CEGEP Linear Algebra — Systems of Linear Equations Worksheet

All 7 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Linear Systems and the Augmented Matrix

   Consider the linear system $$\begin{array}{ccccccc}\hfill 2x& \hfill -\hfill & \hfill y& \hfill +\hfill & \hfill 3z& \hfill =\hfill & \hfill 1\\ \hfill x& \hfill \hfill & \hfill & \hfill +\hfill & \hfill 4z& \hfill =\hfill & \hfill -1\\ \hfill -3x& \hfill +\hfill & \hfill 5y& \hfill \hfill & \hfill & \hfill =\hfill & \hfill 1\end{array}$$

   1. Write its augmented matrix, using the variable order $x,y,z$.
   2. Write out the system whose augmented matrix is $\left[\begin{array}{cccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 0\hfill & \hfill 5\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill -1\hfill & \hfill -4\hfill \end{array}\right]$, using the variables $x_1,x_2,x_3$.
   3. Verify that $(x,y,z)=(3,2,-1)$ is a solution of the system above, and say in one sentence what that verification amounts to, row by row, in the augmented matrix.
2. Q2Row Reduction to Echelon Form

   Reduce $\left[\begin{array}{cccc}\hfill 0\hfill & \hfill 2\hfill & \hfill -4\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 3\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill -1\hfill & \hfill 9\hfill \end{array}\right]$ to a row echelon form, writing each elementary row operation in the form $R_2\to R_2-3R_1$ as you use it. Then state which columns contain the leading entries, and explain why an interchange was needed before any elimination could start.

3. Q3Gaussian Elimination

   Solve $$\begin{array}{ccccccc}\hfill x& \hfill +\hfill & \hfill 2y& \hfill -\hfill & \hfill z& \hfill =\hfill & \hfill 6\\ \hfill 2x& \hfill +\hfill & \hfill 5y& \hfill +\hfill & \hfill z& \hfill =\hfill & \hfill 11\\ \hfill -x& \hfill \hfill & \hfill & \hfill +\hfill & \hfill 3z& \hfill =\hfill & \hfill -4\end{array}$$ by Gaussian elimination: reduce the augmented matrix to row echelon form only, then finish by back-substitution. Do not carry the reduction as far as reduced row echelon form.

4. Q4Gauss-Jordan Elimination

   Solve $$\begin{array}{ccccccc}\hfill x& \hfill +\hfill & \hfill y& \hfill +\hfill & \hfill 2z& \hfill =\hfill & \hfill 6\\ \hfill 2x& \hfill -\hfill & \hfill y& \hfill +\hfill & \hfill z& \hfill =\hfill & \hfill 9\\ \hfill x& \hfill +\hfill & \hfill 2y& \hfill -\hfill & \hfill z& \hfill =\hfill & \hfill -1\end{array}$$ by Gauss-Jordan elimination: carry the augmented matrix all the way to reduced row echelon form and read the solution off. Name each row operation.

5. Q5Interpreting the Type of Solution Set

   Each matrix below is the reduced row echelon form of the augmented matrix of a system in the unknowns $x,y,z$. For each one: state how many free variables the system has, classify it as having a unique solution, infinitely many solutions or no solution, and write the solution set explicitly — with a parameter where one is needed. $$\text{(i) }\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 5\hfill \end{array}\right]\text{(ii) }\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill -2\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\text{(iii) }\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$$

6. Q6Homogeneous Systems

   Consider the homogeneous system $$\begin{array}{ccccccc}\hfill x& \hfill +\hfill & \hfill 2y& \hfill -\hfill & \hfill z& \hfill =\hfill & \hfill 0\\ \hfill 2x& \hfill +\hfill & \hfill 5y& \hfill +\hfill & \hfill z& \hfill =\hfill & \hfill 0\\ \hfill x& \hfill +\hfill & \hfill y& \hfill -\hfill & \hfill 4z& \hfill =\hfill & \hfill 0\end{array}$$

   1. Solve it by Gauss-Jordan elimination and write the solution set with a parameter.
   2. State the trivial solution and give one non-trivial solution with integer entries.
   3. Explain why no homogeneous system can be inconsistent, arguing from what row operations do to the last column of the augmented matrix.
7. Q7Synthesis — drawing on several topics in this set

   Over one morning a caf\'{e} serves $100$ cups of coffee in three sizes: small ($200\text{ mL}$ each), medium ($300\text{ mL}$ each) and large ($400\text{ mL}$ each). The urn readings show that $28$ litres of coffee were poured, and the sachet count shows $180$ sugar sachets used, at $1$ sachet per small cup, $2$ per medium and $3$ per large.

   1. Let $x$, $y$, $z$ be the numbers of small, medium and large cups. Write the system and its augmented matrix, measuring coffee in units of $100\text{ mL}$.
   2. Reduce the matrix to reduced row echelon form and account for what you find in the third row.
   3. Classify the solution set and write it with a parameter.
   4. The three counts must be non-negative whole numbers. Determine which values of the parameter are admissible, how many sales patterns are possible, and give the two extreme ones.

The 6 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

The augmented matrix: one column per variable, and the order is a promise

Q1 is deliberately unglamorous — write a system as a matrix, then write a matrix back out as a system, then verify a given triple. The point is that the translation goes both ways, and that the matrix carries no memory of anything except position.

