Hey guys! Today, we're diving into a fun math problem: solving a system of equations using the echelon method and then figuring out the value of ( x y z ) 2 { (xyz)^2 } ( x yz ) 2
Setting Up the Equations
First, let's lay out the system of equations we need to solve:
{ β 2 x β 4 y β 5 z = 20 β 6 x β 2 y β 3 z = β 2 6 x β y + 5 z = β 13 {
\begin{cases}
-2x - 4y - 5z = 20 \\
-6x - 2y - 3z = -2 \\
6x - y + 5z = -13
\end{cases}
} β© β¨ β§ β β 2 x β 4 y β 5 z = 20 β 6 x β 2 y β 3 z = β 2 6 x β y + 5 z = β 13 β
Our mission is to find the values of x , y , and z that satisfy all three equations. Once we have those, we'll calculate ( x y z ) 2 { (xyz)^2 } ( x yz ) 2
The Echelon Method: Step-by-Step
The echelon method, also known as Gaussian elimination, is a systematic way to solve systems of linear equations. The main idea is to transform the system into an upper triangular form, which makes it easier to solve. Let's get started!
Step 1: Eliminating x from the Second and Third Equations
First, we want to eliminate x from the second and third equations. We can do this by performing row operations. Let's start by eliminating x from the second equation. Multiply the first equation by -3 and add it to the second equation:
Original second equation:
β 6 x β 2 y β 3 z = β 2 {
-6x - 2y - 3z = -2
} β 6 x β 2 y β 3 z = β 2
-3 times the first equation:
β 3 ( β 2 x β 4 y β 5 z ) = β 3 ( 20 ) β 6 x + 12 y + 15 z = β 60 {
-3(-2x - 4y - 5z) = -3(20) \Rightarrow 6x + 12y + 15z = -60
} β 3 ( β 2 x β 4 y β 5 z ) = β 3 ( 20 ) β 6 x + 12 y + 15 z = β 60
Adding these two equations gives us:
( β 6 x β 2 y β 3 z ) + ( 6 x + 12 y + 15 z ) = β 2 + ( β 60 ) β 10 y + 12 z = β 62 {
(-6x - 2y - 3z) + (6x + 12y + 15z) = -2 + (-60) \Rightarrow 10y + 12z = -62
} ( β 6 x β 2 y β 3 z ) + ( 6 x + 12 y + 15 z ) = β 2 + ( β 60 ) β 10 y + 12 z = β 62
So, our new second equation is:
10 y + 12 z = β 62 {
10y + 12z = -62
} 10 y + 12 z = β 62
Now, let's eliminate x from the third equation. To do this, simply add the first equation to the third equation:
Original third equation:
6 x β y + 5 z = β 13 {
6x - y + 5z = -13
} 6 x β y + 5 z = β 13
First equation:
β 2 x β 4 y β 5 z = 20 {
-2x - 4y - 5z = 20
} β 2 x β 4 y β 5 z = 20
Adding these two equations gives us:
( 6 x β y + 5 z ) + ( β 2 x β 4 y β 5 z ) = β 13 + 20 β 4 x β 5 y = 7 {
(6x - y + 5z) + (-2x - 4y - 5z) = -13 + 20 \Rightarrow 4x - 5y = 7
} ( 6 x β y + 5 z ) + ( β 2 x β 4 y β 5 z ) = β 13 + 20 β 4 x β 5 y = 7
So, our new third equation is:
β 5 y = 7 {
-5y = 7
} β 5 y = 7
Step 2: Simplifying the System
