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My question is; How can I find an equation relating $a,b$, and $c$ so that the linear system

$$\begin{cases}2x+y-z=a\\ x-2y-3z=b\\ -3x-y+2z=c\end{cases}$$

is consistent for any values of $a,b$, and $c$ that satisfy that equation.

Thanks,

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3 Answers 3

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Dear MAxcoder, if you're gonna read and use this text as a part of a homework exercise or an exam, you should indicate that you did so.

You effectively want an equation relating $a,b,c$ that is consistent with the three equations you wrote. In most cases, the 3 equations you wrote are independent, so it must be possible to derive the fourth equation involving $a,b,c$ only out of the three equations only.

How do you do it? Note that we want an equation that only includes $a,b,c$ - it doesn't include $x,y,z$. So we must eliminate all variables $x,y,z$. Clearly, the right combination of the 3 equations of yours is linear.

So take $a_1$ times the first equation, $a_2$ times the second, $a_3$ times the third, add these three terms, and you must get an expression such that all $x,y,z$ terms cancel. For a general collection of $3\times 3$ coefficients in front of $x,y,z$ in the three equations (I will avoid the term matrix because I don't feel that you know how to use it), that wouldn't be possible.

However, for the particular equations you wrote, it's possible. You need to find $\vec v=(a_1,a_2,a_3)$ such that the inner product of $\vec v$ with $(2,1,-3)$ as well as with $(1,-2,-1)$ as well as with $(-1,-3,2)$ vanishes. The three vectors are the columns from the 3 equations, and the three conditions are needed to cancel $x,y,z$.

In general, this would be an overdetermined system but in this case it's not. I can get $\vec v$ e.g. as the cross product of $(1,-2,-1)$ and $(-1,-3,2)$ which is $$ (1,-2,-1) \times (-1,-3,2) = (-2\times 2-(-1)(-3), (-1)(-1)-2\cdot 1, 1(-3)-(-2)(-1)) $$ That's $$(-4-3,1-2,-3-2) = (-7,-1,-5) $$ As a check, this vector is also orthogonal to $(2,1,-3)$ - the inner product is $-14-1+15=0$. So all conditions are satisfied. I forgot to say, a cross product is clever because $(\vec u\times \vec v)\cdot \vec u = 0$ automatically.

So now we must take the corresponding combination of the 3 equations you wrote and the result is $$ 0 = -7a -b-5c $$ in which the left hand sides, i.e. $x,y,z$, have completely cancelled, by design. Normal people would probably write it as $7a+b+5c=0$.

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Thanks a lot for your detailed answer. –  MAxcoder May 18 '11 at 13:42
    
It was a pleasure, @MAxcoder. Thanks for your bounty. ;-) –  Luboš Motl May 24 '11 at 7:25
    
you are welcome ;-) –  MAxcoder May 24 '11 at 7:31

Perform Gaussian elimination on the system. For each row with 3 zeros on the left, the corresponding equation on the right is a restriction on $a$, $b$, $c$.

Here are the details: $$ \begin{array}{rrrl} 2 & 1 & -1 & a \\ 1 & -2 & -3 & b \\ -3 & -1 & -2 & c \end{array} \quad\to\quad \begin{array}{rrrl} 1 & -2 & -3 & b \\ 2 & 1 & -1 & a \\ -3 & -1 & -2 & c \end{array} \quad\to\quad \begin{array}{rrrl} 1 & -2 & -3 & b \\ 0 & 5 & 5 & a-2b \\ 0 & -7 & -7 & c+3b \end{array} \quad\to\quad $$ $$ \begin{array}{rrrl} 1 & -2 & -3 & b \\ 0 & 35 & 35 & 7a-14b \\ 0 & -35 & -35 & 5c+15b \end{array} \quad\to\quad \begin{array}{rrrl} 1 & -2 & -3 & b \\ 0 & 35 & 35 & 7a-14b \\ 0 & 0& 0 &7a+b+5c \end{array} $$ Hence we must have $7a+b+5c=0$ if the original system has a solution.

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First of all,thanks for your answer.Can you give me more details? –  MAxcoder May 18 '11 at 12:44
    
@MAxcoder, do you know Gaussian elimination? –  lhf May 18 '11 at 12:46
    
yes,I know.But, I don't exactly understand the logic of the solution. –  MAxcoder May 18 '11 at 12:53
    
@MAxcoder, then do Gaussian elimination as suggested. Ask again if you get stuck. Hint: start by swapping equations 1 and 2. –  lhf May 18 '11 at 13:07

I don't know what tools you have available, but here is how I would consider this problem. If it uses terminology you do not know yet, you might come back to my answer later. This is intentionally not a direct path to the solution, rather the path I went though to solve the problem.

Write this system as a matrix equation:

$$ \left( \begin{array}{ccc} 2 & 1 & -1 \\ 1 & -2 & -3 \\ -3 & -1 & 2 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} a \\ b \\ c \end{array} \right) $$

If the matrix were invertible, we could multiply both sides by the inverse and get relations between $x,y,z$ and $a,b,c$.

$$ \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} 2 & 1 & -1 \\ 1 & -2 & -3 \\ -3 & -1 & 2 \end{array} \right)^{-1} \left( \begin{array}{c} a \\ b \\ c \end{array} \right) $$

But that matrix is not invertible: it has determinant 0, or you can see that the first column plus the third column equals the second column, so that the columns are not linearly independent. There has to be a dependence among the rows as well, but it is not as obvious. The "row reduced echelon form" is a good tool for finding column dependencies, so we use it on the transpose of the coefficient matrix (making the rows we want to relate into columns).

$$\text{rref} \left( \begin{array}{ccc} 2 & 1 & -3 \\ 1 & -2 & -1 \\ -1 & -3 & 2 \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & -7/5 \\ 0 & 1 & -1/5 \\ 0 & 0 & 0 \end{array} \right) $$

For solving $\alpha (2, 1, -1) + \beta(1,-2,-3) + \gamma(-3,-1,2) = 0$, this result means $\alpha - (7/5)\gamma = 0$ and $\beta - (1/5)\gamma = 0$. Since $\gamma$ in independent, make things easy by choosing $\gamma = 5$, which gives $\alpha = 7$ and $\beta = 1$.

So on the left-hand side of the original system, $$7(2x+y-z) + (x-2y-3z) + 5(-3x-y+2z) = 0.$$ Keeping track of the right-hand side gives the dependency you want, $7a + b + 5c = 0$. (It's interesting how row reduction and inner products as used by Luboš lead to the same result.)

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