Linear Algebra and Vector Geometry 201-SN4-RE • CEGEP, Montreal

Revision sheet: applications of linear algebra in science (201-SN4-RE)

Q: How do you turn a physics problem into a linear system?

Three steps, always in this order. First name one unknown per quantity the statement does not give, with its unit and its direction. Then write one equation per constraint: one per component direction in mechanics, one per conserved quantity in chemistry, one per junction and per independent loop in electricity. Finally arrange the coefficients in an augmented matrix, one column per unknown.

Q: What does a system with no solution mean in a science problem?

It is not a computing error, it is an answer. When the rank of the coefficient matrix is smaller than the rank of the augmented matrix, one row says zero equals a non-zero constant: the situation described is physically impossible. You must then name the false assumption, for example horizontal cables that have no vertical component, instead of redoing the row reduction.

Q: How many junction rules should you write for a circuit?

Exactly one fewer than the number of junctions. The sum of all the junction rules of a circuit is the identity zero equals zero, since each branch current leaves one junction and enters another. The last row is therefore a combination of the others and adds no rank. You complete the system with independent loop rules until there are as many pivots as unknowns.

Q: Why can't a redox equation be balanced without conservation of charge?

Because the element conservations alone leave a null space of dimension two instead of one. Two free parameters mean that two independent reactions conserve every atom, and neither one is the reaction you are looking for. Charge is a conserved quantity just like an element: it is added as a row, the rank goes up by one and the equation becomes determined again.

Q: What is the difference between the dot product and the cross product in mechanics?

They select complementary parts of the same force. Work is a dot product, hence a number, and only the component parallel to the displacement counts: the perpendicular component does exactly zero work. Torque is a cross product, hence a vector, and it is the reverse: only the component perpendicular to the arm makes it turn. Where one is zero, the other is at its largest.

Q: Can a system be consistent and still have no usable solution?

Yes, and it is the question that separates the papers. Row reduction knows no sign constraint: it can deliver a line of solutions none of which has all its components positive. A negative mass or concentration does not exist, so the free parameter must be restricted to the physical domain, and you must then check that the resulting interval is not empty.

The fifth element of the competency of Linear Algebra and Vector Geometry, 201-SN4-RE (formerly 201-NYC), asks you to apply the methods of the course to problems in the natural sciences: equilibrium of forces, biomechanics and torques, work, balancing chemical equations, mixing solutions, Kirchhoff's laws, centre of mass. Once the system is set up, Gauss-Jordan is mechanical. The marks are lost before that.

So this sheet does not redo ten solutions. It rests on a single skill, the one nobody teaches explicitly: define the unknowns and COUNT the constraints before computing, then read the rank as a diagnosis. As many pivots as unknowns, the answer is unique; fewer pivots, a constraint is missing; a row $[0\ \cdots\ 0\,|\,c]$ with $c\neq 0$ , and it is the situation described that is impossible. One last check is up to you, because the algebra ignores it: the sign.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Linear algebra does not solve the science problem, it SETS IT UP: marks are lost naming the unknowns and counting the constraints, almost never at the pivot.

This chapter is part of 201-SN4-RE, Linear Algebra and Vector Geometry

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Augmented matrix
Rank of a matrix
Rouché-Capelli theorem
Rank theorem
Null space of a matrix
Dot product
Cross product
Torque of a force
Equilibrium of a rigid body
Kirchhoff's laws
Centre of mass
Convex combination

The essentials

The three steps of the setup, always in this order

• 1. NAME. One unknown per quantity the statement does not give, with its letter, its unit and the chosen orientation: $T$ in newtons, $x_{1}$ in grams, $I_{2}$ in amperes counted in the direction of the arrow. An unknown that is not named is an unknown you will forget.
• 2. COUNT. One equation per CONSTRAINT, not one per number in the statement. A constraint is a component direction in mechanics, a conserved quantity in chemistry, an independent junction or loop in electricity, a condition imposed by the statement.
• 3. ARRANGE. The coefficients in an augmented matrix $[A\,|\,B]$ , one column per unknown, one row per constraint. From then on, no trace of the context remains: Gauss-Jordan does not know whether the letters are newtons or grams.
• The count that matters is never the number of ROWS, it is the RANK: two equations can say only one thing.

Name on the figure:

T_{1}

and

T_{2}

, the cable tensions in newtons, the weight

P=200

N, the axes

x

and

y

. Two unknowns for a point in the plane: two component equations.

Writing out the sentence “ $T$ is the tension in the cable in newtons, $N$ the reaction of the ramp in newtons” is worth half a mark for the setup on almost every marking scheme, and it is the most profitable line of the test.

