I need to find two more $b$'s other than $b=(2,5,7)$, such that the equation can be solved and two more such that the equation can't be solved. $$u\begin{bmatrix}1\\2\\3\end{bmatrix}+v\begin{bmatrix}1\\0\\1\end{bmatrix}+w\begin{bmatrix}1\\3\\4\end{bmatrix}=b$$ How do I find those values? What should I do to find both, values of b that make the system solvable and values that make the system unsolvable?

I tried to set $b$ as $(b_1,b_2,b_3)$, but I ended up with this matrix: $$\left[\begin{array}{l}1&1&1&b_1\\0&2&-1&2b_1-b_2\\0&0&0&b_1-b_3+b_2\end{array}\right]$$ Sorry if it is something obvious, I'm knew with linear algebra.

  • $\begingroup$ You're almost there. The last step is to use the row echelon form matrix you computed to the see if you can pick values of $b_1,b_2,b_3$ so that you do or don't have solutions. For instance, if $b_1=b_2=0$ and $b_3=1$ can you have a solution? $\endgroup$ Commented Jan 26, 2019 at 0:32
  • $\begingroup$ So I just need to give random values to$ b_1$,$ b_2$ and $b_3$? And if $b_1=b_2$ and $b_3=1$, I think that there is no solution, because in the last row there will be a 0=-1, right? $\endgroup$ Commented Jan 26, 2019 at 0:43
  • $\begingroup$ Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't. $\endgroup$ Commented Jan 26, 2019 at 0:46

3 Answers 3


After you apply your row reduction, look at the last row of your matrix.

It shows you that

$0u + 0v + 0 w = b_1 - b_3 + b_2$

This is only possible if $b_1 - b_3 + b_2 = 0$

So if you choose $b_1, b_2, b_3$ that satisfy this condition, your matrix has a solution (in this case infinite), otherwise it has no solutions.

for example if $b = (2, 5, 7)$ as you suggested, then it is solvable because $2 - 7 + 5 = 0$

Since you are new to linear algebra I did not mention determinants, dimensions, ranks etc. You don't need them in this case, just recall how a matrix represents a system of equations.

  • $\begingroup$ Thanks! After reading you answer I was able to find the solution. $\endgroup$ Commented Jan 27, 2019 at 0:34

You can compute $$\det \left(\begin{matrix} 1 & 1 & 1 \\ 2 & 0 & 3 \\ 3 & 1 & 4 \end{matrix} \right) = 9+2-3-8=0$$ which means the three vectors $(1,2,3), (1,0,1), (1,3,4)$ are linearly dependent. However the first two are linearly independent, so they generate a plane in $\mathbb{R}^3$ which contains $(1,3,4)$. The plane is $$t(1,2,3)+s(1,0,1) = (t+s,2t,3t+s)$$ which you can write in Cartesian coordinates as follows: define $x,y,z$ so that $$\begin{cases} x = t+s \\ y = 2t \\ z = 3t+s. \end{cases} $$ Hence $x+y-z=0$, and this is the Cartesian equation of your plane. So every $b=(x,y,z)$ satisfying this equation works. Notice that your $b = (2,5,7)$ works, and also $b=(0,0,0)$ is a trivial choice - which just means that $(1,3,4)$ can be expressed as a linear combination of the first two vectors, as we know. The $b$'s which do not lie on this plane do not work, e.g. $b=(2,5,8)$.

  • $\begingroup$ I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right? $\endgroup$ Commented Jan 26, 2019 at 1:13
  • $\begingroup$ Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $\alpha(1,2,3)+\beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$. $\endgroup$
    – Gibbs
    Commented Jan 26, 2019 at 1:14
  • $\begingroup$ Ok, I get it. How did you realize that the three vectors where linearly dependent? $\endgroup$ Commented Jan 26, 2019 at 2:35
  • $\begingroup$ I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent. $\endgroup$
    – Gibbs
    Commented Jan 26, 2019 at 10:08


Use the criterion for the system to be (not) solvable:

$Ax=b\;$ has a solution if and only if $\operatorname{rank} A\;$ is the same as the rank of the augmented matrix $(A|b)$.

So, as $\operatorname{rank} \,A=2$, you have to find $b$ such that $(A|b)$ has the maximum rank ($3$).

  • $\begingroup$ Sorry, what does $rank$ mean? $\endgroup$ Commented Jan 26, 2019 at 0:41
  • $\begingroup$ The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix. $\endgroup$
    – Bernard
    Commented Jan 26, 2019 at 0:45

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