Now our system looks like this:
{ β 2 x β 4 y β 5 z = 20 10 y + 12 z = β 62 β 5 y = 7 {
\begin{cases}
-2x - 4y - 5z = 20 \\
10y + 12z = -62 \\
-5y = 7
\end{cases}
} β© β¨ β§ β β 2 x β 4 y β 5 z = 20 10 y + 12 z = β 62 β 5 y = 7 β
We can simplify the second equation by dividing by 2:
5 y + 6 z = β 31 {
5y + 6z = -31
} 5 y + 6 z = β 31
So the system becomes:
{ β 2 x β 4 y β 5 z = 20 5 y + 6 z = β 31 β 5 y = 7 {
\begin{cases}
-2x - 4y - 5z = 20 \\
5y + 6z = -31 \\
-5y = 7
\end{cases}
} β© β¨ β§ β β 2 x β 4 y β 5 z = 20 5 y + 6 z = β 31 β 5 y = 7 β
Step 3: Solving for y
From the third equation, we can directly solve for y :
β 5 y = 7 β y = β 7 5 {
-5y = 7 \Rightarrow y = -\frac{7}{5}
} β 5 y = 7 β y = β 5 7 β
Step 4: Solving for z
Now that we have y , we can plug it into the second equation to solve for z :
5 y + 6 z = β 31 β 5 ( β 7 5 ) + 6 z = β 31 β β 7 + 6 z = β 31 β 6 z = β 24 β z = β 4 {
5y + 6z = -31 \Rightarrow 5(-\frac{7}{5}) + 6z = -31 \Rightarrow -7 + 6z = -31 \Rightarrow 6z = -24 \Rightarrow z = -4
} 5 y + 6 z = β 31 β 5 ( β 5 7 β ) + 6 z = β 31 β β 7 + 6 z = β 31 β 6 z = β 24 β z = β 4
Step 5: Solving for x
Finally, we can plug the values of y and z into the first equation to solve for x :
β 2 x β 4 y β 5 z = 20 β β 2 x β 4 ( β 7 5 ) β 5 ( β 4 ) = 20 β β 2 x + 28 5 + 20 = 20 β β 2 x = β 28 5 β x = 14 5 {
-2x - 4y - 5z = 20 \Rightarrow -2x - 4(-\frac{7}{5}) - 5(-4) = 20 \Rightarrow -2x + \frac{28}{5} + 20 = 20 \Rightarrow -2x = -\frac{28}{5} \Rightarrow x = \frac{14}{5}
} β 2 x β 4 y β 5 z = 20 β β 2 x β 4 ( β 5 7 β ) β 5 ( β 4 ) = 20 β β 2 x + 5 28 β + 20 = 20 β β 2 x = β 5 28 β β x = 5 14 β
So, we have:
x = 14 5 , y = β 7 5 , z = β 4 {
x = \frac{14}{5}, \quad y = -\frac{7}{5}, \quad z = -4
} x = 5 14 β , y = β 5 7 β , z = β 4
Calculating ( x y z ) 2 { (xyz)^2 } ( x yz ) 2
Now that we have the values of x , y , and z , we can calculate ( x y z ) 2 { (xyz)^2 } ( x yz ) 2
( x y z ) 2 = ( 14 5 β
β 7 5 β
β 4 ) 2 = ( 14 β
7 β
4 25 ) 2 = ( 392 25 ) 2 = 153664 625 = 245.8624 {
(xyz)^2 = (\frac{14}{5} \cdot -\frac{7}{5} \cdot -4)^2 = (\frac{14 \cdot 7 \cdot 4}{25})^2 = (\frac{392}{25})^2 = \frac{153664}{625} = 245.8624
} ( x yz ) 2 = ( 5 14 β β
β 5 7 β β
β 4 ) 2 = ( 25 14 β
7 β
4 β ) 2 = ( 25 392 β ) 2 = 625 153664 β = 245.8624
Round to Integer Value
The problem gives options that are integers. So, let's verify our calculations and consider if there may be a simpler approach that leads to an integer solution. If a mistake was made in the calculation, we should correct it to align the result with one of the options provided.
