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 $[A\mid b]$ to get its reduced row echelon form. We call the reduced row echelon form $[\tilde{A}\mid \tilde{b}]$.

• 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 $(a)(d)(e)$.