MATH 133 Linear Algebra and Geometry • McGill University, Montreal

Revision sheet: determinants, cofactor expansion and row operations (MATH 133)

This sheet is not a summary of section 3.1 of Nicholson: you already have the course notes. It answers one question only, what makes students lose marks on determinants in MATH 133 at McGill University, and which precise gesture avoids each loss.

Almost every lost mark on this chapter is a bookkeeping error, not a conceptual one: a sign read from the wrong cell of the checkerboard, a swap that was made and never written down, a factor that left the determinant and never came back. The method below is therefore mostly a way of WRITING: choose the line before computing, and record every row operation with its effect on the determinant, one line at a time.

The thread of the chapter

A determinant is never computed by brute force: either you expand along the line that already holds the zeros, or you MAKE the zeros with row operations and keep a ledger of what each one did, a sign for a swap, a factor for a scaling, nothing for a replacement. And the sign of a cofactor belongs to the POSITION, never to the entry.

This chapter is part of MATH 133, Linear Algebra and Geometry (McGill)

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

Determinant of a square matrix
Minor of an entry
Cofactor
Checkerboard of signs
Cofactor expansion
Determinant of a triangular matrix
Elementary row operations
Determinant of a scalar multiple
Determinant of the transpose
Determinant depending on a parameter
Vandermonde determinant

The essentials

Minor, cofactor, and the sign that belongs to the position

• Minor $M_{ij}$ : the determinant of the matrix left once row $i$ and column $j$ are deleted.
• Cofactor $C_{ij} = (-1)^{i+j}M_{ij}$ . The sign depends on $i + j$ only: even gives $+$ , odd gives $-$ . The value of $a_{ij}$ plays no part.
• Expansion along row $i$ : $\det A = a_{i1}C_{i1} + a_{i2}C_{i2} + \dots + a_{in}C_{in}$ . The same formula holds down any column, and every line gives the SAME number.
• $2 \times 2$ : $\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$ . Triangular, upper or lower: the product of the diagonal entries.

Row

2

of every matrix starts with a MINUS, and so does column

2

: reciting

+, -, +

is right only on rows

1

and

3

of a

3 \times 3

Before expanding along a line, write its signs above it, read from the checkerboard. It takes three seconds and removes the most common error of the chapter.

The ledger: what each row operation does to the determinant

• Swap two rows, $R_i \leftrightarrow R_j$ : the determinant is multiplied by $-1$ .
• Scale one row, $R_i \to cR_i$ : multiplied by $c$ . Read backwards, a common factor of one row comes OUT in front.
• Replacement, $R_i \to R_i + cR_j$ with $j \neq i$ : unchanged. The replaced row must keep the coefficient $1$ .
• Zero row, zero column, two equal or proportional rows: the determinant is $0$ .
• $\det(A^{T}) = \det A$ , so every rule above holds for COLUMNS too. And $\det(kA) = k^{n}\det A$ for an $n \times n$ matrix.

Between two consecutive matrices of a reduction, always write an equality sign and the factor, for example $\det A = -\det B$ after a swap. An arrow between matrices tells the marker nothing about the determinant.

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.

What happens to the determinant, operation by operation

Read each line with $\det A = 4$ and $A$ of size $3 \times 3$ . The red lines are the rules students believe and that do not exist: they are not answers, they are the signal to go back to the ledger.

Change made to $A$	Written as	New $\det$
swap two rows	$R_1 \leftrightarrow R_2$	$-\det A$
Example: $\det\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = -2$ and $\det\begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix} = 2$ .
scale one row	$R_2 \to 3R_2$	$3\det A$
Example: $4 \to 12$ ; row $2$ of $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ times $3$ gives $\det = 12 - 18 = -6$ .
replacement	$R_2 \to R_2 + 2R_1$	$\det A$
Example: $\begin{pmatrix} 1 & 2 \\ 5 & 8 \end{pmatrix}$ has $\det = 8 - 10 = -2$ , same as $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ .
scale, then add	$R_2 \to 3R_2 - R_1$	$3\det A$
Example: $\begin{pmatrix} 1 & 2 \\ 8 & 10 \end{pmatrix}$ has $\det = 10 - 16 = -6 = 3 \times (-2)$ .
every row times $k$	$kA$	$k^{n}\det A$
Example: $\det(2A) = 2^{3} \times 4 = 32$ for a $3 \times 3$ .
transpose	$A^{T}$	$\det A$
Example: $\det\begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix} = 4 - 6 = -2 = \det\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ .
sum of two matrices	$A + B$	$\det A + \det B$ no such rule
Example: $\det(I_2 + I_2) = \det(2I_2) = 4$ , while $\det I_2 + \det I_2 = 2$ . What to do: Compute the sum matrix first, then its determinant.
every row times $k$	$kA$	$k\det A$ no such rule
Example: For $A = I_3$ and $k = 2$ : $\det(2I_3) = 8$ , not $2$ . What to do: Count the rows: each one contributes its own factor $k$ , so $k^{n}$ .

