Solve the following pair of linear equations: 62x + 37y = 13 37x + 62y = -112
The problem with the row picture of a $2\times 2$ systems of linear equations is that it does not extend very well to larger $m\times n$ systems of linear equations. To extend our ideas of consistent and independent to larger $m\times n$ systems of linear equations, we need to view the same three systems using the column picture . Recall our three $2\times 2$ systems of linear equations $$\begin{array}{lrcl} \text{1.} & x_1 + x_2 & = &\ \ 3 \\ & x_1 - x_2 & = & -1 \end{array}\qquad \begin{array}{lrcl}\ \ 2. & x_1 - x_2 & = & -1 \\ & x_1 - x_2 & = &\ \ 1 \end{array}\qquad \begin{array}{lrcl}\ \ 3. & x_1 +\ \ x_2 & = & 3 \\ & 3x_1 + 3x_2 & = & 9 \end{array}$$ $$1.\ \ \begin{bmatrix} x_1 + x_2 \\ x_1 - x_2 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_1 \end{bmatrix} + \begin{bmatrix}\ \ x_2 \\ -x_2 \end{bmatrix} = x_1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + x_2\begin{bmatrix} \ \ 1 \\ -1 \end{bmatrix} = \begin{bmatrix}\ \ 3 \\ -1 \end{bmatrix}$$ $$\begin{bmatrix} x_1 + x_2 \\ x_1 - x_2 \end{bmatrix}$$ $$\begin{bmatrix}\ \ 3 \\ -1 \end{bmatrix}$$ $$x_1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + x_2\begin{bmatrix} \ \ 1 \\ -1 \end{bmatrix}$$
This gives us the vector equation $$x_1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + x_2\begin{bmatrix} \ \ 1 \\ -1 \end{bmatrix} = \begin{bmatrix}\ \ 3 \\ -1 \end{bmatrix}$$
If we name our vectors $\mathbf{v}_1$ and $\mathbf{v}_2$, then we have $$ \mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix},\qquad \mathbf{v}_2 = \begin{bmatrix}\ \ 1 \\ -1 \end{bmatrix}$$ $$ x_1\mathbf{v}_1 + x_2\mathbf{v}_2 = \mathbf{b}. $$ Since $1\,\mathbf{v}_1 + 2\,\mathbf{v}_2 = \mathbf{b}$, we have $x_1=1$ and $x_2=2$, so our unique solution is the vector $$\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$
Let us write the second system of linear equations in vector form. $$\begin{bmatrix} x_1-x_2 \\ x_1-x_2 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_1 \end{bmatrix} + \begin{bmatrix} -x_2 \\ -x_2 \end{bmatrix} = x_1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + x_2\begin{bmatrix} -1 \\ -1 \end{bmatrix} = \begin{bmatrix} \ \ 1 \\ -1 \end{bmatrix}$$ These vectors are linear dependent because they are colinear. Since $\mathbf{b}$ is not a vector in their span , there is no linear combination of vectors $\mathbf{v}_1$ and $\mathbf{v}_1$ that will equal $\mathbf{b}$. This system of linear equations has no solution.
Finally, let us write the last linear system in vector form. $$\begin{bmatrix} x_1 + x_2 \\ 3x_1 + 3x_2 \end{bmatrix} = x_1\begin{bmatrix} 1 \\ 3 \end{bmatrix} + x_2\begin{bmatrix} 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 3 \\ 9 \end{bmatrix} = 3\begin{bmatrix} 1 \\ 3 \end{bmatrix} $$ $$ \left(x_1 + x_2\right)\begin{bmatrix} 1 \\ 3 \end{bmatrix} = 3\begin{bmatrix} 1 \\ 3 \end{bmatrix}$$ $$x_1 + x_2 = 3.$$ We obtain an equation for $x_1$ and $x_2$ that has infinitely many solutions, $x_1 + x_2 = 3$. We could plot this equation on a Cartesian plane, and the set of solutions is a line with slope $-1$ and $y$-intercept $3$. All of the points $(x_1, x_2)$ on this line gives us a solution to the linear system. This linear system has infinitely many solutions.
The column picture is more useful to use because we can apply the same ideas to larger systems of $m$ equations and $n$ unknowns. |