Reading the rank: three verdicts, and not one more

• Rouché-Capelli consistency theorem: $AX=B$ has solutions if and only if $\operatorname{rank}(A)=\operatorname{rank}([A\,|\,B])$ .
• Case 1, $\operatorname{rank}=n$ unknowns: a UNIQUE solution, the problem is well posed. For a square matrix, $\det A\neq 0$ says the same thing faster.
• Case 2, $\operatorname{rank}<n$ and the two ranks equal: INFINITELY many solutions, with $n-\operatorname{rank}$ free parameters. A constraint is missing, or the parameter itself has a meaning, like the scale of a chemical reaction.
• Case 3, the two ranks differ: NO solution. A row $[0\ \cdots\ 0\,|\,c]$ with $c\neq 0$ says $0=c$ , and it is the situation described that is impossible, not your row reduction.

Three pivots for three unknowns on the left: a unique answer. In the middle the last row is zero, right-hand side included: a constraint is missing. On the right it carries

5

, so

0=5

: impossible.

The rank is the only tool in the course that DIAGNOSES a problem instead of solving it. That is what linear algebra brings to the sciences and calculus does not.

Dot product or cross product: two complementary selections

• Work done by a constant force along a straight displacement: $W=\vec{F}\cdot\vec{d}=\|\vec{F}\|\,\|\vec{d}\|\cos\theta$ , a NUMBER, in joules. Only the part PARALLEL to the displacement does work.
• Torque of a force: $\vec{M}=\vec{r}\times\vec{F}$ , of magnitude $\|\vec{r}\|\,\|\vec{F}\|\sin\theta$ , a VECTOR, in newton-metres. Only the part PERPENDICULAR to the arm makes it turn.
• Orthogonal projection, the formula used in both cases: $\vec{F_{\parallel}}=\dfrac{\vec{F}\cdot\vec{d}}{\|\vec{d}\|^{2}}\,\vec{d}$ , then $\vec{F_{\perp}}=\vec{F}-\vec{F_{\parallel}}$ . The denominator is $\|\vec{d}\|^{2}$ , never $\|\vec{d}\|$ .
• Immediate consequence: $\cos\theta$ and $\sin\theta$ do not vanish at the same place, so the two products are not merely different, they are complementary. Where one is zero, the other is at its largest.

The sign of the work is read from the sign of the dot product alone, without computing the angle: positive, driving work; negative, resisting work; zero, the force does no work.

What the algebra ignores: the sign

• Gauss-Jordan knows no sign constraint. Yet a mass, a concentration, a length, a one-way flow are positive by nature.
• A system can therefore be perfectly CONSISTENT and have NO admissible solution: the line of solutions exists, but it does not cross the physical domain.
• Conversely, a negative component is not always a mistake: a negative current or joint reaction simply says that the direction assumed on the figure was the opposite of the real one.
• Centre of mass check: the coefficients $\frac{m_{i}}{\sum m_{j}}$ are positive and add up to $1$ , so $G$ is a CONVEX combination of the points and never leaves their convex hull. A $G$ outside the polygon is a wrong answer.

Two consistent systems. The blue line crosses the quadrant of positive quantities (in green): admissible solutions exist. The red one misses it: consistent, but nothing physical.

Every answer is checked twice: first by the rank, then by the physical domain. The second check is the one no calculator will do for you.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

How many INDEPENDENT equations each context provides

Count the rank, never the rows. A red cell marks a row that adds NO rank: there is no point writing it for reassurance, the missing constraint is elsewhere.

