linear algebra
Find the value(s) of a real parameter k so that the system of equations has infinitely many solutions.
$x+y+z=1$
$2x+ky+3z=-2$
$3x+5y+kz=-1$
Please help, I don't know how to go about this. Would you have to try every combination of the equations until you get a contradiction?
An underdetermined system is one in which there are more unknowns than equations. You can use induction on the number $n$ of unknowns; since there is always at most one of them, the base case $n=0$ is vacuously true.
Consider the final unknown $x_n$. If its coefficient in all equations is $0$, then one can give it an arbitrary value, and let all other unknowns be$~0$, to get infinitely many solutions (this is where homogeneity is used). Otherwise single out one equation in which $x_n$ has a nonzero coefficient. Use it to express $x_n$ as a linear combination of the other unknowns. Substitute this expression for $x_n$ into the remaining equations (this is equivalent to eliminating $x_n$ from the other equations by Gauss's method). The remaining equations form an underdetermined system: one less equation (the one singled out) and one less unknown ($x_n$); so by induction it has infinitely many solutions (for the remaining unknowns). For each of these solutions, build one that also gives $x_n$ by using the expression as linear combination of the other unknowns. One easily checks that this gives solutions of the original system.
One can write your system $A x = b$ as augmented matrix and bring it into row echelon form $$ \left[ \begin{array}{rrr|r} 2 & 4 & -2 & 0 \\ 3 & 5 & 0 & 1 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 2 & -1 & 0 \\ 3 & 5 & 0 & 1 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 2 & -1 & 0 \\ 0 & -1 & 3 & 1 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 2 & -1 & 0 \\ 0 & 1 & -3 & -1 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 0 & 5 & 2 \\ 0 & 1 & -3 & -1 \end{array} \right] $$ This translates back into $$ x + 5z = 2 \\ y - 3z = -1 $$ or $x = (2-5z, -1+3z, z)^T$, where $z \in \mathbb{R}$. So there are infinite many solutions.
From a geometric point of view, each equation defines an affine plane in $\mathbb{R}^3$, which is a plane, not necessarily including the origin. $$ (2, 4, -2) \cdot (x,y,z) = 0 \\ (3, 5, 0) \cdot (x, y, z) = 1 $$ The first plane has a normal vector $(2,4,-2)^T$ and includes the origin, the second plane has normal vector $(3,5,0)$ and is $1/\sqrt{3^2 + 5^2}$ away from the origin.
The solution of the system is the intersection of those two planes. And only the cases empty intersection, intersection is a line or intersection is a plane can happen. Here the intersection is a line.
The image shows the two intersecting planes, the intersection line, and the point $P$ which corresponds to the solution with $z = 0$.
Best Answer
Add the first and second and subtract the third equation we have: $(k-4)(y-z) = 0$. Thus we require $k = 4$ to have infinite solutions.