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Is there any clean and pretty way to determine all pairs $(a,b)\in\mathbb R^2$ that make the following systems of linear equations equivalent: $$ S :\left\{ \begin{array}{l} x+ay+z+t=0 \\ 2x-y+bz-t=0 \end{array}\right. \quad S' :\left\{ \begin{array}{l} bx+4z=0 \\ ax-2y+(b-1)z-2t=0 \end{array}\right. $$ I've tried to discuss the system with equations from $S$ and $S'$ but this is infuriating. Thanks in advance.

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I know that I've seen exactly this before ... –  Hagen von Eitzen Jan 6 '13 at 17:35
    
I just searched the site and haven't found anything similar –  V. Galerkin Jan 6 '13 at 17:40
    
If I remember correctly, a more or less similar question appeared a few days ago, but was quickly deleted after being posted. –  user1551 Jan 6 '13 at 17:42
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3 Answers 3

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The coefficient matrices for the two systems of equations are $$ A=\begin{pmatrix} 1&a&1&1\\ 2&-1&b&-1 \end{pmatrix}, \ B=\begin{pmatrix} b&0&4&0\\ a&-2&b-1&-2 \end{pmatrix}. $$ It is easy to see that the two rows of $A$ are linearly independent and the two rows of $B$ are also linearly independent, regardless of the values of $a$ and $b$. Therefore, the two systems of equations are equivalent if and only if the two rows in $B$ lie inside the row space of $A$. Now the first row of $B$ is a linear combination of the two rows of $A$ if and only if every $3\times3$ submatrix of $$ C=\begin{pmatrix} 1&a&1&1\\ 2&-1&b&-1\\ b&0&4&0 \end{pmatrix} $$ has zero determinant. By deleting the third column of $C$, the determinant vanishes if and only if $a=1$. For this $a$, the other three determinants (obtained by deleting resp. the 1st, 2nd and 4th column of $C$) are $0$ and $\pm(b-3)(b+4)$. Hence they are all zero if and only if $b=3$ or $b=-4$. Finally, for these $a,b$, the second row of $B$ is equal to $(-1,1)A$. Hence we conclude that $S$ and $S'$ are equivalent if $a=1$ and $b=3$ or $-4$.

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By adding and subtracting, $S$ is equivalent to $$S'':\begin{cases}3x+(a-1)y+(b+1)z=0\\x-(a+1)y+(b-1)z-2t=0.\end{cases}$$ Comparing witth $S'$ strongly suggests $a=1$, $b=3$. Can you see why $S_1$ and $S_2$ are not equivalent for any other choice?

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For a given $(a,b)$, you have two equations in four unknowns $x,y,z,t$, so you should be able to express any two in terms of the other two. Working with $S'$ we note that $y$ and $t$ do not occur in the first, so let's solve for $y,z$. The first gives $z=-\frac b4 x$. The second gives $y=\frac 12(ax+(b-1)z)-2t)=\frac 12((a-\frac {-b(b-1)}4)x-2t)$. Now solve the first set for $y,z$ the same way and equate the coefficients. The first thing to notice is that the coefficient linking $y$ and $t$ is $-1$, so you must have $a=1$ in $S$. That just leaves $b$.

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