Context	Rows written	Rank added
Point in the plane	2 components	$2$
Example: Light fixture on two cables at $30^{\circ}$ and $60^{\circ}$ : $2$ unknowns, $\det A=-\sin 90^{\circ}=-1$ , so $T_{1}=100$ N and $T_{2}\approx 173.2$ N.
Rigid body in the plane	2 components and 1 torque	$3$
Example: Shoulder of the worked example: $3$ unknowns, $\det A=0.16\sin 16^{\circ}\approx 0.0441\neq 0$ , unique solution.
Torque about a 2nd point	1 more row	$0$ no extra rank
Example: At the deltoid attachment you read $R_{y}=-150$ N again, a value the first three rows already gave. What to do: Choose the point to make an unknown DISAPPEAR from the row, never to gain an equation.
Circuit with $n$ junctions	$n$ junction rules	$n-1$ one row too many
Example: Two junctions: $I_{1}-I_{2}+I_{3}=0$ at $A$ and $-I_{1}+I_{2}-I_{3}=0$ at $B$ , two rows for a rank of $1$ . Same form, other result: The same circuit with $4$ equations keeps a rank of $3$ and still gives $I_{1}=1.50$ A, $I_{2}=2.25$ A, $I_{3}=0.75$ A: more rows than unknowns does not mean overdetermined. What to do: Write $n-1$ junction rules, then complete with independent loop rules.
Chemistry, $k$ elements	$k$ conservations	$\dim\ker=p-r$
Example: $\mathrm{Al}+\mathrm{HCl}\rightarrow\mathrm{AlCl_{3}}+\mathrm{H_{2}}$ : $p=4$ species, $r=3$ , so $\dim\ker=1$ and the coefficients are $(2,\ 6,\ 2,\ 3)$ .
Redox without charge	4 conservations	$\dim\ker=2$ missing constraint
Example: $\mathrm{MnO_{4}^{-}}+\mathrm{Fe^{2+}}+\mathrm{H^{+}}$ : $6$ species, rank $4$ , so $\dim\ker=6-4=2$ . Same form, other result: The two directions of the null space give $\mathrm{MnO_{4}^{-}}+8\,\mathrm{H^{+}}\rightarrow\mathrm{Mn^{2+}}+4\,\mathrm{H_{2}O}$ and $\mathrm{Fe^{2+}}\rightarrow\mathrm{Fe^{3+}}$ : every atom is conserved, and neither one is the reaction. What to do: Add conservation of CHARGE as a fifth row: the rank goes up to $5$ , $\dim\ker$ drops back to $1$ and the coefficients are $(1,\ 5,\ 8,\ 1,\ 5,\ 4)$ .

Two rows of the same kind are independent only if they carry new information. The test is not to look at them, it is to row reduce: a row that vanishes entirely, right-hand side included, was a combination of the others.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Starting to compute without naming the unknowns

1 setup mark, and the whole question if an unknown is missing

What not to write

“Taking components: $F\cos 30^{\circ}+F'\cos 60^{\circ}=0$ and $F\sin 30^{\circ}+F'\sin 60^{\circ}=200$ .”

What to write

“Let $T_{1}$ be the tension in cable $OA$ and $T_{2}$ the tension in cable $OB$ , both in newtons, with the $x$ -axis horizontal to the right and the $y$ -axis vertical and upward.”

Why: Without a declaration, nothing says in which ORDER the columns of the matrix are arranged, and the marker cannot follow the row reduction. It is also the only moment you notice that an unknown quantity has been forgotten.

2. Counting equations instead of counting the rank

2 marks, and a useful piece of data thrown away if the dropped row was not the redundant one

What not to write

“I have four equations for three unknowns, so the system is overdetermined and I drop a row at random.”

What to write

“Row reduction wipes out the second row entirely, right-hand side included: $\operatorname{rank}(A)=\operatorname{rank}([A\,|\,B])=3$ , equal to the number of unknowns, so the solution exists and is unique.”

Why: The number of rows measures what you wrote, the rank measures what you know. Four rows of rank $3$ overdetermine nothing at all.

3. Writing the junction rule at EVERY junction and thinking you gained an equation

the whole question when the system stays underdetermined and you do not understand why

What not to write

“Two junctions, so two junction rules: $I_{1}-I_{2}+I_{3}=0$ and $-I_{1}+I_{2}-I_{3}=0$ . With the two loops, that makes four constraints.”

What to write

“The rule at $B$ is the negative of the rule at $A$ , so it adds no rank: a circuit with $n$ junctions provides $n-1$ independent junction rules. I write one junction rule and two loop rules.”

The three currents are the same in both equations: what enters at

A

is what leaves at

B

. The two rows are negatives of each other term by term, so together they count for a rank of

1

Why: The sum of all the junction rules is the identity $0=0$ : each branch current leaves one junction and enters another, so it appears once with each sign. That is exactly the definition of a linear dependence among those rows.

4. Taking torques about a second point to get one more constraint

the whole question, plus the time lost row reducing rows that cancel each other

What not to write

“The system has three unknowns, so I write torques about the joint, then about the attachment point, then about the hand: that gives me three torque equations.”

What to write

“The equilibrium of a rigid body in the plane provides three constraints, two for the resultant and one for the torque. Torques about another point are a linear combination of the first three, so the rank stays $3$ .”

Why: The choice of point adds no information, it chooses WHICH unknown disappears from the row. With torques about the joint, $R_{x}$ and $R_{y}$ disappear and $T$ is isolated; about the attachment point, $T$ disappears and $R_{y}$ is isolated. It is a choice of pivot, not an extra constraint.