Part (c) then asks what verifying a solution amounts to row by row, and that sentence is worth writing carefully, because it is the definition everything else rests on: multiply the coefficient entries of a row by the corresponding values of the unknowns, and ask whether the total is that row's last entry. A solution passes that test in every row. Passing in one row is not a partial solution; it is not a solution.

Row operations: three of them, and each one gets written down

Q2 asks for a reduction to row echelon form with every operation named, in the form R₂ → R₂ − 3R₁. That notation is not bookkeeping etiquette. It is how you find your own arithmetic slip three steps later, and on a CEGEP paper it is where the method marks live — a correct final matrix with no operations shown is a worse answer than a wrong one with them.

Those three, and nothing else. Each is reversible, which is exactly why none of them changes the solution set: the matrix in front of you and the matrix you started with represent systems with the same solutions, and that fact is the licence for the whole method. Q2 also asks you to say why an interchange had to come first, and the answer is the general rule for pivoting — if the entry you mean to eliminate with is zero, there is nothing to eliminate with, so a row carrying a non-zero entry in that column has to be brought up before any elimination can start.

Gaussian elimination and Gauss-Jordan are two different methods

This is the distinction students blur most, and the worksheet keeps the two apart on purpose: Q3 says Gaussian elimination and explicitly tells you not to carry the reduction as far as reduced row echelon form; Q4 says Gauss-Jordan and tells you to go all the way. Same job, same three row operations, different stopping point and a different finish.

Neither method is the right one in general. Gaussian elimination usually costs less arithmetic and the back-substitution is quick by hand; Gauss-Jordan front-loads the work and hands you a form you can read straight off, which is what you want when a question asks for a solution set rather than a triple — and it is the form the classification below is stated in. When the question names a method, that method is what is marked.

Reading the type off the reduced form

Q5 hands you three matrices already in reduced row echelon form and asks, for each, how many free variables the system has, which of the three types it is, and what the solution set actually is. The three parts are one of each case, on purpose. Take them in this order.

A row of all zeros, augmented column included, is neither a contradiction nor a mistake. It is an equation that told you nothing the other rows had not already said, and it is the ordinary reason a free variable appears.

The same three cases as three planes

Every equation in three unknowns is a plane in ℝ³, and solving the system asks what all the planes have in common. That gives the three cases a picture — the version you can check an answer against, and the version an examiner is fond of asking for in words.

Notice what this rules out. Those three are the complete list: a linear system cannot have exactly two solutions, or exactly seventeen. Two distinct solutions already force a whole line of them.

Homogeneous systems come with one solution already

Q6 is a system with zeros all the way down the augmented column. Part (a) solves it by Gauss-Jordan and asks for the solution set with a parameter, part (b) for the trivial solution and a non-trivial one with integer entries, and part (c) for an argument that no homogeneous system can be inconsistent.

Part (b)'s request for integer entries is a quiet hint about how to parametrise. Choose the parameter so that the general element clears its fractions, and then any convenient value of it gives a clean non-trivial solution. The trivial solution is the parameter set to zero, which is a free check that you parametrised correctly.

The synthesis question, and why it is last

Q7 is a word problem — a morning's cup, volume and sachet counts — and it uses the whole set. Part (a) is the modelling step, and it is worth taking slowly: choose the unknowns explicitly, state the units you are counting in, and write one equation per measurement. The instruction to measure the coffee in hundreds of millilitres is there so that the system comes out in whole numbers; converting units before you build the matrix is much easier than repairing the arithmetic afterwards.

Part (b) asks you to reduce and then to account for what you find in the third row. That is a question about the situation rather than about the matrix: a row that reduces to nothing means one measurement was already implied by the others, and naming the combination of earlier equations that produced it tells the manager that the third reading added no information. Part (c) is the classification and the parametrised set from Q5, in a setting where the parameter means something countable. Part (d) is the step only a modelled problem can ask: the algebraic solution set is infinite, but cups come in non-negative whole numbers, so turn each coordinate's constraint into an inequality on the parameter, intersect them, and keep only the integers. The mathematical solution set and the set of admissible answers are different objects, and holding them apart is the point of the question.

Getting the most out of it

Read the instruction before the first row operation

"By Gaussian elimination" and "by Gauss-Jordan elimination" are different instructions with different stopping points, and a question that names one is marked on that one. Decide at the top of the page whether you are stopping at row echelon form and back-substituting, or going on to reduced row echelon form and reading the answer off. Deciding halfway through is how a page ends up doing neither properly.

Name every operation, in the margin, as you do it

R₃ → R₃ − 2R₁ written beside the new matrix costs a few seconds and buys two things: the method marks, and the ability to find your slip without redoing the reduction. Change one row at a time. Two operations combined in your head and written as one line is the commonest source of an error nobody can locate afterwards.

Check consistency before you count free variables

Scan first for a row whose only non-zero entry is in the augmented column. If one is there, the system has no solution and the free variables are beside the point — there is no set to parametrise. Only once you know the system is consistent does counting the columns without leading entries mean anything.

Answer with the solution set, not with its size

Write a set with a general element and a stated parameter range, or a single triple, or the empty set. "Infinitely many solutions" is the classification, and the classification is half the answer; the set is the other half. Then substitute your general element back into the original equations — a parametrised answer that is right satisfies them for every value of the parameter, and the parameter cancelling out is what shows it does.

Common questions