## Linear Algebra and Its Applications, Exercise 2.5.11

Exercise 2.5.11. Given the incidence matrix from exercise 2.5.10 with the final column removed and a diagonal matrix $C$ with elements 1, 2, 2, and 1, write the equations for the system $\begin{array}{rcrcr} C^{-1}y&+&Ax&=&0 \\ A^Ty&&&=&f \end{array}$

Show that eliminating $y$ gives the system $A^TCAx = -f$ and solve the system for $f = (1, 1, 6)$. If $f = (1, 1, 6)$ represents the currents entering nodes 1, 2, and 3 of the graph, calculate the potentials at each node and the currents on each edge.

Answer: Removing the final column from the matrix in exercise 2.5.10 gives us the new matrix $A = \begin{bmatrix} -1&1&0 \\ -1&0&1 \\ 0&1&0 \\ 0&0&-1 \end{bmatrix}$

The diagonal matrix $C = \begin{bmatrix} 1&0&0&0 \\ 0&2&0&0 \\ 0&0&2&0 \\ 0&0&0&1 \end{bmatrix}$

has as its inverse $C^{-1} = \begin{bmatrix} 1&0&0&0 \\ 0&\frac{1}{2}&0&0 \\ 0&0&\frac{1}{2}&0 \\ 0&0&0&1 \end{bmatrix}$

(Recall from Note 4 of section 1.6, “Inverses and Transposes”, that the inverse of a diagonal matrix $D$ with nonzero entries on the diagonal is itself a diagonal matrix, with the entries on the diagonal of $D^{-1}$ equal to the reciprocals of the corresponding entries on the diagonal of $D$.)

We can thus express the system $C^{-1}y + Ax = 0$ in matrix form as $\begin{bmatrix} 1&0&0&0 \\ 0&\frac{1}{2}&0&0 \\ 0&0&\frac{1}{2}&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \end{bmatrix} + \begin{bmatrix} -1&1&0 \\ -1&0&1 \\ 0&1&0 \\ 0&0&-1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ $= \begin{bmatrix} y_1 \\ \frac{1}{2}y_2 \\ \frac{1}{2}y_3 \\ y_4 \end{bmatrix} + \begin{bmatrix} -x_1+x_2 \\ -x_1+x_3 \\ x_2 \\ -x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

We can express the system $A^Ty = f$ in matrix form as $\begin{bmatrix} -1&-1&0&0 \\ 1&0&1&0 \\ 0&1&0&-1 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \end{bmatrix}$ $= \begin{bmatrix} -y_1-y_2 \\ y_1+y_3 \\ y_2-y_4 \end{bmatrix} = \begin{bmatrix} f_1 \\ f_2 \\ f_3 \end{bmatrix}$

This corresponds to the following system of equations: $\begin{array}{rcrcrcrcrcrcrcr} y_1&&&&&&&-&x_1&+&x_2&&&=&0 \\ &&\frac{1}{2}y_2&&&&&-&x_1&&&+&x_3&=&0 \\ &&&&\frac{1}{2}y_3&&&&&+&x_2&&&=&0 \\ &&&&&&y_4&&&&&-&x_3&=&0 \\ -y_1&-&y_2&&&&&&&&&&&=&f_1 \\ y_1&&&+&y_3&&&&&&&&&=&f_2 \\ &&y_2&&&-&y_4&&&&&&&=&f_3 \end{array}$

Going back to the original two equations $\begin{array}{rcrcr} C^{-1}y&+&Ax&=&0 \\ A^Ty&&&=&f \end{array}$

we can eliminate $y$ by multiplying the first equation by $A^TC$ to get $A^TCC^{-1}y + A^TCAx = 0$ or $A^Ty + A^TCAx = 0$ and then subtracting the second equation from this to get $A^TCAx = -f$.

