms that you'll then have to factor. My solution saves a lot of that effort by factoring-out common multipliers first. (This isn't always possible. It just so happens that the problem allows for it.) To get that the remaining numbers-only determinant has value $37$, use the technique described in that CoolMath link.
– Blue
Sep 17 at 8:47
 |Â
show 3 more comments
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Elements of the first column are all multiples of $x$; elements in the third column are all multiples of $x-2$. You can factor the determinant thusly:
$$0 = left|beginarrayccc
phantom-x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| =
x(x-2) left|beginarrayccc
phantom-1 & -2 & phantom-3 \
phantom-2& phantom-0 & -1 \
-1& phantom-5 & phantom-1
endarrayright| =
37x(x-2) quadimpliesquad x=0,2$$
To add a bit of detail ...
Determinant rules allow you to factor-out common multipliers from an individual colunmn (or row):
$$left|beginarrayccc
pa & b & c \
pd & e & f \
ph & i & j
endarrayright| ;=; pleft|beginarrayccc
a & b & c \
d & e & f \
h & i & j
endarrayright|$$
The justification is immediately clear if you're familiar with the relationship between a determinant and the volume of the parallelepiped determined by the column vectors. But it also makes sense if you pay attention to how determinants are expanded. Here's a quick example with a $2times 2$:
$$left|beginarraycc pa & b \ pc & d endarrayright| = pacdot d -pccdot b = p(ad-cb) = pleft|beginarraycc a & b \ c & d endarrayright|$$
Every term in the expansion includes the multiplier, which can be factored-out.
Anyway ... For the problem in question, I factored-out $x$ and $x-2$ from the first and third columns. Then I calculated the value of the remaining determinant (just to make sure it wasn't zero, in which case $x$ could be anything).
answered Sep 17 at 5:57
Blue
44.8k868142
Elements of the first column are all multiples of $x$; elements in the third column are all multiples of $x-2$. You can factor the determinant thusly:
$$0 = left|beginarrayccc
phantom-x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| =
x(x-2) left|beginarrayccc
phantom-1 & -2 & phantom-3 \
phantom-2& phantom-0 & -1 \
-1& phantom-5 & phantom-1
endarrayright| =
37x(x-2) quadimpliesquad x=0,2$$
To add a bit of detail ...
Determinant rules allow you to factor-out common multipliers from an individual colunmn (or row):
$$left|beginarrayccc
pa & b & c \
pd & e & f \
ph & i & j
endarrayright| ;=; pleft|beginarrayccc
a & b & c \
d & e & f \
h & i & j
endarrayright|$$
The justification is immediately clear if you're familiar with the relationship between a determinant and the volume of the parallelepiped determined by the column vectors. But it also makes sense if you pay attention to how determinants are expanded. Here's a quick example with a $2times 2$:
$$left|beginarraycc pa & b \ pc & d endarrayright| = pacdot d -pccdot b = p(ad-cb) = pleft|beginarraycc a & b \ c & d endarrayright|$$
Every term in the expansion includes the multiplier, which can be factored-out.
Anyway ... For the problem in question, I factored-out $x$ and $x-2$ from the first and third columns. Then I calculated the value of the remaining determinant (just to make sure it wasn't zero, in which case $x$ could be anything).
answered Sep 17 at 5:57
Blue
44.8k868142
edited Sep 17 at 7:34
answered Sep 17 at 5:57
Blue
44.8k868142
answered Sep 17 at 5:57
Blue
44.8k868142
answered Sep 17 at 5:57
Blue
44.8k868142
44.8k868142
can you explain please how you find 37x(x-2)?
– m.s
Sep 17 at 6:25
3
The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
– Mathematician 42
Sep 17 at 6:27
coolmath.com/algebra/14-determinants-cramers-rule/… does this way is useful? For solve?
– m.s
Sep 17 at 8:34
@blue I find 3x3 but not with x inside. If I work as on the link I said will it work?
– m.s
Sep 17 at 8:38
@m.s: The link shows how to expand a $3times 3$ determinant, so it's helpful. If you use it on the original determinant in the problem, you could find yourself multiplying-out a bunch of terms that you'll then have to factor. My solution saves a lot of that effort by factoring-out common multipliers first. (This isn't always possible. It just so happens that the problem allows for it.) To get that the remaining numbers-only determinant has value $37$, use the technique described in that CoolMath link.
– Blue
Sep 17 at 8:47
 |Â
show 3 more comments
can you explain please how you find 37x(x-2)?
– m.s
Sep 17 at 6:25
3
The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
– Mathematician 42
Sep 17 at 6:27
coolmath.com/algebra/14-determinants-cramers-rule/… does this way is useful? For solve?
– m.s
Sep 17 at 8:34
@blue I find 3x3 but not with x inside. If I work as on the link I said will it work?
– m.s
Sep 17 at 8:38
@m.s: The link shows how to expand a $3times 3$ determinant, so it's helpful. If you use it on the original determinant in the problem, you could find yourself multiplying-out a bunch of terms that you'll then have to factor. My solution saves a lot of that effort by factoring-out common multipliers first. (This isn't always possible. It just so happens that the problem allows for it.) To get that the remaining numbers-only determinant has value $37$, use the technique described in that CoolMath link.
