# Does one always use augmented matrices to solve systems of linear equations?

The homework tag is to express that I am a student with no working knowledge of math.

I know how to use elimination to solve systems of linear equations. I set up the matrix, perform row operations until I can get the resulting matrix into row echelon form or reduced row echelon form, and then I back substitute and get the values for each of my variables.

Just some random equations: $a + 3b + c = 5 \\ a + 0 + 5c = -2$

My question is, don't you always want to be using an augmented matrix to solve systems of linear equations? By augmenting the matrix, you're performing row operations to both the left and right side of those equations. By not augmenting the matrix, aren't you missing out on performing row operations on the right side of the equations (the $5$ and $-2$)?

The sort of problems I'm talking about are homework/test-level questions (as opposed to real world harder data and more complex solving methods?) where they give you $Ax = b$ and ask you to solve it using matrix elimination.

Here is what I mean mathematically:

$[A] = \begin{bmatrix} 1 & 3 & 1 \\ 1 & 0 & 5 \\ \end{bmatrix}$

$[A|b] = \left[\begin{array}{ccc|c} 1 & 3 & 1 & 5\\ 1 & 0 & 5 & -2\\ \end{array}\right]$

So, to properly solve the systems of equations above, you want to be using $[A|b]$ to perform Elementary Row Operations and you do not want to only use $[A]$ right?

The answer is: yes, if you use this method of matrices to solve systems of linear equations, you must use the augmented form $[A|b]$ in order to perform EROs to both $A$ and $b$.

Augmenting the matrix is just notational shorthand.

When you do row operations, what you're really doing is multiplying both sides of an equation by some matrix $P_1$:

\begin{align*} Ax &= b \\ P_1Ax &= P_1b \tag{P_1 represents some row operation} \\ P_2P_1Ax &= P_2 P_1 b \tag{P_2 represents some row operation} \\ &\vdots \\ Dx &= P_k \cdots P_2 P_1b \tag{D is now a triangular matrix} \end{align*}

Doing in situ row operations obscures this fact, and augmenting it obscures it even further.

Here is an example of solving a system using augmented matrices:

$$\begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3\end{pmatrix}$$ becomes $$\left(\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 2 \\ 0 & 1 & 1 & 3 \end{array}\right)$$

Then you do the following operations:

$$\left(\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & -1 & 0 & 1 \\ 0 & 1 & 1 & 3 \end{array}\right) \tag{subtract row 1 from row 2}$$

$$\left(\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & -1 & 0 & 1 \\ 0 & 0 & 1 & 4 \end{array}\right) \tag{add row 2 to row 3}$$

And now you have an upper-triangular matrix that you can backsolve.

What you're really doing, however, is this:

\begin{align*} \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} &= \begin{pmatrix} 1 \\ 2 \\ 3\end{pmatrix} \\ \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} &= \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ 3\end{pmatrix} \\ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ 3\end{pmatrix} \end{align*}

Multiply everything out, you get:

$$\begin{pmatrix} 1 & 1 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 4\end{pmatrix}$$

which is exactly what appears in your augmented form.

• I'm afraid I still don't understand. Can't you perform row operations on just $[A]$ instead of $[A|b]$? – Jason Apr 26 '13 at 16:37
• @Jason Of course you can. That's what this answer says. In order to solve the equation, you do the same row operations on $A$ and $b$. The augmented matrix is just "notational shorthand" to make sure you are doing the same operation on both, as Arkamis says above. – Thomas Andrews Apr 26 '13 at 16:42
• @Jason No, because you aren't doing the same thing to both sides. If you multiply the top row by 5, the top coefficient in $b$ must also increase by 5. The $[A|b]$ is just a way of putting everything in one matrix. (I mean, if you do the same thing to $b$ then yes, you can deal with them separately) – Henry Swanson Apr 26 '13 at 16:43
• @HenrySwanson, great, so if you're provided a $A$ and a $B$, you must augment $B$ as $[A|b]$ to properly solve the matrix right? This is just basic linear algebra, as we've just covered basic concepts, without any fancy mathwork. – Jason Apr 26 '13 at 16:45
• @Jason You could either augment it, as is usually done in courses, but if you do what Arkamis typed up, where you just multiply both sides by the same matrix, then you get the same answer. As he said, augmented matrices are just a shorthand. – Henry Swanson Apr 26 '13 at 16:47

I certainly wouldn't use an augmented matrix to solve the following:

$x = 3$

$y - x = 6$

When you solve a system of equations, if doing so correctly, one does indeed perform "elementary row operations", and in any case, when working with any equations, to solve for $y$ above, for example, I would add $x$ to each side of the second equation to get $y = x + 6 = 3 + 6 = 9$

Note: we do treat systems of equations like "row operations" (in fact, row operations are simply modeled after precisely those operations which are legitimate to perform on systems of equations)

$$2x + 2y = 4 \iff x + y = 2\tag{multiply each side by 1/2}$$

$$\qquad\qquad x + y = 1$$ $$\qquad\qquad x - y = 1$$ $$\iff 2x = 2\tag{add equation 1 and 2}$$

$$x+ y + z = 1$$ $$3x + 2y + z = 2$$ $$-x - y + z = 3$$

We can certainly switch "equation 2 and equation 3" to make "adding equation 3 to equation 1" more obvious.

I do believe that using an augmented coefficient matrix is very worthwhile, very often, to "get variables out of the way" temporarily, and for dealing with many equations in many unknowns: and even for $3\times 3$ systems when the associated augmented coefficient matrix has mostly non-zero entries. It just makes more explicit (and easier to tackle) the process one can use with the corresponding system of equations it represents.

• Why not? Are you saying you could rearrange the equations as $x + 0y - 3 = 0$ and $-x + y - 6 = 0$? – Jason Apr 26 '13 at 16:36
• @Jason, I think he's trying to say that some linear systems are soooo easy that using matrices at all, augmented or not, is just a waste of time. – DonAntonio Apr 26 '13 at 16:37
• Oh, I guess I meant, specific to linear algebra homework/test problems where you're given $Ax = b$ and told to solve it using matrix elimination. Sorry! – Jason Apr 26 '13 at 16:38
• Yes, Jason, I am suggesting exactly what @DonAntonio suggest. But I've also added that when one correctly solves a system of equations, one either does use elementary row operations, or solves for variables by adding, subtracting, dividing, or mulitplying both sides of an equation by the same quantity. – Namaste Apr 26 '13 at 16:40
• So basically my question is, if you were forced to use matrices instead of easier equation subtraction/elimination (from grade school), must you augment the $3$ and $6$ (i.e. $[A|b]$ to properly solve the matrix? Otherwise you're kind of not performing EROs to each side, if you only solve $[A]$ right? My question may seen stupidly obvious, but I guess I'm confused... – Jason Apr 26 '13 at 16:42