Given the system:
{ β 2 x β 4 y β 5 z = 20 β 6 x β 2 y β 3 z = β 2 6 x β y + 5 z = β 13 {
\begin{cases}
-2x - 4y - 5z = 20 \\
-6x - 2y - 3z = -2 \\
6x - y + 5z = -13
\end{cases}
} β© β¨ β§ β β 2 x β 4 y β 5 z = 20 β 6 x β 2 y β 3 z = β 2 6 x β y + 5 z = β 13 β
And the values found:
x = 14 5 , y = β 7 5 , z = β 4 {
x = \frac{14}{5}, \quad y = -\frac{7}{5}, \quad z = -4
} x = 5 14 β , y = β 5 7 β , z = β 4
Let's substitute these values back into the original equations to check for correctness:
Equation 1:
β 2 ( 14 5 ) β 4 ( β 7 5 ) β 5 ( β 4 ) = β 28 5 + 28 5 + 20 = 20 {
-2(\frac{14}{5}) - 4(-\frac{7}{5}) - 5(-4) = -\frac{28}{5} + \frac{28}{5} + 20 = 20
} β 2 ( 5 14 β ) β 4 ( β 5 7 β ) β 5 ( β 4 ) = β 5 28 β + 5 28 β + 20 = 20
Equation 2:
β 6 ( 14 5 ) β 2 ( β 7 5 ) β 3 ( β 4 ) = β 84 5 + 14 5 + 12 = β 70 5 + 12 = β 14 + 12 = β 2 {
-6(\frac{14}{5}) - 2(-\frac{7}{5}) - 3(-4) = -\frac{84}{5} + \frac{14}{5} + 12 = -\frac{70}{5} + 12 = -14 + 12 = -2
} β 6 ( 5 14 β ) β 2 ( β 5 7 β ) β 3 ( β 4 ) = β 5 84 β + 5 14 β + 12 = β 5 70 β + 12 = β 14 + 12 = β 2
Equation 3:
6 ( 14 5 ) β ( β 7 5 ) + 5 ( β 4 ) = 84 5 + 7 5 β 20 = 91 5 β 20 = 91 5 β 100 5 = β 9 5 {
6(\frac{14}{5}) - (-\frac{7}{5}) + 5(-4) = \frac{84}{5} + \frac{7}{5} - 20 = \frac{91}{5} - 20 = \frac{91}{5} - \frac{100}{5} = -\frac{9}{5}
} 6 ( 5 14 β ) β ( β 5 7 β ) + 5 ( β 4 ) = 5 84 β + 5 7 β β 20 = 5 91 β β 20 = 5 91 β β 5 100 β = β 5 9 β
There appears to be a mistake. The values do not correctly satisfy Equation 3. Let's recalculate using a different approach.
We have the system:
{ β 2 x β 4 y β 5 z = 20 β 6 x β 2 y β 3 z = β 2 6 x β y + 5 z = β 13 {
\begin{cases}
-2x - 4y - 5z = 20 \\
-6x - 2y - 3z = -2 \\
6x - y + 5z = -13
\end{cases}
} β© β¨ β§ β β 2 x β 4 y β 5 z = 20 β 6 x β 2 y β 3 z = β 2 6 x β y + 5 z = β 13 β
Add equation 2 and 3:
( β 6 x β 2 y β 3 z ) + ( 6 x β y + 5 z ) = β 2 β 13 β β 3 y + 2 z = β 15 {
(-6x - 2y - 3z) + (6x - y + 5z) = -2 - 13 \Rightarrow -3y + 2z = -15
} ( β 6 x β 2 y β 3 z ) + ( 6 x β y + 5 z ) = β 2 β 13 β β 3 y + 2 z = β 15
Multiply equation 1 by 3 and subtract equation 2:
3 ( β 2 x β 4 y β 5 z ) β ( β 6 x β 2 y β 3 z ) = 3 ( 20 ) β ( β 2 ) β β 6 x β 12 y β 15 z + 6 x + 2 y + 3 z = 60 + 2 β β 10 y β 12 z = 62 {
3(-2x - 4y - 5z) - (-6x - 2y - 3z) = 3(20) - (-2) \Rightarrow -6x - 12y - 15z + 6x + 2y + 3z = 60 + 2 \Rightarrow -10y - 12z = 62
} 3 ( β 2 x β 4 y β 5 z ) β ( β 6 x β 2 y β 3 z ) = 3 ( 20 ) β ( β 2 ) β β 6 x β 12 y β 15 z + 6 x + 2 y + 3 z = 60 + 2 β β 10 y β 12 z = 62
Divide by -2:
5 y + 6 z = β 31 {
5y + 6z = -31
} 5 y + 6 z = β 31
Now we have:
{ β 3 y + 2 z = β 15 5 y + 6 z = β 31 {
\begin{cases}
-3y + 2z = -15 \\
5y + 6z = -31
\end{cases}
} { β 3 y + 2 z = β 15 5 y + 6 z = β 31 β
Multiply the first equation by 3:
β 9 y + 6 z = β 45 {
-9y + 6z = -45
} β 9 y + 6 z = β 45
Subtract this from the second equation:
( 5 y + 6 z ) β ( β 9 y + 6 z ) = β 31 β ( β 45 ) β 14 y = 14 β y = 1 {
(5y + 6z) - (-9y + 6z) = -31 - (-45) \Rightarrow 14y = 14 \Rightarrow y = 1
} ( 5 y + 6 z ) β ( β 9 y + 6 z ) = β 31 β ( β 45 ) β 14 y = 14 β y = 1
Now substitute y = 1 into
β 3 y + 2 z = β 15 β β 3 ( 1 ) + 2 z = β 15 β 2 z = β 12 β z = β 6 {
-3y + 2z = -15 \Rightarrow -3(1) + 2z = -15 \Rightarrow 2z = -12 \Rightarrow z = -6
} β 3 y + 2 z = β 15 β β 3 ( 1 ) + 2 z = β 15 β 2 z = β 12 β z = β 6
Now substitute y = 1 and z = -6 into the first original equation:
β 2 x β 4 ( 1 ) β 5 ( β 6 ) = 20 β β 2 x β 4 + 30 = 20 β β 2 x + 26 = 20 β β 2 x = β 6 β x = 3 {
-2x - 4(1) - 5(-6) = 20 \Rightarrow -2x - 4 + 30 = 20 \Rightarrow -2x + 26 = 20 \Rightarrow -2x = -6 \Rightarrow x = 3
} β 2 x β 4 ( 1 ) β 5 ( β 6 ) = 20 β β 2 x β 4 + 30 = 20 β β 2 x + 26 = 20 β β 2 x = β 6 β x = 3
So, now we have x = 3, y = 1, and z = -6.
Now, let's calculate (xyz)Β²:
( x y z ) 2 = ( 3 β
1 β
β 6 ) 2 = ( β 18 ) 2 = 324 {
(xyz)^2 = (3 \cdot 1 \cdot -6)^2 = (-18)^2 = 324
} ( x yz ) 2 = ( 3 β
1 β
β 6 ) 2 = ( β 18 ) 2 = 324
Since 324 isn't among the provided choices, let us reexamine the arithmetic and look for any prior mistakes.
Letβs use the values x = 3, y = 1, z = -6 and check in the original equations:
Equation 1:
β 2 ( 3 ) β 4 ( 1 ) β 5 ( β 6 ) = β 6 β 4 + 30 = 20 β {
-2(3) - 4(1) - 5(-6) = -6 - 4 + 30 = 20 \quad \checkmark
} β 2 ( 3 ) β 4 ( 1 ) β 5 ( β 6 ) = β 6 β 4 + 30 = 20 β β 6 ( 3 ) β 2 ( 1 ) β 3 ( β 6 ) = β 18 β 2 + 18 = β 2 β {
-6(3) - 2(1) - 3(-6) = -18 - 2 + 18 = -2 \quad \checkmark
} β 6 ( 3 ) β 2 ( 1 ) β 3 ( β 6 ) = β 18 β 2 + 18 = β 2 β 6 ( 3 ) β 1 + 5 ( β 6 ) = 18 β 1 β 30 = β 13 β {
6(3) - 1 + 5(-6) = 18 - 1 - 30 = -13 \quad \checkmark
} 6 ( 3 ) β 1 + 5 ( β 6 ) = 18 β 1 β 30 = β 13 β
( x y z ) 2 = ( 3 β
1 β
β 6 ) 2 = ( β 18 ) 2 = 324 {
(xyz)^2 = (3 \cdot 1 \cdot -6)^2 = (-18)^2 = 324
} ( x yz ) 2 = ( 3 β
1 β
β 6 ) 2 = ( β 18 ) 2 = 324
There seems to be no other mistake. Therefore, none of the multiple-choice options are correct.
Final Answer
Thus, ( x y z ) 2 = 324 { (xyz)^2 = 324 } ( x yz ) 2 = 324
Given the options provided, it seems there may have been a slight error in the provided answer choices. Based on our calculations, the correct answer should be 324.
Keep practicing, and you'll nail these problems in no time! Happy math-ing!