Determinant of 3×3 matrix — cofactor expansion: Question bank

This page is a question bank, not a calculator drill. Each line below hides a short answer after the :::. Try to answer out loud before revealing. The goal is to catch the sneaky misconceptions that survive even after you can crunch the numbers — the kind tested in properties and inverse questions.

Parent: Determinant of 3×3 — cofactor expansion.

True or false — justify

TRUE or FALSE: Expanding along row 1 and expanding along column 3 of the same matrix can give different determinants.

FALSE — the determinant is a single fixed number for a matrix; every valid row/column expansion is just a different route to the same answer.

TRUE or FALSE: The minor

M_{ij}

and the cofactor

C_{ij}

are always equal.

FALSE — they are equal only when

i + j

is even (sign

+ 1

); when

i + j

is odd,

C_{ij} = - M_{ij}

TRUE or FALSE: The checkerboard sign of an element depends on the element's value.

FALSE — the sign

(- 1)^{i + j}

depends only on the position

(i, j)

, never on the number sitting there.

TRUE or FALSE: If a matrix has a full row of zeros, its determinant is zero.

TRUE — expand along that row: every term is (zero)×(cofactor), so the whole sum is

0

TRUE or FALSE: You should always expand along row 1.

FALSE — you may expand along any row or column; choosing one rich in zeros kills terms and saves work, so row 1 is rarely the smartest choice.

TRUE or FALSE: The determinant of a triangular 3×3 matrix (all entries above or below the diagonal are zero) equals the product of its diagonal entries.

TRUE — expanding along the zero-heavy row/column repeatedly leaves only the diagonal product, e.g.

det = a_{11} a_{22} a_{33}

TRUE or FALSE: Swapping two rows of a matrix leaves the determinant unchanged.

FALSE — a single row swap flips the sign of the determinant (it reverses orientation); this is a property of determinants.

TRUE or FALSE: If two rows of a 3×3 matrix are identical, the determinant is zero.

TRUE — identical rows are linearly dependent, so the three rows are flat (coplanar) and enclose zero volume.

TRUE or FALSE: The cofactor expansion formula only works if you pick a row, never a column.

FALSE — column expansion is equally valid;

det (A) = \sum_{i} a_{ij} C_{ij}

works down any column

j

Spot the error

A student writes

det = a_{11} M_{11} + a_{12} M_{12} + a_{13} M_{13}

. What's wrong?

The alternating signs are missing — the middle term needs

(- 1)^{1 + 2} = - 1

, so it must be

a_{11} M_{11} - a_{12} M_{12} + a_{13} M_{13}

To find

C_{12}

, a student crosses out row 1 and column 1. What's the mistake?

For

C_{12}

you delete row

1

and column

2

(the indices of the element), not column 1 — they crossed out the wrong column.

A student computes a 2×2 minor as

a c b d = a d + b c

. Fix it.

It should be

a d - b c

(main diagonal minus anti-diagonal); the sign is a subtraction, not addition — see 2×2 determinant.

A student expands along column 2 which contains

0, 5, 0

, but still computes all three cofactors

C_{12}, C_{22}, C_{32}

. Why is that wasteful (not wrong)?

The answer is still correct, but the

0 \cdot C_{12}

and

0 \cdot C_{32}

terms vanish, so computing those two minors was pure wasted effort.

A student says "

C_{23}

has sign

+

because 2 and 3 are both small." Spot the error.

The sign comes from

(- 1)^{2 + 3} = (- 1)^{5} = - 1

, so

C_{23}

is negative; the size of the indices is irrelevant, only whether

i + j

is odd or even.

A student claims that because their expansion gave a negative number, they "must have made a sign error." Is that reasoning valid?

No — determinants are genuinely allowed to be negative (a negative signed volume means the transformation flips orientation), so a negative result is not itself evidence of a mistake.

A student factors out

a_{11}

and keeps rows 1,2 columns 1,2 as the minor

M_{11}

. What went wrong?

The minor

M_{11}

uses the rows and columns that remain after deleting row 1 and column 1 — that is rows 2,3 and columns 2,3, not rows 1,2.

Why questions

Why do the signs alternate

+, -, +

instead of all being

+

They encode orientation: the alternating

(- 1)^{i + j}

pattern comes from the sign of the permutations in the full determinant sum, ensuring volumes add or subtract correctly so the answer respects handedness.

Why is expanding along a row with many zeros a good strategy?

Each zero element multiplies its cofactor to give

0

, deleting that whole term — so more zeros means fewer 2×2 minors to actually compute.

Why does a zero determinant mean the transformation "collapses" 3D space?

Zero signed volume means the three column (or row) vectors are linearly dependent and lie in a plane or line, so the unit cube gets flattened — it loses a dimension and cannot be inverted (relevant to the inverse).

Why can we expand along any row or column and still get the same number?

Because all these expansions are just regroupings of the one underlying permutation-sum definition of the determinant, which produces a single fixed value for the matrix.

Why does the cofactor need a minor (a smaller determinant) inside it rather than just a number?

Because we're slicing a 3D volume along one direction; each slice's contribution is itself a 2D signed area, and a 2×2 determinant is exactly what measures that area.

Why is a

3 \times 3

determinant built from

2 \times 2

determinants and not, say, straight multiplication of entries?

Volume in 3D depends on how the vectors are oriented relative to one another, and only the alternating minor structure captures that; naive entry products ignore orientation and dependence.

Why does the cross product look like a cofactor expansion along the top row?

The cross product is defined as a determinant with

\hat{i}, \hat{j}, \hat{k}

in row 1, and expanding along that row hands you the three components with the exact alternating signs.

Edge cases

What is the determinant of a 3×3 matrix where one column is entirely zero?

Zero — expand along that column and every term is

0 \times C_{ij}

, so the whole sum collapses to

0

If two columns of a 3×3 matrix are proportional (one is

k

times the other), what is the determinant?

Zero — proportional columns are linearly dependent, so the vectors are coplanar and enclose no volume.

What is the determinant of the 3×3 identity matrix, and what does that value mean geometrically?

It is

1

— the identity leaves space unchanged, so the unit cube keeps unit volume and orientation.

For the diagonal matrix

diag (a, b, c)

, what does cofactor expansion give?

It gives

ab c

— expand along any row and only the diagonal term survives, matching the triangular-matrix rule.

If you scale a single row of a 3×3 matrix by a factor

k

, what happens to the determinant?

It gets multiplied by

k

— that row appears once in every surviving term of the expansion, so factoring

k

out multiplies the whole determinant by

k

(a property of determinants).

What determinant does a matrix have if its three rows satisfy

row 3 = row 1 + row 2

Zero — the rows are linearly dependent (row 3 is built from the others), so they lie in a plane and span no 3D volume.

If every entry of a 3×3 matrix is the same nonzero number, what is the determinant?

Zero — all rows are identical, hence linearly dependent, giving no volume.

Recall One-line survival kit

Element ≠ minor ≠ cofactor; sign is position-only; zeros are your friend; and the answer is the same no matter which row or column you march down.