– Blue
Sep 17 at 8:47
can you explain please how you find 37x(x-2)?
– m.s
Sep 17 at 6:25
can you explain please how you find 37x(x-2)?
– m.s
Sep 17 at 6:25
3
3
The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
– Mathematician 42
Sep 17 at 6:27
The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
– Mathematician 42
Sep 17 at 6:27
coolmath.com/algebra/14-determinants-cramers-rule/… does this way is useful? For solve?
– m.s
Sep 17 at 8:34
coolmath.com/algebra/14-determinants-cramers-rule/… does this way is useful? For solve?
– m.s
Sep 17 at 8:34
@blue I find 3x3 but not with x inside. If I work as on the link I said will it work?
– m.s
Sep 17 at 8:38
@blue I find 3x3 but not with x inside. If I work as on the link I said will it work?
– m.s
Sep 17 at 8:38
@m.s: The link shows how to expand a $3times 3$ determinant, so it's helpful. If you use it on the original determinant in the problem, you could find yourself multiplying-out a bunch of terms that you'll then have to factor. My solution saves a lot of that effort by factoring-out common multipliers first. (This isn't always possible. It just so happens that the problem allows for it.) To get that the remaining numbers-only determinant has value $37$, use the technique described in that CoolMath link.
– Blue
Sep 17 at 8:47
@m.s: The link shows how to expand a $3times 3$ determinant, so it's helpful. If you use it on the original determinant in the problem, you could find yourself multiplying-out a bunch of terms that you'll then have to factor. My solution saves a lot of that effort by factoring-out common multipliers first. (This isn't always possible. It just so happens that the problem allows for it.) To get that the remaining numbers-only determinant has value $37$, use the technique described in that CoolMath link.
– Blue
Sep 17 at 8:47
 |Â
show 3 more comments
up vote
1
down vote
This is wrong:
$$x-2+3x-6=0 implies 4x=-8 implies x=-2 $$
This is the right way to do it
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| = 0$$
Apply $R3 rightarrow R3+R1$
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
0 & phantom-3 & 4x-8
endarrayright| = 0$$
Apply $R2rightarrow R2-2R1$
$$left|beginarrayccc
x & -2 & 3x-6 \
0 & phantom-4 & 14-7x \
0 & phantom-3 & 4x-8
endarrayright| = 0$$
$$xleft( 4(4x-8)-3(14-7x)right)=0$$
$$37x(x-2)=0 implies x=0,2$$
answered Sep 17 at 5:45
Deepesh Meena
4,46421025
yes sorry.But is the right way this?
– m.s
Sep 17 at 5:46
I can't understand what you are asking can you edit your question using mathjax
– Deepesh Meena
Sep 17 at 5:48
1
imgur.com/a/qCgZxLj . this is what i ask
– m.s
Sep 17 at 5:48
no , i cant get it ,where you going it.If you could continue i would be greatful
– m.s
Sep 17 at 5:55
see the updated naswer
– Deepesh Meena
Sep 17 at 5:58
 |Â
show 4 more comments
up vote
1
down vote
This is wrong:
$$x-2+3x-6=0 implies 4x=-8 implies x=-2 $$
This is the right way to do it
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| = 0$$
Apply $R3 rightarrow R3+R1$
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
0 & phantom-3 & 4x-8
endarrayright| = 0$$
Apply $R2rightarrow R2-2R1$
$$left|beginarrayccc
x & -2 & 3x-6 \
0 & phantom-4 & 14-7x \
0 & phantom-3 & 4x-8
endarrayright| = 0$$
$$xleft( 4(4x-8)-3(14-7x)right)=0$$
$$37x(x-2)=0 implies x=0,2$$
answered Sep 17 at 5:45
Deepesh Meena
4,46421025
yes sorry.But is the right way this?
– m.s
Sep 17 at 5:46
I can't understand what you are asking can you edit your question using mathjax
– Deepesh Meena
Sep 17 at 5:48
1
imgur.com/a/qCgZxLj . this is what i ask
– m.s
Sep 17 at 5:48
no , i cant get it ,where you going it.If you could continue i would be greatful
– m.s
Sep 17 at 5:55
see the updated naswer
– Deepesh Meena
Sep 17 at 5:58
 |Â
show 4 more comments
up vote
1
down vote
up vote
1
down vote
This is wrong:
$$x-2+3x-6=0 implies 4x=-8 implies x=-2 $$
This is the right way to do it
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| = 0$$
Apply $R3 rightarrow R3+R1$
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
0 & phantom-3 & 4x-8
endarrayright| = 0$$
Apply $R2rightarrow R2-2R1$
$$left|beginarrayccc
x & -2 & 3x-6 \
0 & phantom-4 & 14-7x \
0 & phantom-3 & 4x-8
endarrayright| = 0$$
$$xleft( 4(4x-8)-3(14-7x)right)=0$$
$$37x(x-2)=0 implies x=0,2$$
answered Sep 17 at 5:45
Deepesh Meena
4,46421025
This is wrong:
$$x-2+3x-6=0 implies 4x=-8 implies x=-2 $$
This is the right way to do it
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| = 0$$
Apply $R3 rightarrow R3+R1$
$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
0 & phantom-3 & 4x-8
endarrayright| = 0$$
Apply $R2rightarrow R2-2R1$
$$left|beginarrayccc
x & -2 & 3x-6 \
0 & phantom-4 & 14-7x \
0 & phantom-3 & 4x-8
endarrayright| = 0$$
$$xleft( 4(4x-8)-3(14-7x)right)=0$$
$$37x(x-2)=0 implies x=0,2$$
answered Sep 17 at 5:45
Deepesh Meena
4,46421025
edited Sep 17 at 5:57
answered Sep 17 at 5:45
Deepesh Meena
4,46421025
answered Sep 17 at 5:45
Deepesh Meena
4,46421025
answered Sep 17 at 5:45
Deepesh Meena
4,46421025
4,46421025
yes sorry.But is the right way this?