Only the coefficient of the row being REPLACED matters. In $R_2 \to 3R_2 - R_1$ that coefficient is $3$ , so the ledger says $\times 3$ , however the rest of the line looks.

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. Reading the signs of row 2 as plus, minus, plus

the whole expansion, and every part that uses the value

What not to write

“Along row $2$ : $\det A = a_{21}M_{21} - a_{22}M_{22} + a_{23}M_{23}$ .”

What to write

“Along row $2$ , signs $-, +, -$ : $\det A = -a_{21}M_{21} + a_{22}M_{22} - a_{23}M_{23}$ .”

Why: The sign is $(-1)^{i+j}$ , and on row $2$ , $i + j$ starts at $3$ . The wrong version gives exactly $-\det A$ : on $\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ it returns $+1$ instead of $-1$ , and nothing in the arithmetic looks suspicious.

2. Handing in the minor where the cofactor was asked

1 mark per cofactor, and the whole expansion built on them

What not to write

“ $C_{12} = \det\begin{pmatrix} 0 & 1 \\ 5 & -2 \end{pmatrix} = -5$ .”

What to write

“ $M_{12} = -5$ , and since $1 + 2$ is odd, $C_{12} = -M_{12} = 5$ .”

Why: Minor and cofactor are one determinant and one sign. Writing both, in that order, costs one line and makes the sign visible to the marker. In an expansion, using $M_{12}$ instead of $C_{12}$ shifts the answer by twice the middle term.

3. Making a swap during the reduction and never recording it

the final answer, with every line of arithmetic correct

What not to write

“Pivot $0$ , so I exchange $R_2$ and $R_3$ and continue. Diagonal $1, 2, 3, 1$ , so $\det B = 6$ .”

What to write

“ $R_2 \leftrightarrow R_3$ : $\det B = -\det B'$ . The diagonal of the triangular form gives $6$ , so $\det B = -6$ .”

Why: A swap is always allowed; forgetting its sign is the error. Count the swaps at the end: an odd number flips the sign, an even number does not. Writing $\det B = -\det B'$ on the line of the swap makes the count impossible to get wrong.

4. Pulling a factor out of a row, then losing it

2 marks, and an answer off by a factor of 20

What not to write

“Row $1$ is $4(1, -2, 3)$ and row $3$ is $5(2, 1, 3)$ . After simplifying, $\det M = 34$ .”

What to write

“ $\det M = 4 \times 5 \times \det M' = 20 \times 34 = 680$ .”

Why: Pulling a factor out is a scaling read backwards, so the factor must multiply the outside. Write it in front of the new determinant on the same line and carry it to the end; the arithmetic in between is easier precisely because it is no longer inside.

5. Treating a scaled combination as a replacement

the final answer, off by a factor of 8

What not to write

“ $R_2 \to 4R_2 - 3R_1$ and $R_3 \to 2R_3 - 5R_1$ are row replacements, so $\det M = 5440$ .”

What to write

“ $R_2 \to 4R_2 - 3R_1$ multiplies the determinant by $4$ , and $R_3 \to 2R_3 - 5R_1$ by $2$ : $\det M = 5440 / 8 = 680$ .”

Why: The fraction-free combination is a good idea, but the replaced row is multiplied before anything is added to it. Only $R_i \to R_i + cR_j$ , coefficient $1$ on $R_i$ , leaves the determinant alone.

6. Writing det(2A) as twice det A

the whole question, a classic true or false item

What not to write

“ $A$ is $3 \times 3$ with $\det A = 5$ , so $\det(2A) = 10$ .”

