# How to find values that make a matrix solvable and unsolvable?

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.

• 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? Commented Jan 26, 2019 at 0:32
• 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? Commented Jan 26, 2019 at 0:43
• Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't. Commented Jan 26, 2019 at 0:46

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.

• Thanks! After reading you answer I was able to find the solution. 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. $$b=(2,5,8)$$.

• 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? Commented Jan 26, 2019 at 1:13
• 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 $(1,0,1)$. Commented Jan 26, 2019 at 1:14
• Ok, I get it. How did you realize that the three vectors where linearly dependent? Commented Jan 26, 2019 at 2:35
• 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. Commented Jan 26, 2019 at 10:08

Hint:

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 ($$3$$).

• Sorry, what does $rank$ mean? Commented Jan 26, 2019 at 0:41
• 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. Commented Jan 26, 2019 at 0:45