Corrected exercises: applications of linear algebra in science (201-SN4-RE)

Two unknowns, $T$ and $N$, and two constraints: the equilibrium of a point gives exactly two equations in the plane, one per axis. The cable is parallel to the ramp, so its direction makes $25^{\circ}$ with the horizontal: $\vec{T}=T(\cos 25^{\circ},\ \sin 25^{\circ})$. The force $N$ is perpendicular to the ramp, so at $115^{\circ}$ from the horizontal: $\vec{N}=N(-\sin 25^{\circ},\ \cos 25^{\circ})$. The weight is $(0,\ -400)$. Taking components, $T\cos 25^{\circ}-N\sin 25^{\circ}=0$ along $x$ and $T\sin 25^{\circ}+N\cos 25^{\circ}=400$ along $y$.

The augmented matrix is $\left[\begin{array}{cc|c} \cos 25^{\circ} & -\sin 25^{\circ} & 0 \\ \sin 25^{\circ} & \cos 25^{\circ} & 400 \end{array}\right]$. Its determinant is $\cos^{2}25^{\circ}+\sin^{2}25^{\circ}=1$, by the Pythagorean identity, and not some approximate value from the calculator. This is no accident: the coefficient matrix is a ROTATION matrix, the one that takes the horizontal-vertical frame to the frame attached to the ramp. A determinant equal to $1$, hence non-zero, guarantees a unique solution whatever the right-hand side. In physical terms: whatever the weight, there is one and only one pair $(T, N)$ that holds the cart.

Three unknowns, the masses $x_{1}$, $x_{2}$ and $x_{3}$ taken from the $10\%$, $20\%$ and $50\%$ solutions, in grams. Three constraints: the total mass, the mass of salt, and the lab condition. The total mass gives $x_{1}+x_{2}+x_{3}=600$. The salt brought in is $0.10x_{1}+0.20x_{2}+0.50x_{3}$, and the target contains $0.25\times 600=150$ g. The lab condition reads $x_{2}=2x_{1}$, that is $-2x_{1}+x_{2}=0$. Once the second row is multiplied by $10$ to clear the decimals, the augmented matrix is $\left[\begin{array}{ccc|c} 1 & 1 & 1 & 600 \\ 1 & 2 & 5 & 1500 \\ -2 & 1 & 0 & 0 \end{array}\right]$.

Three unknowns, $I_{1}$, $I_{2}$ and $I_{3}$. The junction gives $I_{1}=I_{2}+I_{3}$, that is $I_{1}-I_{2}-I_{3}=0$. The loop through the source, $R_{1}$ and then $R_{2}$ gives $12-4I_{1}-6I_{2}=0$, that is $4I_{1}+6I_{2}=12$. The loop formed by the two parallel branches gives $6I_{2}-3I_{3}=0$, the two resistors being across the same voltage. Hence $\left[\begin{array}{ccc|c} 1 & -1 & -1 & 0 \\ 4 & 6 & 0 & 12 \\ 0 & 6 & -3 & 0 \end{array}\right]$.

The three setups follow the same protocol, and that is the whole point of the chapter. First name the unknowns, one per quantity the statement does not give. Then write ONE equation per constraint: one per axis in mechanics, one per conserved or imposed quantity in chemistry, one per junction and one per independent loop in electricity. Finally arrange the coefficients in a matrix, and from then on no trace of the context remains: Gauss-Jordan treats the three systems the same way, without knowing whether the letters stand for newtons, grams or amperes. What must be counted before any computation is the number of unknowns on one side and the RANK of the system on the other. As many pivots as unknowns, and the answer is unique; fewer pivots, a constraint is missing and the solution set is infinite; a row of the form $[0\ 0\ 0\,|\,c]$ with $c\neq 0$, and the statement contradicts itself. Counting equations is never enough, because two equations can say only one thing.

