Linear algebra has numerous applications in many diverse fields such as economics, computer science, and engineering. The main reason is that [b]linear models[/b] are commonly used in such fields - many real-life problems can be formulated as [b]systems of linear equations[/b]. In general, a system of linear equations consists of [math]n[/math] variables [math]x_1,x_2,\ldots,x_n[/math] and [math]m[/math] equations as follows:[br][br][math] \left\{\begin{eqnarray} a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n & = & b_1 \\[br]a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n & = & b_2 \\[br]\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots & \cdots & \\[br]a_{m1}x_1+a_{m2}x_2+\cdots+a_{mn}x_n & = & b_m \end{eqnarray} \right.[/math],[br][br]where [math]a_{ij}, \ i=1,\ldots,m, \ j=1,\ldots,n[/math] are real numbers. [br][br]Each equation in the system is a [u]linear equation[/u] - an equation that can be written as equality between a real constant and a sum of terms such that each term is a product of a real number and a variable. Real numbers [math]a_{ij}[/math] attached to the variables are called [b]coefficients[/b]. Usually, the main goal is to find one or more solutions (the values of [math]x_j, \ j=1,\ldots, n[/math]) that satisfy this system. To have a deeper understanding of linear systems, we can first interpret a system of linear equations in terms of vectors. In fact, we can rewrite it more elegantly as follows:[br][br][math] x_1\begin{pmatrix}a_{11} \\ a_{21} \\ \vdots \\ a_{m1}\end{pmatrix}+ x_2\begin{pmatrix}a_{12} \\ a_{22} \\ \vdots \\ a_{m2}\end{pmatrix}+\cdots+x_n\begin{pmatrix}a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn}\end{pmatrix}=\begin{pmatrix}b_1 \\ b_2 \\ \vdots \\ b_m\end{pmatrix}[/math][br][br]It looks familiar, right? The left-hand side is just the linear combination of column vectors formed by the coefficients of the system with those [math]n[/math] variables as weights. And the system will have a solution if the column vector formed by [math]b_i[/math] on the right-hand side is in the span of the column vectors on the left-hand side. Solving the system is equivalent to finding the weights for such a linear combination. We usually call it a [b]vector equation[/b].[br][br]Another even more fruitful interpretation of linear systems is through the use of linear transformations.[br][br]

In the previous section, we already learned that any linear transformation [math]T:\mathbb{R}^n\to\mathbb{R}^m[/math] can be uniquely determined by the column vectors [math]T(e_1),T(e_2),\ldots,T(e_n)[/math] in [math]\mathbb{R}^m[/math]. Now, let us write them down explicitly as column vectors:[br][br][math]T(e_1)=\begin{pmatrix}a_{11} \\ a_{21} \\ \vdots \\ a_{m1}\end{pmatrix}, T(e_2)=\begin{pmatrix}a_{12} \\ a_{22} \\ \vdots \\ a_{m2}\end{pmatrix}, \ldots,[br]T(e_n)=\begin{pmatrix}a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn}\end{pmatrix}[/math][br][br]We can put those column vectors together in order to form a [b]m x n matrix[/b] [math]A[/math]:[br][br][math]A=\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ [br]\vdots & \vdots & \ldots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix}[/math][br][br]This is the [b]matrix of the linear transformation [math]T[/math][/b].[br]Suppose [math]v[/math] is an unknown vector in [math]\mathbb{R}^n[/math]. We write it as a column vector with variable entries: [math]v=\begin{pmatrix}x_1 \\ x_2 \\ \vdots \\ x_n\end{pmatrix}[/math]. In other words, [math]v=x_1e_1+x_2e_2+\cdots+x_ne_n[/math]. Hence [math]T(v)=x_1T(e_1)+x_2T(e_2)+\cdots+x_nT(e_n)[/math]. Explicitly, [math]T(v)[/math] can be computed by "mulitplying" the matrix of [math]T[/math] to the column vector [math]v[/math] as follows:[br][br][math]T(v)=\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ [br]\vdots & \vdots & \ldots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ \vdots \\ v_n\end{pmatrix}=x_1\begin{pmatrix}a_{11} \\ a_{21} \\ \vdots \\ a_{m1}\end{pmatrix}+ x_2\begin{pmatrix}a_{12} \\ a_{22} \\ \vdots \\ a_{m2}\end{pmatrix}+\cdots+x_n\begin{pmatrix}a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn}\end{pmatrix}=\begin{pmatrix}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 \\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ a_{m1}x_1+a_{m2}x_2+\cdots+a_{mn}x_n\end{pmatrix}[/math][br][br]This linear combination of column vectors can be regarded as the definition of the multiplication of an m x n matrix to an n x 1 matrix.[br][br]Suppose [math]b[/math] is a vector in [math]\mathbb{R}^m[/math] that can be written as a column vector with [math]b_i[/math] as entries i.e. [math]b=\begin{pmatrix}b_1 \\ b_2 \\ \vdots \\ b_m\end{pmatrix}[/math]. Then the system of linear equations can be concisely expressed as the [b]matrix equation[/b]: [math]Av=b[/math]. Hence solving the system is equivalent to finding a vector [math]v[/math] in [math]\mathbb{R}^n[/math] such that it is transformed to the given vector [math]b[/math] in [math]\mathbb{R}^m[/math] by the linear transformation [math]T[/math] corressponding to the matrix [math]A[/math]. [br]Now let us consider the following simple example:[br][br][math]\left\{\begin{eqnarray} 2x_1+3x_2 & = & 1 \\ -x_1+2x_2 & = & -4 \end{eqnarray}\right.[/math][br][br]We extract the coefficients of these two linear equations in the system to form a 2 x 2 matrix:[br][br][math]A=\begin{pmatrix} 2 & 3 \\ -1 & 2 \end{pmatrix}[/math][br][br]This is the matrix of the linear transformation [math]T[/math] such that [math]T(\hat{\mathbf{i}})=\begin{pmatrix} 2 \\ -1 \end{pmatrix}[/math] and [math]T(\hat{\mathbf{j}})=\begin{pmatrix} 3 \\ 2 \end{pmatrix}[/math]. Let [math]v=\begin{pmatrix}x_1 \\ x_2\end{pmatrix}[/math] and [math]b=\begin{pmatrix}1 \\ 4\end{pmatrix}[/math]. The system can be written as [math]Av=b[/math], or more explicitly, [math]\begin{pmatrix} 2 & 3 \\ -1 & 2 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}=\begin{pmatrix}1 \\ 4\end{pmatrix}[/math][br][br][br]The following are some questions that test your understanding of the concepts you have just learned:[br]