Linear Algebra | Determine the values of a for which the system has no solutions, exactly one solution, or infinitely many solutions

Question

I am a college student, and this summer I am taking a 15 week course on linear algebra. I was doing my homework today, and I am not sure if my solution is correct. The problem is stated as follows:

"Determine the values of a for which the system has no solutions, exactly one solution, or infinitely many solutions."

The following is my solution for this problem. The text in green is the answer in the back of my textbook. My answer is at the bottom right of the page in black text with red highlights. I am wondering if my solution is also correct or is it just dead wrong? The reason I ask is because my textbook often gives only one example answer for problems which can be solved in a variety of ways. Also, see the text in blue which reads "Is this legal"? Is that operation legal (i.e. can it be done?). Furthermore, if I went wrong somewhere in my procedure, could you specify where?

Thank you for all your help.

-C.I

PS: Sorry about my bad handwriting.

Answer 1

You're given the system of linear equations described by $$\mathbf{A}\mathbf{x}=\mathbf{b}\iff\begin{bmatrix}1&2&-3\\3&-1&5\\4&1&a^2-14\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\2\\a+2\end{bmatrix}$$ Your elimination procedure for writing $$\left[\begin{array}{ccc|c}1&2&-3&4\\3&-1&5&2\\4&1&a^2-14&a+2\end{array}\right]\stackrel{\text{elim}}{\implies}\left[\begin{array}{ccc|c}1&2&-3&4\\0&1&-2&\frac{10}{7}\\0&0&a^2-16&a-4\end{array}\right]$$ is correct.

Let's first assume $a=4$. Under this assumption, the last row is composed entirely of zeros and the matrix $\mathbf{A}$ is of rank $2$ (only two pivots in the $(1,1)$ and $(2,2)$ positions). This means the dimension of the nullspace (the set of vectors $\mathbf{x}$ such that $\mathbf{A}\mathbf{x}=\mathbf{0}$) of $\mathbf{A}$ is $1$, which means that, whether or not $\mathbf{A}\mathbf{x}=\mathbf{b}$ has a "special" solution that is a linear combination of the columns of $\mathbf{A}$, there will be a host (read: infinite number) of solutions determined by the vector in the basis of the nullspace.

With $a=4$, write the augmented matrix in RREF to obtain $$\left[\begin{array}{ccc|c}1&0&1&0\\0&1&-2&0\\0&0&0&0\end{array}\right]$$ Because you know the basis of the nullspace contains one vector, you know that you have one free variable; that is, you can fix any of $x$, $y$, or $z$ and uniquely determine the value of the other two. Suppose $z=1$; then $x=-1$ and $y=2$. Therefore $\mathbf{x}=\begin{bmatrix}-1\\2\\1\end{bmatrix}$ is in the nullspace of $\mathbf{A}$, and furthermore any scalar multiple of this vector is also in the nullspace. This alone is enough to conclude that there is an infinite number of solutions to $\mathbf{A}\mathbf{x}=\mathbf{b}$ precisely when $a=4$.

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Now suppose $a=-4$. The augmented matrix becomes $$\left[\begin{array}{ccc|c}1&2&-3&4\\0&1&-2&\frac{10}{7}\\0&0&0&-8\end{array}\right]$$ The last row is problematic. There is no $\mathbf{x}$ that satisfies this equation.

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Now assume $a\neq\pm4$. You correctly note that for this case, you can multiply the last row by $\frac{1}{a-4}$ to yield $$\left[\begin{array}{ccc|c}1&2&-3&4\\0&1&-2&\frac{10}{7}\\0&0&a+4&1\\\end{array}\right]$$ Because there are three pivots (all sitting along the diagonal), the nullspace has dimension $0$ (so it only contains the zero vector) which means that if there is a solution to $\mathbf{A}\mathbf{x}=\mathbf{b}$, then $\mathbf{b}$ is exactly a linear combination of the columns of $\mathbf{A}$. The last row tells you when this happens. Solving, you get $$\mathbf{x}=\begin{bmatrix}4-2\left(\frac{10}{7}+\frac{2}{a+4}\right)+\frac{3}{a+4}\\\frac{10}{7}+\frac{2}{a+4}\\\frac{1}{a+4}\end{bmatrix}=\begin{bmatrix}\frac{8}{7}-\frac{1}{a+4}\\\frac{10}{7}+\frac{2}{a+4}\\\frac{1}{a+4}\end{bmatrix}$$

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To summarize:

There are infinitely many solutions when $a=4$.

There are no solutions when $a=-4$.

There is exactly one solution otherwise.

To answer your question about the "legal" move - that's absolutely fine so long as $a\neq4$. The problem with your work crops up just below that part: in the case of $a\neq\pm4$, when considering the last row you do not have the equation $a+4=1$, but rather $(a+4)z=1$.

Answer 2

Write your system as $M_aX=S_a$, where $X= \begin{pmatrix} x\\ y\\ z \end{pmatrix}$, $S_a=\begin{pmatrix} 4\\ 2\\ a+2 \end{pmatrix}$ $$M_a=\begin{pmatrix} 1& 2& -3\\ 3& -1& 5\\ 4&1 &a^2-14 \end{pmatrix}$$.

Now this system has exactly one solution whenever $M_a$ is invertible (and then $X=M^{-1}S$), which happens if and only if $\det M_a\neq 0$. Computing this, we find $\det M_a=112-a^2\neq 0 $ for $a\neq \pm4 $. So we are left with two cases to treat, $a=4$ and $a=-4$. In both cases, we are trying to see whether the columns of $M_{4}=M_{-4}$ span a subspace of $\mathbb{R}^3$ which contains $S_4$ or $S_{-4}$, respectively. If they do, then the system has infinitely many solutions, taking $X=X_0+X'$ for any $X_0$ in the nullspace of $M_4$, which is infinite, and for $X'$ a particular solution of the system. If they don't, then the system has no solution, as $S_4$ (resp. $S_{-4}$) will be outside the span of $M_4$. Using reduction on the columns, we find $M_4$ spans the subspace generated by the vectors $V_1=\begin{pmatrix} 1\\ 3\\ 4 \end{pmatrix}$ and $V_2=\begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}$

So we are trying to determine whether $V_1,V_2$, and $S_4$ are linearly dependent (resp. $S_{-4}$), which happens if and only if $d_4=\det (V_1|V_2|M_4)=0$. (resp. $d_{-4}$) computing we find $d_4=0, d_{-4}=-8\neq 0$.
Hence, summing up the system has exactly one solution when $a^2\neq 16$, -infinitely many solutions when $a=4$ and no solutions when $a=-4$.

Answer 3

I expose the method to decide on the existence or no of solutions for this type of system (system that has so many variables than equations). We calculate the determinant of the matrix associated to the system. If this determinant is not zero, then there is existence and uniqueness of the solution (it is a Cramer's system). Otherwise, the system it has infinity solutions or it has no solution; in this case, we calculate the rank r_1 of associated matrix and compare it with the rank r_2 of augmanted matrix of the same system; there will be one of two possibilities: a) they have the same rank r_1 = r_2, in this case, there are infinity solutions. b)the rank r_2 = r_1 + 1, in this case no solutions for the system.

In your example, the determinant of associated matrix is:-7a^2+112. So a) if a is not in {-4,4}, there is existence and uniqueness of the solution. b) If a = 4, then r_1 =_r_2 = 2, there are infinitely many solutions. c) If a = -4 then r_2=r_1+1= 3, there 's no solutions.