What to write

“ $2A$ multiplies each of the $3$ rows by $2$ , so $\det(2A) = 2^{3} \times 5 = 40$ .”

2A

has a factor

2

in EACH row, and each row gives one factor to the determinant: three rows,

2^{3} = 8

Why: Each row contributes its own factor. The same count explains why $\det(-A) = -\det A$ for a $3 \times 3$ but $\det(-A) = \det A$ for a $4 \times 4$ .

7. Cancelling a factor that can be zero, in a determinant with a parameter

half the question, the missing value

What not to write

“ $(x - 1)^{2}(x + 2) = 0$ , divide by $(x - 1)^{2}$ : $x = -2$ is the only value.”

What to write

“ $(x - 1)^{2}(x + 2) = 0$ exactly when $x = 1$ or $x = -2$ .”

Why: Dividing by $x - 1$ assumes $x \neq 1$ , which is precisely one of the values being looked for. Set each factor to zero instead; the factored form is also the proof that there are no other values.

8. Expanding along a full row when a column is nearly empty

rarely the mark, always ten minutes, often a sign

What not to write

“Along row $1$ : $3 \cdot C_{11} + 0 + 2 \cdot C_{13} + (-1) \cdot C_{14}$ ”, three $3 \times 3$ determinants and a sign lost in the second.

What to write

“Column $2$ holds three zeros, so $\det A = 5 \, C_{32} = -5 \, M_{32}$ ”, one $3 \times 3$ determinant.

Why: Every extra small determinant is one more chance to drop a sign. Count the zeros of every row and column before writing anything: it takes ten seconds and decides the length of the whole question.

Which method to choose

Which method, from the shape of the matrix

Look at the zeros and at the size before writing a single product

Left, column

3

holds three zeros: one cofactor to compute. Right, no zero anywhere: row replacements first, to build the column the left matrix already had.

If a zero row, a zero column, or two equal or proportional rows → the determinant is $0$ , no computation

Example: $\det\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 0 & 1 & 5 \end{pmatrix} = 0$ , since $R_2 = 2R_1$
If triangular, upper or lower → product of the diagonal entries

Example: diagonal $2, -1, 3, -2$ gives $12$
If a row or a column with a single non-zero entry, or several zeros → expand along it, signs written from the checkerboard

Example: column $2$ of a $4 \times 4$ holding only a $5$ at $(3,2)$ : $\det = -5M_{32}$
If a row or a column with a common factor → pull the factor out in front, then continue

Example: row $(4, -8, 12) = 4(1, -2, 3)$
If a parameter, or rows that all have the same sum → row operations to extract factors BEFORE expanding

Example: $R_1 \to R_1 + R_2 + R_3$ extracts $x + 2$ from $\det M(x)$
If a dense $4 \times 4$ with no zero → replacements to empty one column, then expand along it

Example: a pivot $1$ clears its column with integer multipliers

The rule of Sarrus, the diagonal trick, exists for $3 \times 3$ ONLY. Applied to a $4 \times 4$ it produces $8$ products where the determinant has $24$ , and a wrong answer with no warning.

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.

A determinant by reduction, with the ledger written out

When to use it: Any determinant of size $4$ or more, or any $3 \times 3$ with a factor or a parameter

1 Count the zeros of each row and column, and say which line you will aim for.
2 Pull out any common factor of a row or a column, writing the factor in front: $\det P = 3\det P_1$ .
3 Perform the operations one line at a time, each named ( $R_3 \to R_3 - 2R_1$ ) and followed by its effect: unchanged, $\times(-1)$ or $\times c$ .
4 Stop as soon as a line has a single non-zero entry, or as soon as the matrix is triangular, and finish by expansion or by the diagonal product.
5 Multiply the ledger back in and give the value in a sentence. Check with a second expansion along another line if time allows.

Concluding sentence

“ $\det P = 3\det P_1$ (factor $3$ from $R_1$ ) $= -3\det P_2$ ( $R_1 \leftrightarrow R_2$ ) $= -3 \times (1)(1)(1)(-29) = 87$ , the remaining operations being replacements.”

The trap: Writing a chain of matrices joined by arrows with no equality sign. The marker cannot tell which factors you tracked, and gives no method marks for a sequence of matrices that only looks like a computation.

