Sistem: x+y+z=d, y+z+t=a, z+t+x=b, t+x+y=c
- Thread starter Adderall
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Let's rewrite the system of equations as follows:
\(\displaystyle 1x+1y+1z+0t=d\)
\(\displaystyle 0x+1y+1z+1t=a\)
\(\displaystyle 1x+0y+1z+1t=b\)
\(\displaystyle 1x+1y+0z+1t=c\)
Next, let's express this system of equations as an augmented matrix:
\(\displaystyle \left[\begin{array}{cccc|c}1 & 1 & 1 & 0 & d \\ 0 & 1 & 1 & 1 & a \\ 1 & 0 & 1 & 1 & b \\ 1 & 1 & 0 & 1 & c \end{array}\right]\)
Now we may use row operations to convert the matrix into reduced row echelon form, a process called Gauss–Jordan elimination.
Now, if we perform the row operation:
\(\displaystyle R_1-2R_2+R_3+R_4\)
We obtain:
\(\displaystyle \left[\begin{array}{cccc|c}3 & 0 & 0 & 0 & -2a+b+c+d \\ 0 & 1 & 1 & 1 & a \\ 1 & 0 & 1 & 1 & b \\ 1 & 1 & 0 & 1 & c \end{array}\right]\)
Next, performing the row operation:
\(\displaystyle \displaystyle \frac{1}{3}R_1\)
We obtain:
\(\displaystyle \left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & \dfrac{-2a+b+c+d}{3} \\ 0 & 1 & 1 & 1 & a \\ 1 & 0 & 1 & 1 & b \\ 1 & 1 & 0 & 1 & c \end{array}\right]\)
Next, we perform the row operation:
\(\displaystyle R_2-R_3+R_1\)
And we have:
\(\displaystyle \left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & \dfrac{-2a+b+c+d}{3} \\ 0 & 1 & 0 & 0 & \dfrac{a-2b+c+d}{3} \\ 1 & 0 & 1 & 1 & b \\ 1 & 1 & 0 & 1 & c \end{array}\right]\)
Can you continue?
\(\displaystyle 1x+1y+1z+0t=d\)
\(\displaystyle 0x+1y+1z+1t=a\)
\(\displaystyle 1x+0y+1z+1t=b\)
\(\displaystyle 1x+1y+0z+1t=c\)
Next, let's express this system of equations as an augmented matrix:
\(\displaystyle \left[\begin{array}{cccc|c}1 & 1 & 1 & 0 & d \\ 0 & 1 & 1 & 1 & a \\ 1 & 0 & 1 & 1 & b \\ 1 & 1 & 0 & 1 & c \end{array}\right]\)
Now we may use row operations to convert the matrix into reduced row echelon form, a process called Gauss–Jordan elimination.
Now, if we perform the row operation:
\(\displaystyle R_1-2R_2+R_3+R_4\)
We obtain:
\(\displaystyle \left[\begin{array}{cccc|c}3 & 0 & 0 & 0 & -2a+b+c+d \\ 0 & 1 & 1 & 1 & a \\ 1 & 0 & 1 & 1 & b \\ 1 & 1 & 0 & 1 & c \end{array}\right]\)
Next, performing the row operation:
\(\displaystyle \displaystyle \frac{1}{3}R_1\)
We obtain:
\(\displaystyle \left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & \dfrac{-2a+b+c+d}{3} \\ 0 & 1 & 1 & 1 & a \\ 1 & 0 & 1 & 1 & b \\ 1 & 1 & 0 & 1 & c \end{array}\right]\)
Next, we perform the row operation:
\(\displaystyle R_2-R_3+R_1\)
And we have:
\(\displaystyle \left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & \dfrac{-2a+b+c+d}{3} \\ 0 & 1 & 0 & 0 & \dfrac{a-2b+c+d}{3} \\ 1 & 0 & 1 & 1 & b \\ 1 & 1 & 0 & 1 & c \end{array}\right]\)
Can you continue?
HallsofIvy
Elite Member
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- Jan 27, 2012
- Messages
- 7,763
Ah, poor z. It never gets any respect.
If you don't like using matrices, here is a more "basic" way of doing the same thing:
You have the equations
x+ y+ z= d
y+ z+ t= a
z+ t+ x= b
t+ x+ y= c
Four equations in four unknowns (I presume that a, b, c, and d are "given" numbers and that x, y, z, and t are the "unknowns" we are to solve for.)
We start by "eliminating" one unknown at a time. I will, pretty arbitrarily, start by eliminating t. Subtracting the third equation from the second, -x+ y= a- b. Subtracting the fourth equation from the third -y+ z= b- c.
Since there was no t in the first equation we now have
x+ y+ z= d
-x+ y= a- b
-y+ z= b- c Three equations in three unknowns.
Now eliminate z by subtracting the third of these equations from the first: x+ 2y= d- b+ c
Since there was no z in the second equation we now have
x+ 2y= d- b+ c
-x+ y= a- b Two equations in two unknowns.
Now we can eliminate x by adding those two equations:
3y= a- 2b+ c+ d. One equation in one unknown.
We can solve that last equation for y by dividing both sides by 3.
Then "back substitute" that value of y in either of the two equations, x+ 2y= d- b+ c or -x+ y= a- b, to get an equation to solve for x.
Then substitute those two values for x and y in any of the previous three equations to get an equation to solve for z.
Finally, substitute those values for x, y, and z into any of the four original equation to get an equation to solve for t.
If you don't like using matrices, here is a more "basic" way of doing the same thing:
You have the equations
x+ y+ z= d
y+ z+ t= a
z+ t+ x= b
t+ x+ y= c
Four equations in four unknowns (I presume that a, b, c, and d are "given" numbers and that x, y, z, and t are the "unknowns" we are to solve for.)
We start by "eliminating" one unknown at a time. I will, pretty arbitrarily, start by eliminating t. Subtracting the third equation from the second, -x+ y= a- b. Subtracting the fourth equation from the third -y+ z= b- c.
Since there was no t in the first equation we now have
x+ y+ z= d
-x+ y= a- b
-y+ z= b- c Three equations in three unknowns.
Now eliminate z by subtracting the third of these equations from the first: x+ 2y= d- b+ c
Since there was no z in the second equation we now have
x+ 2y= d- b+ c
-x+ y= a- b Two equations in two unknowns.
Now we can eliminate x by adding those two equations:
3y= a- 2b+ c+ d. One equation in one unknown.
We can solve that last equation for y by dividing both sides by 3.
Then "back substitute" that value of y in either of the two equations, x+ 2y= d- b+ c or -x+ y= a- b, to get an equation to solve for x.
Then substitute those two values for x and y in any of the previous three equations to get an equation to solve for z.
Finally, substitute those values for x, y, and z into any of the four original equation to get an equation to solve for t.
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