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4.E Systems of linear equations

One of the main uses of linear algebra is to assist in finding the solutions to systems of linear equations. Recall that a system of linear equations is a collection of equations of the form:

$a_{11}x_{1}+\cdots+a_{1n}x_{n}=b_{1}$
$\vdots$
$a_{m1}x_{1}+\cdots+a_{mn}x_{n}=b_{m}$

Here $a_{ij},b_{i}\in F$ are considered fixed, and the symbols $x_{i}$ are viewed as variables for which we would like to find solutions. So if we define the matrix $A:=[a_{ij}]$ , and the vectors $X:=\begin{bmatrix}x_{1}&\cdots&x_{n}\end{bmatrix}^{T}$ , $B:=\begin{bmatrix}b_{1}&\cdots&b_{m}\end{bmatrix}^{T}$ , then the system of equations is equivalent to the following matrix equation:

AX=B.

In MATH105 a lot of effort went into solving various systems of equations like this, in particular using the “augmented matrix method”. We will use the symbol $[A|B]$ to refer to such an augmented matrix. Is consists of concatenating the matrix $A$ with the column vector $B$ . Using the concept of rank we can summarize the various situations:

Theorem 4.28.

Let $A, X, B$ be defined as above, a system of $m$ equations in $n$ variables.

•

If $\operatorname{rank}A\neq\operatorname{rank}[A|B]$ , then the system has no solutions.
•
If $\operatorname{rank}A=\operatorname{rank}[A|B]$ , then the system has solutions, and in particular:
- –
  
  If $\operatorname{rank}A=n$ then there is a unique solution.
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  If $\operatorname{rank}A<n$ then there are infinitely many solutions.

If we assumed $A$ is an invertible $n\times n$ matrix, then it is clear how to find the unique solution: $AX=B$ implies $X=A^{-1}AX=A^{-1}B$ .

But in general, the number of equations might not match the number of variables. So, if $\operatorname{rank}A=\operatorname{rank}[A|B]=n$ , we can perform row operations on $[A|B]$ until $[A^{\prime}|B^{\prime}]$ has $n$ non-zero rows. There is no harm in discarding the zero rows of $[A^{\prime}|B^{\prime}]$ , since they correspond to the equation $0=0$ . The resulting truncated $A^{\prime}$ will be an invertible $n\times n$ matrix, and we can use its inverse to find the unique solution, as above.

Example 4.29.

How many solutions does the following system of linear equations have:

$x+y=0$
$x-y+z=1$
$2x-y-z=1$

Solution: Perform row operations on the augmented matrix:

[A|B]=\left[\begin{array}[]{ccc|c}1&1&0&0\\ 1&-1&1&1\\ 2&-1&-1&1\end{array}\right]\rightarrow\cdots\rightarrow\left[\begin{array}[]{% ccc|c}1&-1&1&1\\ 0&1&-3&-1\\ 0&0&5&1\end{array}\right].

Therefore, $\operatorname{rank}A=\operatorname{rank}[A|B]=3$ , and so there is a unique solution.

To find this solution (which we weren’t asked to do), one multiplies the matrix equation $AX=B$ on the left by $A^{-1}$ :

X=A^{-1}B=\frac{1}{5}\begin{bmatrix}2&1&1\\ 3&-1&-1\\ 1&3&-2\end{bmatrix}\begin{bmatrix}0\\ 1\\ 1\end{bmatrix}=\begin{bmatrix}2/5\\ -2/5\\ 1/5\end{bmatrix}.

It is also easy to check that $(x,y,z)=(2/5,-2/5,1/5)$ satisfies the above equations.

Exercise 4.30:

Consider the following system of linear equations:

$x+y+z=1$
$ax+ay+z=2-a$
$ax+2y+z=2$

Here $a\in{\mathbb{R}}$ is treated as a fixed number, and $x, y, z$ are treated as variables.

i.

Write this system as a matrix equation $AX=B$ .
ii.

For which values of $a\in{\mathbb{R}}$ is $A$ invertible?
iii.

For all values of $a\in{\mathbb{R}}$ , calculate $\operatorname{rank}A$ and $\operatorname{rank}[A|B]$ .
iv.

For each $a\in{\mathbb{R}}$ with infinitely many solutions, describe the solution set.

[End of Exercise]

A key connection that you are expected to make, that links this section with the previous ones is the following fact: Using notation as above, the system of linear equations $AX=B$ has a solution if and only if $B$ is in the image of $A$ .