# Why does the reduced matrix $\left(\begin{array}{ccc|c}0&1&0&-7\\ 0&0&1&10\end{array}\right)$ have infinitely many solutions?

$$\left(\begin{array}{ccc|c}0&1&0&-7\\ 0&0&1&10\end{array}\right)$$

I thought the requirement for a matrix to have a unique solution was that when every variable is leading. It seems like both 1's in the above matrix are leading and the other variables are 0. So why doesn't this matrix have only a unique solution?

This is simple yet entertaining question. The system of linear equations is basically one of the most important things to understand in Linear Algebra.

Lets begin with basics. As it was pointed out this is system of 3 variables - let it be $(a, b ,c)$. Therefore your equation can be rewritten from your:

$$\left(\begin{array}{ccc|c}0&1&0&-7\\ 0&0&1&10\end{array}\right)$$

to

$$\begin{eqnarray*} {0a + 1b + 0c}&=&{-7} \\ {0a + 0b + 1c}&=&{10} \end{eqnarray*}$$

The solution is a vector formed with exactly those three variables.

The most obvious is that you do not care about $a$ and therefore $a$ could be anything. This leads us to any vector corresponding to $(a, b, c) = (a, -7, 10)$ as an solution.

Finding solutions

Saying that you can see if there is a solution using the Frobenius rule. It says basically that solution exists exactly when the system of equations $Ax=b$ equals $rank(A) = rank(A | b)$ .

Number of solutions

To determine how many solutions (for linear system $Ax = b$) there are 3 simple rules (using rank of the matrix):

Where $n$ is number of variables and we expect Frobenius rule to be in place => for $Ax=b$ equals $rank(A) = rank(A | b)$ . Further more matrix $A$ is in space $n \times m$, where $n$ is number of columns (as dedscribed earlier) and $m$ is number of rows.

1. If $rank(A | b)$ is less then $n$ => There is infinite number of solutions (as the system of linear equations is underdetermined system)
2. If $rank(A | b)$ is equal to $n$

a) $n = m$ => There is *only one unique solution** (as the system of linear equations is nonsingular/invertible system - though in our school we called it regular)

b) $n < m$, simply said more (linearly independent) rows (or equations if you will) => There is no exact solution (as the system of linear equations is overdetermined system), but it can be fitted with least squares to closed possible fit.

This equation

There is $n = 3$ (number of variables) and $rank(A) = rank(A | b) = 2$

$$\left(\begin{array}{ccc|c}0&1&0&-7\\ 0&0&1&10\end{array}\right)$$

• How can $\operatorname{rank}(A\mid b)$ be greater than $n$ if $n$ is the width of $A$ and $\operatorname{rank}(A)=\operatorname{rank}(A\mid b)$? – Henning Makholm Oct 30 '17 at 12:15
• Well it is shown in the link - overdetermined system. Simple example could be determining some trend for the future .. Check on Diamond Prices in Singapore – Polostor Oct 30 '17 at 14:55
• Oh I see what you mean. My mistake, I will fix that when I get home. Sorry for that – Polostor Oct 30 '17 at 15:10

Because $(x_1, -7, 10)$ is a solution to the system.

To have unique solution, we can't have any non-pivot columns, in this case, the first column is a non-pivot column.

That's because the solutions are $\;y=-7,\enspace z=10$ and $x$ can have any value, so the set of solutions in vector form is $$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}x\\-7\\10\end{bmatrix}= x\begin{bmatrix}1\\0\\0\end{bmatrix}+\begin{bmatrix}0\\-7\\10\end{bmatrix}.$$

Think intuitively: the system represented here is 2 equations in three unknowns. If you have a solution at all, you will have a line's worth of solutions (here it is quick to see the $x$ variable is unconstrained).

This is equivalent to the system:

$$\begin{eqnarray*} {0x + y + 0z}&=&{-7} \\ {0x + 0y+ z}&=&{10} \end{eqnarray*}$$

Whatever $x$ you choose will work because $0x=0$ for all $x$. If the system were:

$$\begin{eqnarray*} { y + 0z}&=&{-7} \\ { 0y+ z}&=&{10} \end{eqnarray*}$$

Then there is only one solution. It is very common for people to forget the $0x_i$ matter, a lot of people answer a different problem because they write the first system as the second system. The first system has infinite solutions, the second system as one solution.

I suggest you to work on the equivalence between matrices and systems of linear equations and see what each result in matrices means in terms of results of systems of linear equations and vice-versa.