Marking: Typically 1 mark for a sensible choice of line or first operation, 2 for the operations with their effects recorded, 1 for the final expansion or diagonal product, and 1 for the value. A correct number without the ledger is usually worth 1 out of 5.

Check before you hand in

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

Integer entries give an integer determinant

The determinant is a sum of products of entries. If every entry is an integer and your answer is a fraction, a factor of the ledger has been divided instead of multiplied.

A $3 \times 3$ integer matrix with announced determinant $\frac{85}{2}$ : a pivot of $2$ was divided out and never multiplied back.
A second expansion, along another line

Every row and every column gives the same determinant. Re-expanding the last small determinant along a different line costs two 2 by 2 computations and catches almost every sign error.

$M_{32} = -43$ along row $2$ , and $3(-12) - 1(7) = -43$ along column $1$ : consistent.
Plug a value into a parametric answer

When the answer depends on $x$ , substitute a simple value such as $x = 0$ in BOTH the matrix and your formula. Two different numbers mean the factorisation is wrong.

$\det M(0) = 2$ and $(0 + 2)(0 - 1)^{2} = 2$ : consistent.
Count the swaps and the factors at the end

Read the ledger once more before the final line: the number of swaps decides the sign, the product of the extracted factors decides the size. Most wrong answers on this chapter are right up to one of these two.

Two swaps and a factor $2$ : the answer is $+2$ times the diagonal product.

The typical problem, taken apart

A 4 by 4 with a factor, a zero pivot and a replacement chain

Compute $\det P$ for $P = \begin{pmatrix} 0 & 3 & -6 & 9 \\ 1 & 2 & 0 & -1 \\ 2 & 4 & 1 & 1 \\ -1 & 1 & 2 & 5 \end{pmatrix}$ , showing every row operation and its effect on the determinant.

Step 1

Row $1$ is $3(0, 1, -2, 3)$ : $\det P = 3\det P_1$ with $P_1 = \begin{pmatrix} 0 & 1 & -2 & 3 \\ 1 & 2 & 0 & -1 \\ 2 & 4 & 1 & 1 \\ -1 & 1 & 2 & 5 \end{pmatrix}$ .

Why

The factor goes out first and is written in front, so that every later number is smaller and nothing has to be remembered.

Step 2

The entry at $(1,1)$ is $0$ , so $R_1 \leftrightarrow R_2$ : $\det P = -3\det P_2$ with $P_2 = \begin{pmatrix} 1 & 2 & 0 & -1 \\ 0 & 1 & -2 & 3 \\ 2 & 4 & 1 & 1 \\ -1 & 1 & 2 & 5 \end{pmatrix}$ .

Why

A zero cannot be a pivot. The swap brings up a $1$ , which clears column $1$ with integer multipliers, and its sign is recorded on the same line.

Step 3

$R_3 \to R_3 - 2R_1$ gives $(0, 0, 1, 3)$ and $R_4 \to R_4 + R_1$ gives $(0, 3, 2, 4)$ . Replacements: still $\det P = -3\det P_3$ .

Why

Column 1 is now empty below the pivot. Here one could already expand along column 1; continuing to a triangle shows the other route.

Step 4

$R_4 \to R_4 - 3R_2$ gives $(0, 0, 8, -5)$ , then $R_4 \to R_4 - 8R_3$ gives $(0, 0, 0, -29)$ . Replacements again.

Why

Each pivot so far equals $1$ , so every multiplier is an integer and no fraction ever appears, which is what the exam numbers are designed for.

Step 5

The matrix is upper triangular with diagonal $1, 1, 1, -29$ : $\det P = -3 \times (-29) = 87$ .

Why

The ledger has exactly two entries, the factor 3 and one swap. Multiplying them back in is the last line, not an afterthought.

Step 6

Check: after step 3, expand along column 1 instead: $\det P_3 = \det\begin{pmatrix} 1 & -2 & 3 \\ 0 & 1 & 3 \\ 3 & 2 & 4 \end{pmatrix} = -2 - 18 - 9 = -29$ .

Why

A different route to the same small determinant: the two agree, so the sign and the factor are confirmed.

The conclusion, written out

“ $\det P = 3 \times (-1) \times (-29) = 87$ .”