– m.s
Sep 17 at 5:46
I can't understand what you are asking can you edit your question using mathjax
– Deepesh Meena
Sep 17 at 5:48
1
imgur.com/a/qCgZxLj . this is what i ask
– m.s
Sep 17 at 5:48
no , i cant get it ,where you going it.If you could continue i would be greatful
– m.s
Sep 17 at 5:55
see the updated naswer
– Deepesh Meena
Sep 17 at 5:58
 |Â
show 4 more comments
yes sorry.But is the right way this?
– m.s
Sep 17 at 5:46
I can't understand what you are asking can you edit your question using mathjax
– Deepesh Meena
Sep 17 at 5:48
1
imgur.com/a/qCgZxLj . this is what i ask
– m.s
Sep 17 at 5:48
no , i cant get it ,where you going it.If you could continue i would be greatful
– m.s
Sep 17 at 5:55
see the updated naswer
– Deepesh Meena
Sep 17 at 5:58
yes sorry.But is the right way this?
– m.s
Sep 17 at 5:46
yes sorry.But is the right way this?
– m.s
Sep 17 at 5:46
I can't understand what you are asking can you edit your question using mathjax
– Deepesh Meena
Sep 17 at 5:48
I can't understand what you are asking can you edit your question using mathjax
– Deepesh Meena
Sep 17 at 5:48
1
1
imgur.com/a/qCgZxLj . this is what i ask
– m.s
Sep 17 at 5:48
imgur.com/a/qCgZxLj . this is what i ask
– m.s
Sep 17 at 5:48
no , i cant get it ,where you going it.If you could continue i would be greatful
– m.s
Sep 17 at 5:55
no , i cant get it ,where you going it.If you could continue i would be greatful
– m.s
Sep 17 at 5:55
see the updated naswer
– Deepesh Meena
Sep 17 at 5:58
see the updated naswer
– Deepesh Meena
Sep 17 at 5:58
 |Â
show 4 more comments
up vote
1
down vote
You can also use the Laplace formula on the second column and get
$$beginvmatrix
x & -2 & 3x-6 \
2x & 0 & 2-x \
-x & 5 & x-2
endvmatrix = 2 beginvmatrix 2x & 2-x \ -x & x-2 endvmatrix - 5 beginvmatrix x & 3x-6 \ 2x & 2-x endvmatrix=37x(x-2) $$
Hence $x in 0,2$.
answered Sep 17 at 9:26
Marvin
2,3783920
add a comment |Â
up vote
1
down vote
You can also use the Laplace formula on the second column and get
$$beginvmatrix
x & -2 & 3x-6 \
2x & 0 & 2-x \
-x & 5 & x-2
endvmatrix = 2 beginvmatrix 2x & 2-x \ -x & x-2 endvmatrix - 5 beginvmatrix x & 3x-6 \ 2x & 2-x endvmatrix=37x(x-2) $$
Hence $x in 0,2$.
answered Sep 17 at 9:26
Marvin
2,3783920
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You can also use the Laplace formula on the second column and get
$$beginvmatrix
x & -2 & 3x-6 \
2x & 0 & 2-x \
-x & 5 & x-2
endvmatrix = 2 beginvmatrix 2x & 2-x \ -x & x-2 endvmatrix - 5 beginvmatrix x & 3x-6 \ 2x & 2-x endvmatrix=37x(x-2) $$
Hence $x in 0,2$.
answered Sep 17 at 9:26
Marvin
2,3783920
You can also use the Laplace formula on the second column and get
$$beginvmatrix
x & -2 & 3x-6 \
2x & 0 & 2-x \
-x & 5 & x-2
endvmatrix = 2 beginvmatrix 2x & 2-x \ -x & x-2 endvmatrix - 5 beginvmatrix x & 3x-6 \ 2x & 2-x endvmatrix=37x(x-2) $$
Hence $x in 0,2$.
answered Sep 17 at 9:26
Marvin
2,3783920
answered Sep 17 at 9:26
Marvin
2,3783920
answered Sep 17 at 9:26
Marvin
2,3783920
answered Sep 17 at 9:26
Marvin
2,3783920
2,3783920
add a comment |Â
add a comment |Â
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