GRE multiple choice question: Solving a linear system 4 equations in 4 unknown.

Question

What is the quickest yet systematic way to solve this question?

Consider the system of linear equations:

$$w + 3x + 2y +2z = 0$$

$$w + 4x + y = 0$$

$$3w + 5x + 10y + 4z = 0$$

$$2w+ 5x + 5y + 6z = 0$$

with solutions of the form $(w,x,y,z)$, where $w,x,y,z$ are real. Which of the following statements is false:

A. The system is consistent.

B. The system has infinitely many solutions.

C. The sum of any two solutions is a solution.

D. $(-5,1,1,0)$ is a solution.

E. Every solution is a scalar multiple of $(-5,1,1,0)$

A related question is if we have a large $(4\times 4$ or $5\times 5$) matrix, and we can't tell if the rows/columns are linearly independent just by looking at them, how do we tell if the matrix is invertible or not?

Answer 1

Start with (D). Obviously, it's true. This also means that the determinant of this homogeneous system is zero.

Jump to (A). The system has at least one solution, so it's consistent. In fact all homogeneous systems are consistent because a vector of zeros is always a solution.

Jump to (B). Multiply the solution from (D) with $k$ and you will see that $(-5k,k,k,0)$ is also a solution. The number of solutions is infinite.

Jump to (C). If you have two different solutions $X,Y$ such that $AX=0,AY=0$, then $A(X+Y)=0$. So a linear combination of solutions is also a solution.

Jump to (E). You'll have to prove that the matrix rank is 3 (by Gaussian elimination, for example). With a 4x4 matrix it should be fairly simple exercise. This proves the fact that (E) is actually true.

Answer 2

The fastest way, subjectively speaking, is putting it in matrix form and reducing it. Putting the equations in matrix form gives us: $$ \begin{pmatrix} 1 & 3 & 2 & 2 \\ 1 & 4 & 1 & 0 \\ 3 & 5 & 10 & 4 \\ 2 & 5 & 5 & 6 \\ \end{pmatrix} $$

Which reduces to:$$ \begin{pmatrix} 1 & 0 & 5 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} $$

A: The system is consistent, which we could have checked even before reducing it since the system of equation is homogenous. So (A) can not be false.

B: Translating the matrix back into equation form we get $w=-5y$, $x=y$, $z=0$ and as we can see, $y$ is a free variable. Therefore, It has infinitely many solutions. so (B), can not be false either.

C: Since the equations are homogenous any linear combination of the solution is also a solution, so (C) is not false either.

D: $(-5,1,1,0)$ is indeed a solution. Therefore, (D) is not false either.

E: This one is not false either, since $w=-5y$, $x=y$, and $z=0$, the only solution set possible for the matrix is a multiple of $-5$, signified by $w=-5y$. the 1:1 ratio of $x$ to $y$ shows that the $x$ and $y$ component of the solution vectors must be a multiple of 1, and $z$ can not be anything but $0$. Therefore, All the solution sets to this matrix are indeed multiples of $(-5,1,1,0)$.

Given a $(m\times m)$ Matrix, showing that it is non-singular is enough to prove it is indeed invertible.