5. Forgetting the direction of a force when taking components

2 marks, and negative tensions handed in without comment

What not to write

“Both cables pull upward, so $T_{1}\cos 30^{\circ}+T_{2}\cos 60^{\circ}=0$ along the $x$ -axis.”

What to write

“A tension PULLS the point toward its attachment: $\vec{T_{1}}=T_{1}(-\cos 30^{\circ},\ \sin 30^{\circ})$ and $\vec{T_{2}}=T_{2}(\cos 60^{\circ},\ \sin 60^{\circ})$ , hence $-T_{1}\cos 30^{\circ}+T_{2}\cos 60^{\circ}=0$ .”

\vec{T_{1}}

pulls the point toward ITS attachment, at the upper left: its horizontal component is

-T_{1}\cos 30^{\circ}

(in green). The wrong projection (in red, dashed) makes it pull to the right.

Why: A sum of components of the same sign cannot vanish: a horizontal component equation whose terms are all positive is wrong before any computation, and that takes three seconds to see.

6. Confusing the dot product and the cross product

the whole question, since the result is even of the wrong KIND, a number instead of a vector

What not to write

“The torque is $\vec{r}\cdot\vec{F}=rF\cos\theta$ , that is $0.30\times 150\times\cos 60^{\circ}=22.5$ N $\cdot$ m.”

What to write

“Torque is a CROSS product: $\vec{M}=\vec{r}\times\vec{F}$ , of magnitude $rF\sin\theta=0.30\times 150\times\sin 60^{\circ}\approx 38.97$ N $\cdot$ m.”

Same geometry on both sides, opposite selection: on the left only

F_{\parallel}

does work and

F_{\perp}

gives

W=0

; on the right only

F_{\perp}

turns and

F_{\parallel}

gives

M=0

Why: The two products select complementary parts. With $\vec{r}=(0.30,\ 0,\ 0)$ and $\vec{F}=(0,\ 200,\ 0)$ , the dot product is exactly $0$ while the torque is at its LARGEST: the dot product keeps the parallel part, the cross product keeps the perpendicular part.

7. Believing that a force perpendicular to the displacement does a little work

1 mark, and twice that if the value is carried into an energy balance

What not to write

“The normal force is $N\approx 225$ N and the displacement $8.0$ m, so it does a small amount of work that we neglect.”

What to write

“ $\vec{N}$ is vertical and $\vec{d}$ horizontal, so $W_{N}=\vec{N}\cdot\vec{d}=0$ J exactly, whatever the value of $N$ . There is no need to compute $N$ to say so.”

The normal force

\vec{N}

is vertical, the displacement

\vec{d}

horizontal: the right angle makes

\vec{N}\cdot\vec{d}=0

, so the work is EXACTLY zero, whatever the value of

N

Why: The work done by a force orthogonal to the displacement is zero, not small, and it depends neither on the magnitude nor on the distance travelled. But the perpendicular component is not useless: it relieves the floor, so it changes the friction, so it changes the balance.

8. Concluding “no solution” and stopping there

2 marks out of 3, the physical conclusion being exactly the question asked

What not to write

“The two ranks differ, the system is inconsistent.” Then the row reduction is redone three times, looking for the mistake.

What to write

“ $\operatorname{rank}(A)=1$ and $\operatorname{rank}([A\,|\,B])=2$ : the system is inconsistent. Two horizontal cables have no vertical component, so no tension can balance the weight: the situation described is physically impossible.”

Two horizontal cables:

T_{1}

and

T_{2}

have no vertical component, nothing balances the weight

P

. The inconsistency says the situation is physically impossible.

Why: In a science problem, inconsistency is an ANSWER, not a failure: it says that an assumption of the statement is false. That is how troubleshooting is done in the lab, when an ammeter reading contradicts the stated component values.

9. Taking a consistent system for a solved problem

the whole last question, the one that separates the papers

What not to write

“Both ranks are $2$ , so the system is consistent: $600$ g of a $60\%$ solution can be made from the three stock solutions.”

What to write

“The system is consistent, with solutions $(3t-2400,\ 3000-4t,\ t)$ . But a mass is positive: $3t-2400\geq 0$ requires $t\geq 800$ , whereas $t\leq 600$ . No solution is ADMISSIBLE, which was predictable, the most concentrated stock solution being at $50\%$ .”

x_{1}=3t-2400\geq 0

requires

t\geq 800

(in orange), the total mass imposes

t\leq 600

(in blue): the two intervals do not meet, no solution is admissible.

Why: Algebra decides consistency, never admissibility: sign constraints do not enter the matrix. Two checks are therefore always needed, the rank first, then the physical domain.

