Determinants — cofactor expansion along any row - column: Question bank

Before we start, one word we lean on constantly: a minor $M_{ij}$ is the determinant of what's left after you delete row $i$ and column $j$ ; a cofactor $C_{ij} = (- 1)^{i + j} M_{ij}$ is that minor with a checkerboard sign glued on. If either feels shaky, reread §1 of the parent first — everything below assumes those two live in your head as pictures (cover a row and a column with your hand; the leftover grid is the minor).

True or false — justify

The verdict alone earns nothing — the reason is the whole point.

T/F. You may expand a determinant along any row or any column and always get the same number.

True. The determinant is the single Leibniz permutation sum (every term picks one entry per row and per column), and grouping that sum by "which column the row-

i

entry hit" gives row-

i

expansion — the grouping choice is free, so all lines agree.

T/F. Expanding along a row of all zeros gives determinant

0

True. Every term in that expansion is (a zero entry)

\times

(its cofactor)

= 0

, so the sum is

0

— which matches the fact that a zero row makes the matrix singular.

T/F. The minor

M_{ij}

and the cofactor

C_{ij}

are always different numbers.

False. They differ only by the sign

(- 1)^{i + j}

, which equals

+ 1

whenever

i + j

is even (e.g. positions

(1, 1), (2, 2), (1, 3)

), so there minor and cofactor are identical.

T/F. Swapping two rows before expanding leaves the determinant unchanged, so it doesn't matter which order you write the rows.

False. A row swap multiplies the determinant by

- 1

(it flips the parity of every permutation in the Leibniz sum). Only adding a multiple of one row to another leaves

det

untouched.

T/F. If you pick a "wrong" row to expand along, you may get a different (larger) number because that row has more nonzeros.

False. More nonzeros means more arithmetic, not a different answer. The final number is invariant; only your effort changes — that's the whole 80/20 point.

T/F. For a

3 \times 3

matrix, cofactor expansion produces three

2 \times 2

determinants no matter which line you choose.

True in count, not in cost. You always get three cofactor terms, but any term whose entry is

0

contributes nothing to compute — so a line with two zeros needs only one real

2 \times 2

evaluation.

T/F. The sign

(- 1)^{i + j}

depends on whether you call the sweep a "row expansion" or a "column expansion."

False. The sign depends only on the position

(i, j)

of the entry in the original matrix. Read

i

and

j

off the grid; the direction you walk is irrelevant.

T/F.

det A = det A^{T}

, so column expansion of

A

equals row expansion of

A^{T}

True. Transposing swaps rows and columns but leaves the Leibniz sum invariant (each permutation maps to its inverse, same sign), so column-

j

expansion of

A

is literally row-

j

expansion of

A^{T}

T/F. Multiplying one row of a

3 \times 3

matrix by

5

multiplies its determinant by

5

True. By multilinearity, the determinant is linear in each row separately, so scaling one row by

k

scales

det

by exactly

k

(not

k^{3}

T/F. If a matrix has two identical rows, some clever choice of expansion line will give a nonzero determinant.

False. Two equal rows force

det = 0

for every line — swapping them changes nothing yet must flip the sign, so

det = - det \Rightarrow det = 0

. No expansion can escape this.

Spot the error

Each statement contains one specific mistake. Name it.

A student writes

C_{12} = + M_{12}

because "the checkerboard is symmetric."

The error is the sign: position

(1, 2)

has

i + j = 3

(odd), so

C_{12} = (- 1)^{3} M_{12} = - M_{12}

. The checkerboard alternates, it is not all

+

"To expand along column 2, I multiply each entry by the column-2 signs

-, +, -

reading top to bottom."

The signs

-, +, -

down column 2 are correct here, but the reason given is wrong-headed: they come from

(- 1)^{i + 2}

i

runs

1, 2, 3

, i.e. the entry's actual row index — not from a memorised "column pattern." Track

(i, j)

, not a direction.

det

of a

3 \times 3

needs

3 \times 3 = 9

cofactors."

A single expansion uses

3

cofactors (one per entry of the chosen line), each itself a

2 \times 2

determinant. There is no

9

; the recursion is

n

terms at each level, not

n^{2}

A student expands along row 2 but forgets the row is

(040)

and dutifully computes all three cofactors.