Cable $OA$ leaves $O$ toward the upper left, so $\vec{T_{1}}=T_{1}(-\cos 30^{\circ},\ \sin 30^{\circ})$. Cable $OB$ leaves toward the upper right, so $\vec{T_{2}}=T_{2}(\cos 60^{\circ},\ \sin 60^{\circ})$. The weight is $\vec{P}=(0,\ -200)$. A tension always PULLS the point toward the attachment: that is the first mistake to avoid, the sign mistake. The equilibrium $\vec{T_{1}}+\vec{T_{2}}+\vec{P}=\vec{0}$ gives $-T_{1}\cos 30^{\circ}+T_{2}\cos 60^{\circ}=0$ along $x$ and $T_{1}\sin 30^{\circ}+T_{2}\sin 60^{\circ}=200$ along $y$.

$A=\begin{pmatrix} -\cos 30^{\circ} & \cos 60^{\circ} \\ \sin 30^{\circ} & \sin 60^{\circ} \end{pmatrix}$ and $B=\begin{pmatrix} 0 \\ 200 \end{pmatrix}$. Then $\det A=-\cos 30^{\circ}\sin 60^{\circ}-\cos 60^{\circ}\sin 30^{\circ}=-(\sin 60^{\circ}\cos 30^{\circ}+\cos 60^{\circ}\sin 30^{\circ})=-\sin(60^{\circ}+30^{\circ})=-\sin 90^{\circ}=-1$. The addition formula avoids every decimal, and the exact result $-1$ comes from the two cables being PERPENDICULAR, $30^{\circ}+60^{\circ}=90^{\circ}$.

Cramer gives $T_{1}=\dfrac{1}{\det A}\begin{vmatrix} 0 & \cos 60^{\circ} \\ 200 & \sin 60^{\circ} \end{vmatrix}=\dfrac{-200\cos 60^{\circ}}{-1}=200\cos 60^{\circ}=100$ N, and $T_{2}=\dfrac{1}{\det A}\begin{vmatrix} -\cos 30^{\circ} & 0 \\ \sin 30^{\circ} & 200 \end{vmatrix}=\dfrac{-200\cos 30^{\circ}}{-1}=200\cos 30^{\circ}=100\sqrt{3}\approx 173.2$ N. The cable closer to the vertical carries more, which matches intuition. The most costly mistake here is to believe the two tensions share the $200$ N: their sum is $273.2$ N, well above the weight, because they are VECTORS and their horizontal components cancel each other.

Along $x$: $-100\cos 30^{\circ}+173.205\cos 60^{\circ}=-86.603+86.603=0$. Along $y$: $100\sin 30^{\circ}+173.205\sin 60^{\circ}=50+150=200$, which exactly balances the weight. The term $100\sqrt{3}\times\frac{\sqrt{3}}{2}=150$ comes out exact, a sign that the exact computation was worth more than the rounded value.

With two horizontal cables, the directions become $(-1,\ 0)$ and $(1,\ 0)$. The system is $-T_{1}+T_{2}=0$ along $x$ and $0\,T_{1}+0\,T_{2}=200$ along $y$. The matrix $A=\begin{pmatrix} -1 & 1 \\ 0 & 0 \end{pmatrix}$ has a zero row, so $\operatorname{rank}(A)=1$, while the augmented matrix $\left[\begin{array}{cc|c} -1 & 1 & 0 \\ 0 & 0 & 200 \end{array}\right]$ has rank $2$: its second row says $0=200$. The two ranks differ, so the system is INCONSISTENT, and the Rouché-Capelli theorem concludes that there is no solution. This is not a computing mistake, it is physical information: two horizontal cables have no vertical component, so no tension, however large, can balance a weight. A system with no solution in a science problem does not signal an algebra error, it signals that the situation described is impossible.

$W=\vec{F}\cdot\vec{d}=40\times 6+(-15)\times 2+25\times(-3)=240-30-75=135$ J. The dot product is computed component by component, without any angle, and this is the safest form: it needs neither a norm nor a cosine.