10. Balancing a redox equation with the element conservations only

the whole question, the equation being wrongly balanced

What not to write

“Four elements, so four equations, and I solve.” You get two free parameters and fix one at random.

What to write

“ $\dim\ker=6-4=2$ , so a conservation law is missing: CHARGE is conserved just like an element. The row $(-1,\ 2,\ 1,\ -2,\ -3,\ 0)$ brings the rank to $5$ and $\dim\ker$ to $1$ .”

Why: A null space of dimension $1$ signals a well-posed reaction, the parameter simply being the scale. A dimension of $2$ or more is not an embarrassment of riches, it is the diagnosis of a forgotten conservation law.

Which method to choose

How many equations to write, depending on the form of the problem

Look at what is in equilibrium or what is conserved, never at the topic of the exercise. The list of constraints is read on the object, not on the chapter.

If a POINT (knot of cables, ring, hanging mass) in equilibrium in the plane → 2 equations, one per component direction

Example: light fixture on two cables: $2$ unknowns, $2$ equations, $\det A=-1$

in space, 3 equations
If a RIGID BODY (bar, forearm, beam) in equilibrium in the plane → 3 equations: 2 for the resultant and 1 for the torque

Example: shoulder and deltoid: $T$ , $R_{x}$ , $R_{y}$ , three rows

a 4th torque equation never raises the rank
If a CIRCUIT with $n$ junctions and $b$ branches → $n-1$ junction rules, completed by $b-n+1$ independent loop rules

Example: $2$ junctions and $3$ branches: $1$ junction rule and $2$ loops, so $3$ equations

a loop is independent if it contains a new branch
If a CHEMICAL EQUATION to balance, $p$ species and $k$ elements → $k$ element conservations, plus charge as soon as an ion appears; you look for $\ker A$

Example: $\mathrm{Al}+\mathrm{HCl}$ : $k=3$ , $p=4$ , $\dim\ker=1$

the system is HOMOGENEOUS, the right-hand side is always zero
If a MIXTURE or a distribution of masses → 1 equation per conserved quantity (total mass, mass of solute) plus 1 per condition imposed by the statement

Example: three stock solutions, one target mass, one protocol: $3$ equations for $3$ unknowns

“twice as much of the $20\%$ solution” is a constraint, not a piece of data

If no branch applies, the object has not been identified: first ask yourself whether the problem is about a point, a rigid body, a network or a conservation law.

The row reduction is done: what do you say?

Look at the reduced augmented matrix, count the pivots, and read the last non-zero row.

If as many pivots as unknowns → unique solution: back substitute, then check the SIGN of each component

Example: $3$ pivots, $3$ unknowns: $I_{1}=1.50$ A, $I_{2}=2.25$ A, $I_{3}=0.75$ A
If fewer pivots than unknowns, the two ranks equal → infinitely many solutions: parametrize by the free unknowns, THEN restrict by the sign constraints

Example: $(3t-300,\ 900-4t,\ t)$ with $t\in[100, 225]$ , a segment and not a line

in chemistry, the free parameter is the scale of the reaction and is chosen as the smallest integer
If a row $[0\ \cdots\ 0\,|\,c]$ with $c\neq 0$ → inconsistent system: name the assumption of the statement that is false

Example: $0=200$ for two horizontal cables: no vertical component available
If the system is consistent but a component that must be positive comes out negative → no ADMISSIBLE solution: conclude on the physical domain, not on the rank

Example: $600$ g at $60\%$ from stock solutions at $50\%$ at most

a negative current or joint reaction, on the other hand, is admissible: it reverses a direction

Only one thing is forbidden here: redoing the row reduction because you do not like the result. A row $0=c$ is checked once, then interpreted.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Turning a science problem into a linear system

When to use it: As soon as the statement describes a physical or chemical situation and asks for numerical quantities, even if the word “system” appears nowhere.

1 Declare the unknowns, with the letter, the quantity, the unit and the chosen direction or orientation. One line, in words.
2 Set the frame or the convention: axes $x$ and $y$ , positive direction of the currents, order of the species in the columns.
3 List the CONSTRAINTS before translating them, and count them out loud: “two components”, “one conservation per element plus charge”, “one junction and two loops”.
4 Translate each constraint into one equation, in the order of the list, then arrange the coefficients in $[A\,|\,B]$ following the announced column order.
5 Row reduce, state $\operatorname{rank}(A)$ and $\operatorname{rank}([A\,|\,B])$ , conclude with the Rouché-Capelli theorem, then solve.
6 Check the physical domain of each component, and comment on any negative sign.