Not a sign error but a wasted-effort error: two entries are

0

, so two terms vanish. Only

4 \cdot C_{22}

survives — "chase the zeros" before lifting the pen.

"After doing

R_{2} \to R_{2} - 2 R_{1}

I must multiply the determinant by

\frac{1}{2}

to undo it."

There is nothing to undo. Adding a multiple of one row to another does not change

det

at all, so no correction factor is needed. (Scaling a whole row is the operation that would need a factor.)

"The cofactor of a

1 \times 1

block

[a]

(- 1)^{1 + 1} \cdot det (empty) = 0

The base case is

det [a] = a

, not

0

. The minor of the only entry is the determinant of the empty matrix, defined as

1

, so the recursion bottoms out correctly at

a

, never at

0

Why questions

The parent note asserts these; here you must explain them.

Why does the sign

(- 1)^{i + j}

appear at all, in one sentence?

It is the parity (sign) of the permutation needed to shuffle entry

(i, j)

into the top-left pivot corner — the Leibniz sum already carries that sign, and

(- 1)^{i + j}

just packages it.

Why can grouping the Leibniz sum by

σ (i)

never lose or duplicate a term?

Because for each permutation,

σ (i)

is a single well-defined column, so every one of the

n!

terms lands in exactly one bucket

j = σ (i)

— a clean partition.

Why is expanding along a line with many zeros legitimate and not a "shortcut cheat"?

The theorem guarantees every line gives the same value, so choosing the zero-rich line is just using a true equality that happens to be cheap — the answer is identical, only the labour drops.

Why does

det A = det A^{T}

let us treat rows and columns on equal footing?

Transposing leaves the determinant invariant, so any statement proven for row expansion transfers verbatim to columns by applying it to

A^{T}

— no separate proof needed.

Why do the cofactors feed directly into the inverse of a matrix?

The adjugate is the transpose of the cofactor matrix, and

A^{- 1} = \frac{1}{d e t A} adj (A)

— so each cofactor is literally a numerator in an entry of

A^{- 1}

Why must a matrix with

det = 0

fail to have an inverse?

The formula

A^{- 1} = \frac{1}{d e t A} adj (A)

divides by

det A

; when

det A = 0

this is undefined, mirroring the geometry — the map collapses a dimension and can't be reversed.

Edge cases

The degenerate inputs the recipe still has to survive.

What is the determinant of a

1 \times 1

matrix

[a]

, and does cofactor expansion apply?

It is just

a

— this is the recursion's base case, below which there are no more minors to take.

What does cofactor expansion give for a

2 \times 2

matrix, and why is it the true base workhorse?

Expanding along row 1:

a_{11} a_{22} - a_{12} a_{21}

, i.e.

a d - b c

. Every larger determinant bottoms out into these, so memorising it is non-negotiable.

If an entire row is zero, is the matrix invertible, and what does every expansion return?

Not invertible: expanding along that zero row gives

0

instantly, and

det = 0

means singular.

For the

n \times n

zero matrix, what is the determinant?

0

for any

n \geq 1

, since expanding along any (all-zero) line sums zeros; geometrically the map crushes everything to a point.

Does cofactor expansion give the right sign for a lower-triangular matrix expanded along column 1?

Yes — only the top-left entry is a "free choice" with cofactor

+ M_{11}

, and recursing peels off the diagonal, so

det

= product of diagonal entries, sign and all.

Is a

0 \times 0

(empty) matrix's determinant defined, and what is it?

By convention it is

1

(the empty product), which is exactly what makes the

1 \times 1

minor recursion consistent.

Recall One-line self test

"Can I expand along a column of all zeros and trust the $0$ ?" ::: Yes — every term is zero, the answer is genuinely $0$ , and the matrix is singular.

Connections

Determinants — cofactor expansion along any row - column — the parent recipe these traps stress-test.
Leibniz formula for determinants — the definition every "why" answer traces back to.
Properties of determinants (multilinearity, alternation) — source of the swap/scale/add rules.
Adjugate matrix and inverse — where cofactors go to live.
Invertibility and singular matrices — what $det = 0$ actually means.
Cramer's rule — cofactors solving linear systems.