CEGEP Linear Algebra — Determinants 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. Q1Determinants by Cofactor Expansion

   Let $$A=\left[\begin{array}{ccc}3& 1& -2\\ 0& 4& 1\\ 5& -1& 2\end{array}\right].$$

   1. Compute $det(A)$ by cofactor expansion along the first row, writing out the three cofactors with their signs.
   2. Recompute $det(A)$ by expanding along the first column, and state what the agreement of the two values illustrates.
2. Q2Determinants by Row Reduction

   Evaluate $det(A)$ for $$A=\left[\begin{array}{cccc}0& 1& 3& 1\\ 1& 2& -1& 3\\ 2& 5& 0& 4\\ 6& 14& -4& 16\end{array}\right]$$ by reducing $A$ to upper triangular form. State, beside each operation, its effect on the determinant, and give the value.

3. Q3Properties of the Determinant

   $A$ and $B$ are $3\times 3$ matrices with $det(A)=5$ and $det(B)=-2$. Evaluate each of the following, naming the property used, or state that the information given is not enough.

   1. $det(A^{T}B)$
   2. $det(2A)$
   3. $det(A^{-1}{B}^2)$
   4. $det((AB{)}^{-1})$
   5. $det(A+B)$
4. Q4The Adjoint and the Inverse

   For $$A=\left[\begin{array}{ccc}1& 2& 2\\ 1& 3& 1\\ 2& 4& 5\end{array}\right],$$ compute $det(A)$, build the matrix of cofactors, write down $adj(A)$, and hence obtain $A^{-1}$ from $A^{-1}={\displaystyle \frac{1}{det(A)}}adj(A)$. Verify your inverse by computing one column of $A{A}^{-1}$.

5. Q5Cramer's Rule

   Solve $$\begin{gathered} 2x+y-z\&=3 \\ x-y+2z\&=1 \\ 3x+2y+z\&=8 \end{gathered}$$ by Cramer's rule. Show the four determinants you use, and check your solution in the second equation.

6. Q6Determinants, Invertibility and Solution Type

   Let $$A(k)=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& k\\ 1& 4& k^2\end{array}\right].$$

   1. Show that $det(A(k))=(k-1)(k-2)$, and state all $k$ for which $A(k)$ is invertible.
   2. Take a value of $k$ making $det(A(k))=0$. What does that tell you about the number of solutions of the homogeneous system $A(k)𝐱=\mathbf{0}$? And about $A(k)𝐱=𝐛$ for an arbitrary $𝐛$?
   3. For $k=1$, exhibit a nonzero solution of $A(1)𝐱=\mathbf{0}$.
7. Q7Synthesis — drawing on several topics in this set

   Let $$A=\left[\begin{array}{ccc}1& 2& -1\\ 2& 3& 1\\ 3& 5& 2\end{array}\right],𝐛=\left[\begin{array}{c}6\\ 7\\ 11\end{array}\right].$$

   1. Find $det(A)$ by row reduction, stating the effect of each operation.
   2. State, with the reason, whether $A𝐱=𝐛$ has a unique solution.
   3. Use Cramer's rule to find $x$ alone.
   4. Evaluate $det(A^{-1})$, $det(2A^T)$ and $det(A^3)$.

The 5 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.

One number, and it settles the questions the whole course has been asking

Up to now, deciding whether a square matrix is invertible has meant reducing it and seeing whether the identity appeared; deciding how many solutions a system has has meant reducing the augmented matrix and reading the last row. Determinants do not replace that work — they compress the part of it that concerns a square coefficient matrix into a single scalar.

Q6(b) is where the limits of the tool show up, and it is the part most worth being careful about. A non-zero determinant is a complete answer: unique solution, every time. A zero determinant is only half an answer. It rules "unique" out and leaves two possibilities standing — no solutions at all, or infinitely many — and which of the two you are in depends on the right-hand side, not on the matrix. Nothing about the determinant distinguishes them; only row-reducing the augmented matrix does. The one case where the ambiguity disappears is the homogeneous system in that same part, which is always consistent because the zero vector solves it, so "not unique" there can only mean infinitely many. Q6(c) then asks you to produce one of those non-zero solutions by hand, so that "infinitely many" stops being a phrase and becomes something you have written down.

