Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 17/Solution 1

First write the linear system in the form of $Ax=b$ where $A$ is the coefficient matrix with $7$ rows and $8$ columns, $x$ is a $8\times 1$ vector representing the unknowns while $b$ is a $8\times 1$vector representing the right-hand side. Now we perform Gaussian Elimination to the augmented matrix $\left[A\mid b\right]$ to get its reduced row echelon form. We call the reduced row echelon form $\left[{\tilde {A}}\mid {\tilde {b}}\right]$.

• If there is a zero row in ${\tilde {A}}$ but the entry of ${\tilde {b}}$ on this row is non-zero, then there is no solution. So (a) is possible.

• The system has a unique solution if the corresponding homogeneous system $Ax=0$ has zero solution. Note that the matrix $A$ has $8$ columns and $7$ rows, its column spaces are linear dependent. This implies that there is a nonzero vector in the null space, so (b) is wrong.

• (c) is not possible. If we assume for the sake of contradiction that the linear system has exactly eight solutions ${\mathbf {x}}^{1},{\mathbf {x}}^{2},{\mathbf {x}}^{3},{\mathbf {x}}^{4},{\mathbf {x}}^{5},{\mathbf {x}}^{6},{\mathbf {x}}^{7},{\mathbf {x}}^{8}$ , then $y={\frac {1}{8}}({\mathbf {x}}^{1}+{\mathbf {x}}^{2}+{\mathbf {x}}^{3}+{\mathbf {x}}^{4}+{\mathbf {x}}^{5}+{\mathbf {x}}^{6}+{\mathbf {x}}^{7}+{\mathbf {x}}^{8})$ also solves the linear system, yielding a contradiction.

• The matrix $A$ has $8$ columns and $7$ rows, the maximal possible rank of matrix $A$ is $7.$ If $A$ has rank $7$, that is, no zero rows in ${\tilde {A}}$ , it is possible for the linear system to have one parameter family of solutions.

• If there is one zero row in the reduced row echelon form $[{\tilde {A}}\mid {\tilde {b}}]$ , then the matrix $A$ has rank $6$. In this case, it is possible for the system to have two parameter family of solutions.

Thus, the possible results are $\color {blue}{(a)(d)(e)}$.