$\|\vec{F}\|=\sqrt{1600+225+625}=\sqrt{2450}\approx 49.497$ N and $\|\vec{d}\|=\sqrt{36+4+9}=\sqrt{49}=7$ m. So $\cos\theta=\dfrac{135}{49.497\times 7}\approx 0.38963$ and $\theta\approx 67.1^{\circ}$. The angle is acute, the dot product is positive, the work is POSITIVE (driving): the force goes along with the displacement. The sign of the work is read from the sign of the dot product alone, without computing the angle, and that is the reflex to keep on a test.

The projection of $\vec{F}$ onto $\vec{d}$ is $\vec{F_{\parallel}}=\dfrac{\vec{F}\cdot\vec{d}}{\|\vec{d}\|^{2}}\,\vec{d}=\dfrac{135}{49}(6,\ 2,\ -3)\approx(16.531,\ 5.510,\ -8.265)$. By subtraction, $\vec{F_{\perp}}=\vec{F}-\vec{F_{\parallel}}\approx(23.469,\ -20.510,\ 33.265)$. Check: $\vec{F_{\perp}}\cdot\vec{d}\approx 140.82-41.02-99.80=0$, up to rounding, and the exact computation gives zero. Finally $\vec{F_{\parallel}}\cdot\vec{d}=\dfrac{135}{49}\times 49=135$ J, the same work as in a). The classic trap is to divide by $\|\vec{d}\|$ instead of $\|\vec{d}\|^{2}$: you then get a vector seven times too long, and everything after it is wrong.

Work done by the tension: $W_{T}=120\times 8.0\times\cos 35^{\circ}\approx 786.4$ J. Work done by the weight: the weight is vertical and the displacement horizontal, so the angle is $90^{\circ}$ and $W_{P}=0$ J. Work done by the normal force: it is vertical too, so $W_{N}=0$ J, whatever its value, and its value is not even needed to say so. On a test, computing the normal force before concluding that it does no work is wasted time; the dot product answers before the computation.

Because the dot product of two orthogonal vectors is zero by definition: $\vec{F_{\perp}}\cdot\vec{d}=0$, so $W=\vec{F}\cdot\vec{d}=(\vec{F_{\parallel}}+\vec{F_{\perp}})\cdot\vec{d}=\vec{F_{\parallel}}\cdot\vec{d}$: only the part parallel to the displacement counts, and it counts in full. This is not a small amount of work, it is EXACTLY zero. But the perpendicular component is not useless: for the crate, $F_{\perp}=120\sin 35^{\circ}\approx 68.8$ N relieves the floor, which now exerts only $N=30\times 9.8-68.8\approx 225.2$ N. Friction, proportional to $N$, decreases accordingly. The perpendicular component supplies no energy, but it changes the other forces, hence the final balance. Confusing the two, saying that a force that does no work is useless, is the most costly reasoning error of the chapter.

$\|\vec{M}\|=r\,F\sin\theta=0.30\times 150\times\sin 60^{\circ}=45\times 0.866025\approx 38.97$ N$\cdot$m. Another reading, the lever arm: the distance from $O$ to the line of action of the force is $0.30\sin 60^{\circ}\approx 0.2598$ m, and $0.2598\times 150\approx 38.97$ N$\cdot$m. Both readings give the same number because the sine can be attached either to the length or to the force.

Perpendicular, $\sin 90^{\circ}=1$ and $\|\vec{M}\|=0.30\times 150=45$ N$\cdot$m. The ratio is $\sin 60^{\circ}\approx 0.8660$, so $13.4\%$ of the useful effect is missing. This is exactly why you pull perpendicular to the handle: any departure from $90^{\circ}$ is muscle spent for nothing.