The classic mistake on this problem: Forgetting the swap gives $-87$ , forgetting the factor gives $29$ , forgetting both gives $-29$ . All three come with perfectly correct arithmetic, which is why the ledger has to be written, not remembered.

Learn by heart

• $C_{ij} = (-1)^{i+j}M_{ij}$ : the sign belongs to the POSITION. Row $2$ starts with a minus.
• Expand along ANY row or column: same result every time. Choose the line with the most zeros.
• Triangular: product of the diagonal. Zero row or two equal rows: $0$ .
• Swap: $\times(-1)$ . Scale one row by $c$ : $\times c$ . $R_i \to R_i + cR_j$ : unchanged.
• $R_i \to aR_i + cR_j$ multiplies the determinant by $a$ , NOT by $1$ .
• $\det(kA) = k^{n}\det A$ and $\det(A^{T}) = \det A$ . There is NO rule for $\det(A + B)$ .
• Vandermonde $3 \times 3$ : $(b - a)(c - a)(c - b)$ , zero exactly when two nodes coincide.

Frequently asked questions

Which row should I expand along to compute a determinant?

Along the row or column with the most zeros, because each zero removes a whole smaller determinant from the work. If no line has zeros, use row replacements first to create them in one column, since a replacement does not change the determinant, and then expand along that column.

What is the difference between a minor and a cofactor?

The minor of an entry is the determinant of the matrix left after deleting that entry's row and column. The cofactor is the same number with a sign attached: plus when the row number plus the column number is even, minus when it is odd. The sign depends only on the position, never on the value of the entry.

How do row operations change the determinant?

Swapping two rows changes its sign. Multiplying one row by a number multiplies the determinant by that number. Adding a multiple of another row to a row leaves it unchanged. Record each operation with its effect as you go, and multiply the factors back in at the end, because a forgotten swap or factor is the most common wrong answer.

Is the determinant of 2A equal to twice the determinant of A?

No. Multiplying the matrix by 2 multiplies every row by 2, and each row gives its own factor, so for an n by n matrix the determinant is multiplied by 2 to the power n. For a 3 by 3 matrix that is a factor of 8, and for a 4 by 4 matrix a factor of 16.

Can I use the rule of Sarrus on a 4 by 4 determinant?

No. The diagonal rule of Sarrus is a shortcut valid for 3 by 3 determinants only. On a 4 by 4 it produces eight products, while the true determinant is a sum of twenty-four, so the answer is simply wrong. For larger matrices, use row reduction to a triangular form or a cofactor expansion along a line with zeros.

Practise it

Corrected exercises: Determinants: cofactor expansion and row operations, MATH 133 at McGill

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
150 minutes

Do the exercises

Previous sheet Matrix inverses Next sheet Determinants, adjugate and Cramer's rule

Looking for a MATH 133 tutor in Montreal?

Get in touch for a first session. The determinant chapter is where linear algebra becomes a matter of method: choose the line, keep the ledger, check with a second expansion.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: determinants, cofactor expansion and row operations (MATH 133)

The essentials

Minor, cofactor, and the sign that belongs to the position

The ledger: what each row operation does to the determinant

The rules in table form

What happens to the determinant, operation by operation

The mistakes that cost marks

1. Reading the signs of row 2 as plus, minus, plus

2. Handing in the minor where the cofactor was asked

3. Making a swap during the reduction and never recording it

4. Pulling a factor out of a row, then losing it

5. Treating a scaled combination as a replacement

6. Writing det(2A) as twice det A

7. Cancelling a factor that can be zero, in a determinant with a parameter

8. Expanding along a full row when a column is nearly empty

Which method to choose

Which method, from the shape of the matrix

How the answer is expected to be written

A determinant by reduction, with the ledger written out

Check before you hand in

The typical problem, taken apart

A 4 by 4 with a factor, a zero pivot and a replacement chain

Learn by heart

Frequently asked questions

Which row should I expand along to compute a determinant?

What is the difference between a minor and a cofactor?

How do row operations change the determinant?

Is the determinant of 2A equal to twice the determinant of A?

Can I use the rule of Sarrus on a 4 by 4 determinant?

Corrected exercises: Determinants: cofactor expansion and row operations, MATH 133 at McGill

See also

Looking for a MATH 133 tutor in Montreal?