Systems of equations

Mar 2019
196
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TTF area
3x + y - 2z = -7 equation one

-az = x + y + 6 equation 2

2x + 2y + z = 9 equation 3


Find the values of a for which the system is consistent. Give algebraic reasons for your answer and interpret the solution geometrically.

I know the system has a unique solution for a≠1/2.
Now the question is: what happens when a=1/2? In that case, the system has no solutions. But I don't understand it - can someone share their working + explanation? Thanks.
 

skipjack

Forum Staff
Dec 2006
21,478
2,470
Doubling the second equation and adding the result to the third equation leads to
(1 - 2a)z = 21, which reduces to 0 = 21 when a = 1/2.

When a = 1/2, equations 2 and 3 are the equations of (distinct) parallel planes. As these planes do not intersect, there are no solutions of the system of equations.
 
Dec 2015
1,078
166
Earth
Express z in terms of a , now find the domain.
1-2a≠0 or a≠1/2.
 

mathman

Forum Staff
May 2007
6,932
774
Express as matrix times (x,y,z)= (-7,6,9) and evaluate determinant as function of a. Det = 0 is bad a, otherwise good.
 
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Mar 2019
196
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I meant how do you solve it to get the values of a?? not matrix
 
Aug 2019
87
22
India
Equations are
$$ 3x + y -2z =-7 \\
-x -y -az = 6 \\
2x + 2y + z =9 $$

Now, for the above system of questions to have a unique solution we have to have
$$ \begin {vmatrix}

3 & 1& -3 \\
-1 & -1 & -a \\
2 & 2 & 1 \\

\end {vmatrix} \neq 0 \\

\implies 3 (-1 +2a) + 1( -2a +1) -2(-2+2) \neq 0 \\

6a - 3 -2a+1 \neq 0 \\
4a-2 \neq 0 \\
a \neq 1/2 $$

Now, when $a =1/2$ we will get an unconsistent system of equations.
 
Dec 2015
1,078
166
Earth
From equation1 and equation 3, you know the value of x+y, then plug it into equation 2.
Now leave z alone and find for which values of a the variable z makes no-sense.
The variable makes no-sense only when it equals \(\displaystyle \pm \infty\).:)
\(\displaystyle \frac{21}{1-2a}\neq \infty \) or \(\displaystyle 1-2a \neq 0\) or \(\displaystyle a\neq 1/2\).
 
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SDK

Sep 2016
797
541
USA
Generally speaking, you should NEVER compute the determinant of anything. The determinant is essentially a theoretical tool. It's almost never the best thing to compute, even when it is actually equivalent to the questions being asked, and in this case it isn't even "morally" the right thing to do. By that, I mean the question (in matrix form) is literally asking when a fixed vector is in the span of 3 other vectors. I can't imagine anything more absurd than computing the determinant of a matrix to determine this. Not only is this far more complicated than simple row reduction, but it doesn't even properly answer the question, since the equations can be consistent even if the determinant is 0 e.g. if the right-hand side is in the span on the columns for any choice of $A$. Instead, you just row reduce. For this example, you do the usual trick of augmenting the right hand side so that all row operations are carried out automatically. This gives you the matrix
\[
\begin{pmatrix}
3 & 1 & -2 & -7 \\
1 & 1 & a & -6 \\
2 & 2 & 1 & 9
\end{pmatrix}
\]
After row reducing, you are left with the equivalent matrix
\[
\begin{pmatrix}
3 & 1 & -2 & -7 \\
0 & \frac{2}{3} & a + \frac{2}{3} & -\frac{11}{3} \\
0 & 0 & 1 - 2a & 21
\end{pmatrix}
\]
From this matrix, it is trivial to see that all 3 equations are consistent if and only if the last equation is consistent. The last equation reads $(1-2a)z = 21$, which is consistent provided $1-2a \neq 0$.
 
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Mar 2019
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What does the working out look like equation style? NO matrix