$\vec{F}=(75,\ 129.904,\ 0)$. The cross product is $\vec{r}\times\vec{F}=\big(0\times 0-0\times 129.904,\ 0\times 75-0.30\times 0,\ 0.30\times 129.904-0\times 75\big)=(0,\ 0,\ 38.97)$. Only the $z$-component survives, as expected: $\vec{r}$ and $\vec{F}$ both lie in the $xy$-plane, so their cross product is perpendicular to that plane. Its magnitude is indeed $38.97$ N$\cdot$m, and its positive sign says that the bolt turns counterclockwise as seen from the $z$-axis.

$\vec{F_{\parallel}}=(75,\ 0,\ 0)$ and $\vec{F_{\perp}}=(0,\ 129.904,\ 0)$. The torque of the parallel part is $\vec{r}\times\vec{F_{\parallel}}=(0.30,\ 0,\ 0)\times(75,\ 0,\ 0)=\vec{0}$, since two parallel vectors have a zero cross product. The torque of the perpendicular part is $(0,\ 0,\ 0.30\times 129.904)=(0,\ 0,\ 38.97)$, that is, the whole torque. The parallel with the previous exercise is exact and worth remembering: for work, only the component PARALLEL to the displacement counts and the perpendicular one gives nothing; for torque it is the reverse, only the component PERPENDICULAR to the arm counts and the parallel one gives nothing. The dot product and the cross product select complementary parts, and that is why confusing the two costs the whole question.

$\vec{M}=(0.25,\ 0,\ 0.10)\times(0,\ 200,\ 0)=\big(0\times 0-0.10\times 200,\ 0.10\times 0-0.25\times 0,\ 0.25\times 200-0\times 0\big)=(-20,\ 0,\ 50)$ N$\cdot$m. Its magnitude is $\sqrt{400+2500}=\sqrt{2900}\approx 53.85$ N$\cdot$m. Two non-zero components mean that the force tends to turn the system about TWO axes at once: about $z$, which actually tightens the bolt, and about $x$, which tends to tip the wrench out of its plane. The component $M_{z}=50$ is the only useful one; the component $M_{x}=-20$ is a parasitic load that the assembly must absorb. The torque vector therefore carries more information than a number: its DIRECTION is the axis of rotation, and when the torque is oblique, part of it works against the assembly.

The rows are the ELEMENTS, the columns the SPECIES, in the order $\mathrm{Al}$, $\mathrm{HCl}$, $\mathrm{AlCl_{3}}$, $\mathrm{H_{2}}$. Aluminum: $x_{1}-x_{3}=0$. Hydrogen: $x_{2}-2x_{4}=0$. Chlorine: $x_{2}-3x_{3}=0$. Hence $A=\begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -2 \\ 0 & 1 & -3 & 0 \end{pmatrix}$, of size $3\times 4$: three elements, four species. The right-hand side is the zero vector, always, because a conservation law says that what goes in equals what comes out.

$R_{3}\to R_{3}-R_{2}$ gives $(0,\ 0,\ -3,\ 2)$. The echelon form is $\begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & -3 & 2 \end{pmatrix}$, with three pivots: $\operatorname{rank}(A)=3$. The rank theorem gives $\dim\ker A=4-3=1$. So there is exactly one direction of solutions, a line in $\mathbb{R}^{4}$, and that is why a chemical equation balances in only one way UP TO SCALE.

Let $x_{4}=t$. The third row gives $-3x_{3}+2t=0$, so $x_{3}=\frac{2t}{3}$; the second gives $x_{2}=2t$; the first $x_{1}=x_{3}=\frac{2t}{3}$. The null space is spanned by $\left(\frac{2}{3},\ 2,\ \frac{2}{3},\ 1\right)$, or more conveniently by $(2,\ 6,\ 2,\ 3)$, taking $t=3$. The smallest positive integers are therefore $x_{1}=2$, $x_{2}=6$, $x_{3}=2$, $x_{4}=3$, and the balanced equation is $2\,\mathrm{Al}+6\,\mathrm{HCl}\rightarrow 2\,\mathrm{AlCl_{3}}+3\,\mathrm{H_{2}}$. Check: aluminum, $2$ on the left and $2$ on the right; hydrogen, $6$ and $6$; chlorine, $6$ and $6$. The trap is to take $t=1$ by reflex: you get thirds, and the whole family then has to be multiplied. The free parameter need not be $1$; the common denominator decides.

