Science:Math Exam Resources/Courses/MATH152/April 2022/Question A12/Solution 1

Following the first hint, we see that option 1 is possible: consider the possibility that $A$ is the 0 matrix and $b$ is not zero. If, instead, $b$ is the zero vector, then every $x\in \mathbb {R} ^{7}$ solves the equation.

Now let us consider if option 2 is possible. We need to be in a situation where we have at least one vector $x_{1}$ that satisfies $Ax_{1}=b.$

Following the second hint, consider the homogenous equation

$$Ax=0,$$

which corresponds to a system of equations

$${\begin{cases}a_{1,1}x_{1}+a_{1,2}x_{2}+a_{1,3}x_{3}+\dots +a_{1,7}x_{7}&=0\\a_{2,1}x_{1}+a_{2,2}x_{2}+a_{2,3}x_{3}+\dots +a_{2,7}x_{7}&=0\\\quad \vdots &\\a_{5,1}x_{1}+a_{5,2}x_{2}+a_{5,3}x_{3}+\dots +a_{5,7}x_{7}&=0\end{cases}}$$

This system of equations has more variables than equations. Since there are two fewer equations than variables, there will be (at least) two free variables when solving the system, so the vectors $x_{0}$ that solve the homogeneous equation form a subspace of dimension at least 2.

Now, given a solution $x_{1}$ of the original equation and a solution $x_{0}$ of the homogenous equation, $x_{1}+x_{0}$ is another solution of the original equation. Since we have a subspace of homogenous solutions of dimension at least 2, it follows that we also have at least a 2-dimensional set of solutions to the original equation. This shows that options 2, 3, 4, 5 are not possible, so only options 1 and 7 can describe the solutions to $Ax=b$.