# Determinant and linear system

• July 1st 2012, 01:18 PM
loui1410
Determinant and linear system
Let A be a 4x4 matrix and Ax = b such that x = (x1, x2, x3, x4) and b = (b1, b2, b3, b4).
For each i let us mark the i-column of A with ai.
It is given that b = a2 + 2a3 + 3a4.
(1) Prove, using Cramer's rule or any other way, that if det(A) does not equal zero then the linear system has one solution in which x1 = 0.
(2) Is it true that there is a solution to the linear system in which x1 = 0 also when det(A) = 0? Prove or refute.

I already solved (1), but have no idea how to prove (2). Any suggestions?
• July 1st 2012, 06:40 PM
HallsofIvy
Re: Determinant and linear system
Why not just make up an example of four equations in four variables in which $x_1= 0$?
Suppose $x_1= 0$, $x_2= 1$, $x_2= 1$, $x_4= 1$
what are
$2x_1+ x_2- 3x_3+ x_4$
$x_1- 3x_2+ 2x_3- x_4$
$x_1+ 2x_2- x_3- x_4$
$2x_1- 3x_2+ 3x_3+ x_4$?

What four equations does that give you? What linear system and determinant?
• July 2nd 2012, 12:58 AM
loui1410
Re: Determinant and linear system
But one example the meets the criteria is not a prove, is it?
• July 2nd 2012, 05:30 AM
HallsofIvy
Re: Determinant and linear system
One counter-example is enough to prove that a general statement is NOT true.
• July 2nd 2012, 11:47 AM
BobP
Re: Determinant and linear system
Before talking about (2), can I say that I dislike the wording of (1). The wording 'the linear system has one solution in which x1=0' suggests that there are solutions within which x1 is not equal to zero when in fact all solutions will have x1=0.

As to (2), if det A = 0 then the system of equations will have either none or an infinite number of solutions. The situation, as far as Cramer's rule is concerned, is that if one of the determinants A1,A2,A3,A4 is not zero then there will be no solutions, if all are zero there will be an infinite number of solutions. What has to be done then is to check out each of these determinants and prove or otherwise that each one is zero.
If it turns out that the system does have an infinite number of solutions then provided that A is not identically zero there will be at least one free variable, we can let that (or one of them) be x1 and we are free to make that zero.