Cramer's rule: Question bank — Maths deep dive

Before we start, three words we will lean on, each pinned to a picture in your head:

Why questions

Why does building

X_{i}

(identity with column

i

replaced by

x

) give

det X_{i} = x_{i}

All other columns are basis vectors contributing factor

1

; expanding leaves only the diagonal entry

x_{i}

. It's multilinearity of

det

doing the work.

Why is it a column of

A

that gets replaced by

b

, not a row?

Because

A x

is a combination of the columns of

A

; when

X_{i}

carries

x

in column

i

, the product

A X_{i}

swaps exactly that column for

A x = b

Why must

det A \neq = 0

appear as a condition, geometrically?

det A = 0

means the column arrows are squashed into a lower dimension (a flat box). Then

b

either can't be reached or can be reached many ways — no single recipe, so no unique

x_{i}

Why does a tiny

det A

make the answer numerically dangerous?

You divide by a near-zero number, so small errors in

det A_{i}

det A

get hugely amplified — the solution is ill-conditioned, a key reason to prefer Gaussian Elimination in practice.

Why does

det (M N) = det M det N

show up in the proof?

Taking determinants of

A X_{i} = A_{i}

needs to split

det (A X_{i})

into

det A \cdot det X_{i}

. Multiplicativity is exactly that split, turning the matrix identity into a number equation

det A \cdot x_{i} = det A_{i}

Why is Cramer's rule better suited to theory than to computation?

It gives each unknown as a clean closed-form ratio — perfect for proving facts (like smoothness of solutions in the entries). But computing many determinants is costly, so it loses to elimination on real numbers.

Edge cases

b

equals one of the columns of

A

, say

b = a_{1}

— what does Cramer give?

Then

x_{1} = 1

and all other

x_{i} = 0

: for

i \neq = 1

A_{i}

has two equal columns (

a_{1}

appears twice) so

det A_{i} = 0

; for

i = 1

A_{1} = A

giving

x_{1} = 1

A

is the identity matrix — what is the solution?

A_{i}

becomes the identity with column

i

set to

b

, so

det A_{i} = b_{i}

and

det A = 1

; thus

x_{i} = b_{i}

, i.e.

x = b

, exactly as

I x = b

demands.

1 \times 1

system

a x = b

— does Cramer's rule still make sense?

Yes:

det A = a

A_{1} = (b)

det A_{1} = b

, giving

x = b / a

— the ordinary division, valid precisely when

a \neq = 0

Two columns of

A

are identical — what happens?

det A = 0

(repeated column ⇒ flat box), so Cramer's rule is void. Geometrically two ingredients are the same, so the recipe for reaching

b

isn't unique (or is impossible).

The system has no solution (inconsistent) — will Cramer detect it?

Only indirectly: inconsistency requires

det A = 0

, so the formula already refuses to run. Cramer never returns a fake answer for a genuinely square inconsistent system, because it stops at the division-by-zero gate.

One entry of

b

is zero — is that a problem?

No. A zero entry of

b

is fine; only a zero determinant

det A

breaks the rule. The column

b

can freely contain zeros.

Recall One-line summary of every trap

The single guardrail behind almost all of these ::: Replace a column with $b$ , and only run the rule when $det A \neq = 0$ — everything else follows from $det$ being a signed, multilinear, alternating volume.

Connections

Cramer's rule — the parent this bank stress-tests.
Determinants — signed volume, the source of every sign trap.
Matrix Inverse — why Cramer and $A^{- 1} b$ must agree.
Cofactor Expansion — where the alternating signs bite.
Gaussian Elimination — the practical alternative when $det A$ is tiny or $n$ is large.
Multilinear and Alternating Maps — the abstract engine.
Invertible Matrix Theorem — $det A \neq = 0 ⟺$ unique solution.

Question bank — Cramer's rule

True or false — justify

Spot the error

Why questions

Edge cases

Connections