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If there exists a solution for the linear equation $A x = \left( \begin{array}{lll} 1 \\ 1 \\ 1 \end{array} \right)$, then there also exists a solution for $Ax = \left( \begin{array}{lll} 1 \\ 2 \\ 1 \end{array} \right)$.

I have a list of statements, and I need to prove or disprove them.

Here is an example of such a statement.

I don't understand the format of the question.

I know a contradictory example to the statement I'm attaching is a square matrix $3 \times 3$ of all $1$'s. Why?

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2 Answers 2

up vote 4 down vote accepted

Here’s what this particular question is asking:

Suppose that $A$ is a $3\times 3$ matrix such that the equation $$Ax=\left[\matrix{1\\1\\1}\right]\tag{1}$$ has a solution; is it then necessarily true that the equation $$Ax=\left[\matrix{1\\2\\1}\right]\tag{2}$$ also has a solution?

The answer is no, because when $$A=\left[\matrix{1&1&1\\1&1&1\\1&1&1}\right]\;,$$ equation $(1)$ has a solution, but equation $(2)$ does not.

$$\left[\matrix{1&1&1\\1&1&1\\1&1&1}\right]\left[\matrix{1\\0\\0}\right]=\left[\matrix{1\\1\\1}\right]\;,$$ but

$$\left[\matrix{1&1&1\\1&1&1\\1&1&1}\right]\left[\matrix{x\\y\\z}\right]=\left[\matrix{x+y+z\\x+y+z\\x+y+z}\right]\;,$$ and no matter what $x,y$, and $z$ you choose, $$\left[\matrix{x+y+z\\x+y+z\\x+y+z}\right]\ne\left[\matrix{1\\2\\1}\right]\;,$$ because all three of its components are the same.


In general, if your asked to prove or disprove a statement of the form $$\text{if }P\text{ is true then }Q\text{ is true}\;,$$ you first have to decide whether it is true: if the mathematical objects mentioned in the statement are chosen in such a way that $P$ is true, does that automatically make $Q$ true as well? If the answer is yes, you’ll need to find a reason, a logical argument leading from the truth of $P$ to the truth of $Q$. For example, suppose that the statement is:

if the $3\times 3$ matrix $A$ is invertible, then the equation $$Ax=\left[\matrix{a\\b\\c}\right]$$ has a solution for every possible choice of real numbers $a,b,c$,

you can show that it’s true by pointing to the solution $$x=Ix=(A^{-1}A)x=A^{-1}(Ax)=A^{-1}\left[\matrix{a\\b\\c}\right]\;.$$

If the statement is not always true, your job is in some ways easier: you just have to demonstrate one example in which $P$ is true and $Q$ is not. That’s what I did in the specific example that you gave: the matrix $$A=\left[\matrix{1&1&1\\1&1&1\\1&1&1}\right]$$ shows that equation $(1)$ can have a solution without $(2)$ also having a solution.

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@Srivatsan: Ouch! Nothing like cutting and pasting the wrong matrix. Thanks. –  Brian M. Scott Dec 24 '11 at 22:31

They're sometimes not as clear about this as they should be: $A$ is considered to be a (specific, fixed, constant) 3x3 matrix and $\vec x$ is a variable vector. So when they say "If there exists a solution for the linear equation...then..." they mean, if $A$ is some 3x3 matrix for which there exists an $x$ making $Ax = [1,1,1],$ then..."

It's only considered true if it's always the case. Now consider the matrix you supplied, the matrix of all 1s. Notice that if we let $x_1,x_2,$ and $x_3$ be the entries of $\vec x,$ then calculating $A\vec x,$ we get $x_1+x_2+x_3$ in each entry. Thus, if $x_1=x_2=x_3=1/3,$ for example, or if $x_1=1$ and $x_2=x_3=0,$ then $A\vec x=[1,1,1].$ But no matter what the entries of $\vec x$ are, the entries of $A\vec x$ will always be equal since they're all $x_1+x_2+x_3.$ Thus there is no solution $\vec x$ to the equation $A\vec x=[1,2,1].$

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