Concluding sentence

“The unknowns are $T$ , $R_{x}$ and $R_{y}$ , in newtons. The equilibrium of a rigid body in the plane provides three constraints: two components and one torque equation. The resulting system satisfies $\operatorname{rank}(A)=\operatorname{rank}([A\,|\,B])=3$ , equal to the number of unknowns, so the solution exists and is unique.”

The trap: Step 3. Almost everyone translates the constraints one by one as they discover them, and realizes too late that one is missing or that the same one was written twice.

Marking: Typically 1 mark for the unknowns and the convention, 1 mark per correct equation, 1 mark for reading the rank, 1 mark for the physical conclusion.

Concluding on a system that has no unique solution

When to use it: As soon as the row reduction produces a zero row or a row $[0\ \cdots\ 0\,|\,c]$ . It is almost always the last question, and it is the one worth the most.

1 Give both ranks explicitly, that of $A$ and that of $[A\,|\,B]$ , and cite the Rouché-Capelli theorem by name.
2 Say which case it is: infinitely many solutions with $n-\operatorname{rank}$ parameters, or no solution.
3 Translate into PHYSICS: which constraint is missing, or which assumption of the statement cannot be true.
4 If the system is consistent, parametrize, then restrict the parameter by the sign constraints, and give the admissible interval.

Concluding sentence

“ $\operatorname{rank}(A)=1$ and $\operatorname{rank}([A\,|\,B])=2$ : the two ranks differ, so by the consistency theorem the system has no solution. Physically, two horizontal cables have no vertical component: no tension, however large, can balance the weight. The situation described is impossible.”

The trap: Stopping at “inconsistent”. The mathematical word earns only one mark out of three: the question asked what it means for the physical system.

Marking: Usually 1 mark for the two ranks, 1 mark for the mathematical conclusion, 1 mark for the physical interpretation.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Substitute back into the ORIGINAL equations

Take the equations as they were written in step 4, not the reduced matrix, and plug in the values found. A pivot error shows up immediately; an error in SETTING UP the equations would never show on the echelon form.

Left loop: $3\times 1.50+6\times 2.25=4.5+13.5=18$ , the source voltage. Right loop: $6\times 2.25+2\times 0.75=15$ . If either one does not come out right, it is the row that is wrong, not the computation.
The sign check, quantity by quantity

Go through the answer and ask, for each component, whether a negative value makes sense. Mass, concentration, length, magnitude of a cable tension: no, there is a mistake. Current, support reaction, force component: yes, the assumed direction was the opposite, and you must SAY so.

$R_{y}=-150$ N is not a mistake: the joint pushes the arm down, because the deltoid already pulls upward harder than the total weight. A mass of $-80$ g, on the other hand, is a mistake.
Count the pivots before handing in

Reread the top of your work: how many unknowns declared, how many pivots found. The two numbers must be equal, or the answer must contain a parameter. A unique answer coming out of a rank- $2$ system in three unknowns is wrong by construction.

Three unknowns and two pivots, but a single numerical answer on the paper: a constraint was used mentally without being written, and the marker cannot guess it.
Order of magnitude and convex hull

Compare the result with the geometry. A centre of mass must fall INSIDE the polygon of the points; a muscle tension must be large compared with the load when the lever arm is short; a mixture percentage cannot exceed that of the most concentrated solution.

Masses $2$ , $3$ and $5$ kg at $(0, 0)$ , $(8, 0)$ and $(4, 5)$ : $G(4.4,\ 2.5)$ is indeed in the triangle, and the coefficients $0.2$ , $0.3$ and $0.5$ add up to $1$ . A $G$ with $y$ -coordinate $6$ would have been impossible.

The typical problem, taken apart

The outstretched arm: three unknowns, three constraints, and the choice of point

An arm held out horizontally carries a $50$ N load in the hand, $0.55$ m from the shoulder joint. The arm itself weighs $32$ N, its centre of gravity being $0.30$ m from the joint. The deltoid attaches $0.16$ m from the joint and pulls upward and toward the shoulder, in a direction making $16^{\circ}$ with the bone.

Origin at the joint, $x$ -axis along the arm toward the hand, $y$ -axis upward. Find the deltoid tension $T$ and the two components $R_{x}$ and $R_{y}$ of the force the joint exerts on the arm.

The three distances are all measured from the joint

O

, never from one another, and the

16^{\circ}

angle is between the tension and the BONE, not the vertical.

Step 1

Unknowns: $T$ the deltoid tension, $R_{x}$ and $R_{y}$ the components of the joint reaction, all three in newtons. The arm is a RIGID BODY, so equilibrium in the plane provides three constraints: two components and one torque equation.

