Find the general solution to the homogeneous ystem

It can be convenient to write our solutions in vector form. For later work with matrices, we use column vectors. Instead of giving solutions for \(x_1, x_2, \ldots, x_n\) separately, we collect things into a vector \(\bbm x_1\\x_2\\ \vdots\\ x_n\ebm\text{.}\)

For homogeneous systems this lets us easily determine the basic solutions. If our general solution has parameters \(t_1,t_2,\ldots, t_k\text{,}\) the basic solutions are obtained by setting one parameter equal to 1, and the others to 0.

This can be notationally convenient. A system

We want to write our system in the form \(A\vec{x}=\vec{b}\text{.}\) How do we define the product \(A\vec{x}\text{?}\)

• As a linear combination:

  \begin{equation*} \bbm a_{11} \amp a_{12} \amp \cdots \amp a_{1n}\\ a_{21} \amp a_{22} \amp \cdots \amp a_{2n}\\ \vdots \amp \vdots \amp \ddots \amp \vdots\\ a_{m1} \amp a_{m2} \amp \cdots \amp a_{mn}\ebm \bbm x_1\\x_2\\ \vdots \\x_n\ebm = x_1\bbm a_{11}\\ a_{21}\\ \vdots \\ a_{m1}\ebm + x_2\bbm a_{12}\\a_{22}\\ \vdots \\ a_{m2}\ebm + \cdots + x_n \bbm x_{1n}\\ x_{2n}\\ \vdots \\x_{mn}\ebm\text{.} \end{equation*}

• Using row-times-column “dot products”:

  \begin{equation*} \bbm a_{11} \amp a_{12} \amp \cdots \amp a_{1n}\\ a_{21} \amp a_{22} \amp \cdots \amp a_{2n}\\ \vdots \amp \vdots \amp \ddots \amp \vdots\\ a_{m1} \amp a_{m2} \amp \cdots \amp a_{mn}\ebm \bbm x_1\\x_2\\ \vdots \\ x_n\ebm = \bbm a_{11}x_1 +a_{12}x_2 +\cdots + a_{1n}x_n\\ a_{21}x_1+a_{22}x_2+\cdots + a_{2n}x_n\\ \vdots \\ a_{m1}x_1+a_{m2}x_2+\cdots a_{mn}x_n\ebm\text{.} \end{equation*}

• Both produce the same result! (Let's try an example or two.)

Let \(A\) be an \(m\times n\) matrix, and let \(\vec{x}\) be an \(k\times 1\) column vector.

1. The product \(A\vec{x}\) is only defined if \(k=n\)

2. The result is an \(m\times 1\) column vector.

3. For \(n\times 1\) vectors \(\vec{v},\vec{w}\text{,}\) \(A(\vec{v}+\vec{w})=A\vec{v}+A\vec{w}\text{.}\)

4. For any \(n\times 1\) vector \(\vec{v}\) and scalar \(c\text{,}\) \(A(c\vec{v})=c(A\vec{v})\text{.}\)

With our new notation, we can write a system of equations as \(A\vec{x}=\vec{b}\text{.}\)

• This is notationally convenient: it's an easy way to refer to a general system of equations.

• It's meant to remind you of the case of one equation with one variable: \(ax=b\text{.}\)

• It can be used to quickly establish facts about systems.

The rank of a matrix is the number of leading ones in reduced row-echelon form. Given a system \(A\vec{x}=\vec{b}\) where \(A\) is \(m\times n\text{,}\) note:

• There are \(m\) equations.

• There are \(n\) variables.

• If \(\rank(A) = r\text{,}\) then the general solution to \(A\vec{x}=\vec{0}\) has \(k=n-r\) parameters.

• If \(\rank\left(\!\begin{array}{c|c}A \amp \vec{b}\end{array}\!\right)=\rank(A)\text{,}\) the system \(A\vec{x}=\vec{b}\) is consistent, and the general solution can be written as \(\vec{x}=\vec{x}_p+\vec{x}_h\text{,}\) where \(\vec{x}_p\) is a particular solution to \(A\vec{x}=\vec{b}\text{,}\) and \(\vec{x}_h\) is the general solution to the homogeneous system \(A\vec{x}=\vec{0}\text{.}\)

• If \(\rank\left(\!\begin{array}{c|c}A\amp \vec{b}\end{array}\!\right)\gt\rank(A)\text{,}\) then the system is inconsistent.

Determine the rank of \(A\text{,}\) and solve \(A\vec{x}=\vec{b}\text{,}\) where

1. \(A = \bbm -2 \amp 1\amp 3\\1\amp 0\amp -2\\ 0\amp 1\amp -1\ebm\text{,}\) \(\vec{b}=\bbm 5\\-2\\1\ebm\)

2. \(A = \bbm 1 \amp -2 \amp 2\amp 0\\2\amp -3 \amp 7\amp -2\ebm\text{,}\) \(\vec{b}=\bbm 2\\-1\ebm\)

By now, we've seen plenty of matrices. A matrix is just a rectangular array of numbers, arranged into rows and columns. If a matrix has \(m\) rows and \(n\) columns, we say it has size (or shape) \(m\times n\text{.}\)

Some examples of matrices:

Particular entries in a matrix are referenced by row and column. We write \(a_{ij}\) for the \((i,j)\)-entry of a matrix \(A\text{.}\)

This works exactly like you'd expect. Given matrices \(A = [a_{ij}], B=[b_{ij}]\) of the same size, define

Aside: what do we mean by “=”?

Let \(A\text{,}\) \(B\text{,}\) and \(C\) be matrices of the same size. Then:

1. \(\displaystyle A+B=B+A\)
2. \(\displaystyle A+(B+C)=(A+B)+C\)
3. \(A+0=A\) (here \(0\) is the zero matrix)
4. \(\displaystyle c(A+B)=cA+cB\)
5. \(\displaystyle (c+d)A = cA+dA\)
6. \(\displaystyle c(dA)=(cd)A\)