Column order: $\mathrm{MnO_{4}^{-}}$, $\mathrm{Fe^{2+}}$, $\mathrm{H^{+}}$, $\mathrm{Mn^{2+}}$, $\mathrm{Fe^{3+}}$, $\mathrm{H_{2}O}$. Manganese: $x_{1}-x_{4}=0$. Oxygen: $4x_{1}-x_{6}=0$. Iron: $x_{2}-x_{5}=0$. Hydrogen: $x_{3}-2x_{6}=0$. The matrix $\begin{pmatrix} 1 & 0 & 0 & -1 & 0 & 0 \\ 4 & 0 & 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & 0 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 & -2 \end{pmatrix}$ is $4\times 6$. A single operation, $R_{2}\to R_{2}-4R_{1}$, puts it in echelon form with pivots in columns $1$, $2$, $3$ and $4$: $\operatorname{rank}=4$, so $\dim\ker=6-4=2$, and the free unknowns are $x_{5}$ and $x_{6}$. Setting $x_{6}=4a$ and $x_{5}=b$ gives $x_{1}=a$, $x_{4}=a$, $x_{3}=8a$, $x_{2}=b$, that is, the family $(a,\ b,\ 8a,\ a,\ b,\ 4a)$. Two free parameters, so the equation is NOT determined: $a=1,\ b=0$ gives $\mathrm{MnO_{4}^{-}}+8\,\mathrm{H^{+}}\rightarrow\mathrm{Mn^{2+}}+4\,\mathrm{H_{2}O}$ and $a=0,\ b=1$ gives $\mathrm{Fe^{2+}}\rightarrow\mathrm{Fe^{3+}}$, two equations that conserve every atom and that no chemist would accept. The dimension of the null space has just said exactly what is missing: a constraint.

Charge is a conserved quantity just like an element, so it is one more ROW. Charges, in column order: $-1$, $+2$, $+1$ for the reactants, $+2$, $+3$, $0$ for the products, hence the row $(-1,\ 2,\ 1,\ -2,\ -3,\ 0)$. Applied to the previous family, it gives $-a+2b+8a-2a-3b=0$, that is $5a-b=0$ and $b=5a$. The rank goes up to $5$, the dimension of the null space drops to $6-5=1$, and the equation is determined again. With $a=1$: $x_{1}=1$, $x_{2}=5$, $x_{3}=8$, $x_{4}=1$, $x_{5}=5$, $x_{6}=4$, that is $\mathrm{MnO_{4}^{-}}+5\,\mathrm{Fe^{2+}}+8\,\mathrm{H^{+}}\rightarrow\mathrm{Mn^{2+}}+5\,\mathrm{Fe^{3+}}+4\,\mathrm{H_{2}O}$. Charge check: $-1+10+8=17$ on the left, $2+15=17$ on the right. This is the lesson of the whole chapter: the dimension of the null space does not merely count the solutions, it DIAGNOSES the statement. A dimension of $1$ signals a well-posed reaction; a dimension of $2$ or more signals that a conservation law was forgotten, here charge; and an inconsistent system would signal an impossible reaction.

Junction rule at $A$: the currents $I_{1}$ and $I_{3}$ come in, $I_{2}$ goes out, so $I_{1}-I_{2}+I_{3}=0$. Junction rule at $B$: $I_{2}$ comes in, $I_{1}$ and $I_{3}$ go out, so $-I_{1}+I_{2}-I_{3}=0$. Left loop, traversed from $B$ to $A$ through branch $1$ and then from $A$ to $B$ through the middle branch: $18-3I_{1}-6I_{2}=0$, that is $3I_{1}+6I_{2}=18$. Right loop, in the same way: $15-2I_{3}-6I_{2}=0$, that is $6I_{2}+2I_{3}=15$.