Cofactor expansion: the sign depends on both indices

Q1 asks for the same 3 × 3 determinant twice — once expanding along the first row, once along the first column — and then asks what the agreement of the two values illustrates. The point of doing it twice is that the value belongs to the matrix, not to the line you chose, so any row or column is allowed. That freedom is worth something practical.

The second habit Q1 is training is choosing the line. Every zero entry in the line you expand along kills a minor outright, so a column containing a zero is strictly less work than a full row — one fewer 2 × 2 determinant, and the same value at the end. Scan for zeros first, expand second.

Row reduction is the cheap route, and each operation has its own price

Q2 is a 4 × 4, and it is a 4 × 4 on purpose: expanding it by cofactors means four 3 × 3 determinants, which is twelve 2 × 2 ones. Reducing it to upper triangular form instead costs a handful of row operations, and the determinant of a triangular matrix is just the product down its diagonal. Past 3 × 3 the two methods stop being comparable.

The catch is that the reduction does not preserve the determinant. Two of the three elementary operations change it, in different ways, and the question requires you to state the effect beside each operation as you use it — because that running record is the calculation.

The property that multiplies, and the one that does not exist

Q3 gives you nothing but det(A) and det(B) for two 3 × 3 matrices, then asks for five expressions, naming the property used — or stating that the information given is not enough. That last option is the whole reason the question is built this way.

The other four parts are the properties you are expected to name out loud rather than merely use: a transpose leaves the determinant alone, a product multiplies, an inverse reciprocates, a power raises to that power. One of them has a trap of its own.

The adjoint: a formula for the inverse, and one step everybody skips

Q4 walks the whole adjoint route on a 3 × 3: the determinant first, then the nine cofactors, then the adjoint, then the inverse, then a verification. It is deliberately one long single-thread computation, because every error available in it is a bookkeeping error and the only defence is a tidy layout.

The verification the question asks for is worth doing properly. Computing one column of AA⁻¹ is three dot products, and it either produces a column of the identity or it does not — a couple of lines that catch a transposed adjoint, a dropped sign, or an arithmetic slip in the determinant. This formula also makes visible why a zero determinant blocks an inverse: the whole expression is divided by det(A).

Cramer's rule: one unknown at a time, and when not to use it

Q5 solves a 3 × 3 system by Cramer's rule and asks for the four determinants explicitly. The shape of the rule is easy: take the determinant of the coefficient matrix, then for each unknown replace that unknown's column by the constants and divide.

The synthesis question, and why it is last

Q7 is the set closing its own loop, and nothing in it is new. Part (a) is Q2's method on a 3 × 3: reduce, state each operation's effect, read the diagonal. Part (b) turns that number into a statement about solutions, with the reason attached — Q6's logic applied to one specific matrix. Part (c) is Cramer's rule used the way it should be, for one unknown rather than three. Part (d) is Q3's properties again, this time stacked: an inverse, a scaled transpose and a cube, each of which you should be able to evaluate from the one determinant you already found in part (a), without touching the matrix again.

That is the test worth setting yourself on this set. Once det(A) is known, the determinant of every matrix built from A by transposing, scaling, inverting or raising to a power follows immediately — and det(A + B) still does not.

Getting the most out of it

Look for the zeros before you choose a method

On a 3 × 3, expand along whichever row or column carries the most zeros — each zero removes a minor, and the value is the same whatever line you pick. On a 4 × 4 or larger, do not expand at all: reduce to triangular form and multiply down the diagonal. The size at which the second method wins is smaller than most people expect.

Write the effect beside every row operation, as you do it

A swap negates, a scaling multiplies by that scalar, adding a multiple of a row changes nothing. Keep the accumulated factor as a visible running product in the margin and unwind it at the end. Reconstructing which operations you performed after the fact is where this question is lost, and the loss looks like a perfectly good reduction with the wrong sign on it.

Name the property, not just the number

Several questions here award the mark for the justification: which property of the determinant you are appealing to, why the matrix is invertible, why Cramer's rule applies at all. Practise writing the reason as a clause — "by det(AB) = det(A)det(B)", "the coefficient determinant is non-zero, so the solution is unique". The devis asks for rigorous mathematical reasoning, and on a CEGEP final that clause is where it is looked for.

Check by substituting, not by re-doing

Recomputing a determinant the same way twice reproduces the same mistake. Expand along a different line, multiply one column of A by your inverse, or put your solution back into an equation you have not used yet. Every question in this set has a cheap independent check available, and two lines of checking is worth more than a second pass.

Common questions