Why

This is the step that decides everything else. Three unknowns and three constraints announce from the first line that you are aiming for a unique solution. A particle would have given only two, and the system would have stayed underdetermined without your understanding why.

Step 2

The tension is along $(-\cos 16^{\circ},\ \sin 16^{\circ})$ , the weight of the arm is $(0,\ -32)$ and the load $(0,\ -50)$ . Component along $x$ : $-T\cos 16^{\circ}+R_{x}=0$ . Component along $y$ : $T\sin 16^{\circ}+R_{y}-32-50=0$ , that is $T\sin 16^{\circ}+R_{y}=82$ .

Why

The minus sign comes from the deltoid pulling toward the SHOULDER, so in the direction of decreasing $x$ . Writing the components BEFORE projecting avoids the costliest sign error of the chapter.

Step 3

Torques about the JOINT: $\vec{R}$ is applied there, so its torque is zero; the component $-T\cos 16^{\circ}$ lies along the $x$ -axis, which passes through the origin, so it gives nothing either. What remains is $0.16\,T\sin 16^{\circ}-0.30\times 32-0.55\times 50=0$ , that is $0.16\,T\sin 16^{\circ}=9.6+27.5=37.1$ .

Why

The point is not chosen at random: at the joint, TWO unknowns disappear at once and the row isolates $T$ . It is a choice of pivot made in advance, and it is what turns a $3\times 3$ system into three successive computations.

Step 4

$A=\begin{pmatrix} -\cos 16^{\circ} & 1 & 0 \\ \sin 16^{\circ} & 0 & 1 \\ 0.16\sin 16^{\circ} & 0 & 0 \end{pmatrix}$ , and expanding along the third row, $\det A=0.16\sin 16^{\circ}\approx 0.0441\neq 0$ : $\operatorname{rank}(A)=\operatorname{rank}([A\,|\,B])=3$ , equal to the number of unknowns, so the solution is unique.

Why

For a square matrix, the determinant answers the rank question in one line. Since the third row has only one non-zero entry, the expansion is done mentally: looking for a zero before expanding is always better than running through the rule of Sarrus.

Step 5

$T\sin 16^{\circ}=\dfrac{37.1}{0.16}=231.875$ N, so $T=\dfrac{231.875}{\sin 16^{\circ}}\approx 841$ N. Then $R_{x}=T\cos 16^{\circ}\approx 809$ N and $R_{y}=82-231.875\approx -150$ N.

Why

You back substitute in the reverse order of the reduction, each row containing only one new unknown. Keeping $T\sin 16^{\circ}$ as a block until the end avoids rounding the same quantity twice.

Step 6

Checks. Sign: $R_{y}<0$ is admissible, the joint PUSHES the arm down. Magnitude: $\|\vec{R}\|=\sqrt{809^{2}+150^{2}}\approx 822$ N, about $16$ times the load held. Torques about the attachment point, as a check: $-0.16\,R_{y}-0.14\times 32-0.39\times 50=0$ gives $-0.16\,R_{y}=4.48+19.5=23.98$ , so $R_{y}\approx -150$ N.

Why

The third check is the most instructive: it gives $R_{y}$ again but adds no rank, since this row is a linear combination of the three previous ones. It therefore serves as a check, never as a fourth constraint.

The conclusion, written out

“The three unknowns are $T$ , $R_{x}$ and $R_{y}$ , in newtons, and the equilibrium of the rigid body provides exactly three constraints. The determinant of the system is $0.16\sin 16^{\circ}\approx 0.0441\neq 0$ , so the solution is unique: $T\approx 841$ N, $R_{x}\approx 809$ N and $R_{y}\approx -150$ N. The negative sign of $R_{y}$ means that the joint pushes the arm down, the deltoid already pulling upward harder than the total weight. The joint force, of magnitude about $822$ N, is more than sixteen times the load held.”

The classic mistake on this problem: Taking torques about the deltoid attachment point FROM THE START, then taking them again about the joint in the belief that this gives a fourth constraint. The two rows are equivalent modulo the component equations, the rank stays $3$ , and the row reduction produces an entirely zero row that looks like a computing error.