We have $A^TCA = \begin{bmatrix} -1&-1&0&0 \\ 1&0&1&0 \\ 0&1&0&-1 \end{bmatrix} \begin{bmatrix} 1&0&0&0 \\ 0&2&0&0 \\ 0&0&2&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} -1&1&0 \\ -1&0&1 \\ 0&1&0 \\ 0&0&-1 \end{bmatrix}$ $= \begin{bmatrix} -1&-2&0&0 \\ 1&0&2&0 \\ 0&2&0&-1 \end{bmatrix} \begin{bmatrix} -1&1&0 \\ -1&0&1 \\ 0&1&0 \\ 0&0&-1 \end{bmatrix} = \begin{bmatrix} 3&-1&-2 \\ -1&3&0 \\ -2&0&3 \end{bmatrix}$

If $f = (1, 1, 6)$ the system $A^TCAx = -f$ can therefore be expressed in matrix form as $\begin{bmatrix} 3&-1&-2 \\ -1&3&0 \\ -2&0&3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \\ -6 \end{bmatrix}$

or as the system of equations $\begin{array}{rcrcrcr} 3x_1&-&x_2&-&2x_3&=&-1 \\ -x_1&+&3x_2&&&=&-1 \\ -2x_1&&&+&3x_3&=&-6 \end{array}$

We begin elimination by multiplying the first equation by -1/3 and subtracting it from the second equation, and multiplying the first equation by -2/3 and subtracting it from the third equation. The result is the following system: $\begin{array}{rcrcrcr} 3x_1&-&x_2&-&2x_3&=&-1 \\ &&\frac{8}{3}x_2&-&\frac{2}{3}x_3&=&-\frac{4}{3} \\ &&-\frac{2}{3}x_2&+&\frac{5}{3}x_3&=&-\frac{20}{3} \end{array}$

We then multiply the second equation by -1/4 and subtract it from the third equation: $\begin{array}{rcrcrcr} 3x_1&-&x_2&-&2x_3&=&-1 \\ &&\frac{8}{3}x_2&-&\frac{2}{3}x_3&=&-\frac{4}{3} \\ &&&&\frac{3}{2}x_3&=&-7 \end{array}$

Solving for $x_3$ we have $x_3 = \frac{2}{3}(-7) = -\frac{14}{3}$. Substituting the value of $x_3$ into the second equation we have $\frac{8}{3}x_2 - \frac{2}{3}(-\frac{14}{3}) = -\frac{4}{3}$ or $\frac{8}{3}x_2 = -\frac{4}{3} - \frac{28}{9} = -\frac{40}{9}$ so that $x_2 = \frac{3}{8}(-\frac{40}{9}) = -\frac{5}{3}$. Finally we substitute the values of $x_2$ and $x_3$ into the first equation to obtain $3x_1 -(-\frac{5}{3})-2(-\frac{14}{3}) = -1$ or $3x_1 + 11 = -1$ so that $x_1 = \frac{1}{3}(-12) = -4$.

The solution to the system $A^TCAx = -f$ when $f = (1, 1, 6)$ is thus $x = (-4, -\frac{5}{3}, -\frac{14}{3})$. If $f = (1, 1, 6)$ represents the currents into each of nodes 1, 2, and 3 respectively then $x = (-4, -\frac{5}{3}, -\frac{14}{3})$ represents the potentials at each of those nodes.

What are the currents along the edges? To determine that we must solve for $y$. Since $C^{-1}y + Ax = 0$ we have $y = -CAx$ or $y = \begin{bmatrix} -1&0&0&0 \\ 0&-2&0&0 \\ 0&0&-2&0 \\ 0&0&0&-1 \end{bmatrix} \begin{bmatrix} -1&1&0 \\ -1&0&1 \\ 0&1&0 \\ 0&0&-1 \end{bmatrix} \begin{bmatrix} -4 \\ -\frac{5}{3} \\ -\frac{14}{3} \end{bmatrix}$ $= \begin{bmatrix} 1&-1&0 \\ 2&0&-2 \\ 0&-2&0 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} -4 \\ -\frac{5}{3} \\ -\frac{14}{3} \end{bmatrix} = \begin{bmatrix} -\frac{7}{3} \\ \frac{4}{3} \\ \frac{10}{3} \\ -\frac{14}{3} \end{bmatrix}$

We can check this answer using the equation $A^Ty = f$ $A^Ty = \begin{bmatrix} -1&-1&0&0 \\ 1&0&1&0 \\ 0&1&0&-1 \end{bmatrix} \begin{bmatrix} -\frac{7}{3} \\ \frac{4}{3} \\ \frac{10}{3} \\ -\frac{14}{3} \end{bmatrix}$ $= \begin{bmatrix} \frac{3}{3} \\ \frac{3}{3} \\ \frac{18}{3} \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 6 \end{bmatrix} = f$

So $y = (-\frac{7}{3}, \frac{4}{3}, \frac{10}{3}, -\frac{14}{3})$ represents the currents flowing along edges 1, 2, 3, and 4 respectively.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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