QS-016 · Linear Algebra · Easy

Unique and Infinite Solutions in Ax=b

Let $A$ be an $m\times n$ matrix and $x$ and $b$ be column vectors. Which of the following is/are TRUE? 1. If $Ax=0$ has only the trivial solution $x=0$, then $Ax=b$ always has a unique solution. 2. There are vectors $b_1,b_2,b_3\in\mathbb{R}^m$ such that $Ax=b_1$ has no solutions, $Ax=b_2$ has one solution, and $Ax=b_3$ has infinitely many solutions. 3. If the columns of $A$ are linearly independent, then the equation $Ax=b$ has exactly one solution for every $b$. 4. If a system $Ax=b$ has more than one solution, then so does the system $Ax=0$.

Solution Let $A$ be an $m\times n$ matrix. Now, checking each statement: 1. If $Ax=0$ has only the trivial solution $x=0$, then any solution of $Ax=b$, if it exists, is unique. However, $Ax=b$ may have no solution for some $b$ when $b$ is not in the column space of $A$. So, A is false. 2. If $Ax=b$ is consistent and $Ax=0$ has a nontrivial solution, then $Ax=b$ has infinitely many solutions. If $Ax=0$ has only the trivial solution, then every consistent $Ax=b$ has exactly one solution. Therefore, with the same matrix $A$, we cannot have one solution for one $b$ and infinitely many solutions for another $b$. So, B is false. 3. Linear independence of the columns of $A$ guarantees uniqueness when a solution exists. It does not guarantee that a solution exists for every $b$. Existence for every $b$ would require the columns of $A$ to span $\mathbb{R}^m$. So, C is false. 4. Suppose $x_1$ and $x_2$ are two different solutions of $Ax=b$. Then, $$ Ax_1=b $$ and $$ Ax_2=b. $$ Subtracting, $$ Ax_1-Ax_2=b-b=0 $$ which gives $$ A(x_1-x_2)=0. $$ Since $x_1\neq x_2$, $$ x_1-x_2\neq0. $$ Thus, $Ax=0$ has a nontrivial solution, so it has more than one solution. Therefore, D is true.

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