Learn by heart

• Three steps, never two: NAME the unknowns, COUNT the constraints, ARRANGE them in $[A\,|\,B]$ .
• Rouché-Capelli: there are solutions if and only if $\operatorname{rank}(A)=\operatorname{rank}([A\,|\,B])$ , and they are unique when this common rank EQUALS the number of unknowns.
• Point in the plane: $2$ constraints. Rigid body in the plane: $3$ constraints, two for the resultant and one for the torque. Torque about another point NEVER adds rank.
• Circuit with $n$ junctions: $n-1$ independent junction rules, the sum of all of them being $0=0$ .
• $W=\vec{F}\cdot\vec{d}$ keeps the PARALLEL part; $\vec{M}=\vec{r}\times\vec{F}$ keeps the PERPENDICULAR part. Where one vanishes, the other is at its largest.
• Balancing: species in columns, elements in rows, zero right-hand side, and $\dim\ker A=p-\operatorname{rank}(A)$ . $\dim\ker=1$ , well-posed reaction; $\dim\ker=2$ , CHARGE was forgotten.
• Orthogonal projection: $\vec{F_{\parallel}}=\dfrac{\vec{F}\cdot\vec{d}}{\|\vec{d}\|^{2}}\,\vec{d}$ , with $\|\vec{d}\|^{2}$ in the denominator and never $\|\vec{d}\|$ .
• The rank decides CONSISTENCY, the sign decides ADMISSIBILITY. Both checks are required.

Frequently asked questions

How do you turn a physics problem into a linear system?

Three steps, always in this order. First name one unknown per quantity the statement does not give, with its unit and its direction. Then write one equation per constraint: one per component direction in mechanics, one per conserved quantity in chemistry, one per junction and per independent loop in electricity. Finally arrange the coefficients in an augmented matrix, one column per unknown.

What does a system with no solution mean in a science problem?

It is not a computing error, it is an answer. When the rank of the coefficient matrix is smaller than the rank of the augmented matrix, one row says zero equals a non-zero constant: the situation described is physically impossible. You must then name the false assumption, for example horizontal cables that have no vertical component, instead of redoing the row reduction.

How many junction rules should you write for a circuit?

Exactly one fewer than the number of junctions. The sum of all the junction rules of a circuit is the identity zero equals zero, since each branch current leaves one junction and enters another. The last row is therefore a combination of the others and adds no rank. You complete the system with independent loop rules until there are as many pivots as unknowns.

Why can't a redox equation be balanced without conservation of charge?

Because the element conservations alone leave a null space of dimension two instead of one. Two free parameters mean that two independent reactions conserve every atom, and neither one is the reaction you are looking for. Charge is a conserved quantity just like an element: it is added as a row, the rank goes up by one and the equation becomes determined again.

What is the difference between the dot product and the cross product in mechanics?

They select complementary parts of the same force. Work is a dot product, hence a number, and only the component parallel to the displacement counts: the perpendicular component does exactly zero work. Torque is a cross product, hence a vector, and it is the reverse: only the component perpendicular to the arm makes it turn. Where one is zero, the other is at its largest.

Can a system be consistent and still have no usable solution?

Yes, and it is the question that separates the papers. Row reduction knows no sign constraint: it can deliver a line of solutions none of which has all its components positive. A negative mass or concentration does not exist, so the free parameter must be restricted to the physical domain, and you must then check that the resulting interval is not empty.

Practise it

Corrected exercises: Applications in science, 201-SN4-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

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Previous sheet Complex numbers

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-sn4-applications-science. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: applications of linear algebra in science (201-SN4-RE)

The essentials

The three steps of the setup, always in this order

Reading the rank: three verdicts, and not one more

Dot product or cross product: two complementary selections

What the algebra ignores: the sign

The rules in table form

How many INDEPENDENT equations each context provides

The mistakes that cost marks

1. Starting to compute without naming the unknowns

2. Counting equations instead of counting the rank

3. Writing the junction rule at EVERY junction and thinking you gained an equation

4. Taking torques about a second point to get one more constraint

5. Forgetting the direction of a force when taking components

6. Confusing the dot product and the cross product

7. Believing that a force perpendicular to the displacement does a little work

8. Concluding “no solution” and stopping there

9. Taking a consistent system for a solved problem

10. Balancing a redox equation with the element conservations only

Which method to choose

How many equations to write, depending on the form of the problem

The row reduction is done: what do you say?

How the answer is expected to be written

Turning a science problem into a linear system

Concluding on a system that has no unique solution

Check before you hand in

The typical problem, taken apart

The outstretched arm: three unknowns, three constraints, and the choice of point

Learn by heart

Frequently asked questions

How do you turn a physics problem into a linear system?

What does a system with no solution mean in a science problem?

How many junction rules should you write for a circuit?

Why can't a redox equation be balanced without conservation of charge?

What is the difference between the dot product and the cross product in mechanics?

Can a system be consistent and still have no usable solution?

Corrected exercises: Applications in science, 201-SN4-RE

See also

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