Is det(2A) equal to 2·det A?

No. Scaling a matrix by a number k multiplies every row by k, and the determinant picks up one factor of k per row. For an n×n matrix this gives det(kA) = kⁿ·det A. So for a 3×3 matrix det(2A) = 2³·det A = 8·det A, not 2·det A. The single factor of 2 only counts one row when all n rows were scaled.

Is det(A + B) = det A + det B?

No, the determinant is not additive. A clean counterexample: take A = B = I, the identity. Then A + B = 2I, so det(A + B) = det(2I) = 2ⁿ, while det A + det B = 1 + 1 = 2. For a 3×3 that is 8 ≠ 2. In general det(A + B) cannot be recovered from det A and det B at all.

Which determinant rules actually hold?

The determinant multiplies, it does not add. The rules that hold are: det(AB) = det A·det B (multiplicative), det(kA) = kⁿ·det A (row-scaling), and det(adj A) = (det A)ⁿ⁻¹ (the adjugate scaling identity, from A·adj A = (det A)·I). The traps that do not hold are det(A + B) = det A + det B, det(kA) = k·det A, and adj A = Aᵀ.

What is det(adj A) for a 3×3 matrix?

det(adj A) = (det A)ⁿ⁻¹, so for a 3×3 matrix det(adj A) = (det A)². It comes from the identity A·adj A = (det A)·I: take determinants of both sides, det A·det(adj A) = (det A)ⁿ, and divide by det A. This is JEE Advanced 2020 Paper 1 Q8, which tests the det/adj power identities as always-true statements. Note the adjugate is the transpose of the cofactor matrix, not the transpose of A.

JEE Advanced Maths · Matrices & Determinants

Does det(2A) = 2·det A? Is the determinant linear?

No. The determinant does not scale or add like a number. Scaling a matrix by $k$ multiplies every row by $k$ , and the determinant collects one factor of $k$ per row, so $\det(kA) = k^n \det A$ for an $n \times n$ matrix. For a $3\times3$ that means $\det(2A) = 2^3 \det A = 8\,\det A$ , not $2\,\det A$ . And the determinant is not additive: $\det(A + B) \ne \det A + \det B$ in general.

The one rule that genuinely holds is multiplication: $\det(AB) = \det A \cdot \det B$ . The clean statement to hold is: the determinant multiplies, it does not add; scaling by $k$ gives $k^n$ , not $k$ . JEE Advanced punishes the linear instinct through det/adj power identities (2020 Paper 1 Q8), trace-versus-determinant (2020 Paper 2 Q4), and count-by-determinant questions (2024 Paper 1 Q10).

Ready to fix this? The Matrices & Determinants lesson works through this misconception and the others in Matrices & Determinants, one altitude at a time.

How to spot it in your own work

You wrote " $\det(2A) = 2\,\det A$ " — pulling a single factor out when every row was scaled.
You wrote " $\det(A + B) = \det A + \det B$ ", treating the determinant as additive.
You treated the adjugate as the transpose ( $\operatorname{adj} A = A^{T}$ ), so you read $\det(\operatorname{adj} A)$ as $\det A$ .
You tried to count matrices by a fixed determinant as if the determinant were a linear function of the entries.

An exam question that triggers it

Here is a question that tests this misconception head-on, in the spirit of JEE Advanced 2020 Paper 1 Q8:

$A$ is a $3\times3$ matrix with $\det A = 4$ . What is $\det(2A)$ ?

The misconception answer reasons "pull the 2 out front, so $\det(2A) = 2 \times 4 = 8$ ." The correct move is to notice that $2A$ multiplies all three rows by 2, and the determinant is linear in each row separately, so each scaled row contributes its own factor of 2:

\det(2A) = 2 \cdot 2 \cdot 2 \cdot \det A = 2^3 \cdot 4 = 32.

Three rows, three factors of 2 — $2^3 = 8$ , giving 32, not 8. The exponent is the size of the matrix.

Why students fall for this

The rule " $\det(AB) = \det A \cdot \det B$ " is genuinely true, so students absorb a vague "the determinant plays nicely with the algebra" and quietly extend it into the linear-looking cousins: $\det(kA) = k\,\det A$ and $\det(A + B) = \det A + \det B$ . These look like the distributive and scaling laws that hold for genuinely linear maps, which is exactly why they survive.

Kazunga and Bansilal (2018), studying students' understanding of the determinant, found that $\det(A + B) = \det A + \det B$ and $\det(kA) = k\,\det A$ are among the most persistent determinant misconceptions. Being told the determinant is "multilinear in the rows" rarely dislodges them. What dislodges it is scaling a matrix by $k$ one row at a time and watching the factor of $k$ appear once per row, so an $n\times n$ matrix collects $k^n$ — and then meeting a counterexample to additivity by hand.

The fix: The determinant multiplies, it does not add

The determinant is linear in each row separately, so scaling all $n$ rows multiplies it by $k^n$ : $\det(kA) = k^n \det A$ . It distributes over products but never over sums: $\det(AB) = \det A \cdot \det B$ always, while $\det(A + B) \ne \det A + \det B$ in general.

The adjugate gives a third face of the same scaling. From $A \cdot \operatorname{adj} A = (\det A)\,I$ , taking determinants of both sides gives $\det A \cdot \det(\operatorname{adj} A) = (\det A)^n$ , so $\det(\operatorname{adj} A) = (\det A)^{n-1}$ — a power, not a linear factor, and certainly not $\det A$ (the adjugate is the transpose of the cofactor matrix, not the transpose of $A$ ).

Worked example

Scale the $3\times3$ identity by 2 one row at a time, watching the determinant.

Start. $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ , so $\det A = 1$ .
Scale row 1 by 2. The determinant is multiplied by 2: $\det = 2 \cdot 1 = 2$ .
Scale row 2 by 2. Another factor of 2: $\det = 2 \cdot 2 = 4$ — already past the "single factor" answer.
Scale row 3 by 2. All three rows are now doubled, which is exactly $2A$ . A third factor of 2:
$\det(2A) = 2 \cdot 2 \cdot 2 \cdot \det A = 2^3 \cdot 1 = 8.$
Conclude. Three rows, three factors of 2, so $\det(kA) = k^n \det A$ — for a $3\times3$ , $\det(2A) = 8\,\det A$ , never $2\,\det A$ .

Now break additivity. Take $A = B = I$ , so $\det A = \det B = 1$ . Then $A + B = 2I$ , and by the scaling rule just derived $\det(A + B) = \det(2I) = 2^3 = 8$ , while $\det A + \det B = 1 + 1 = 2$ . Same determinants going in, $8 \ne 2$ coming out — so no addition rule can hold. The same $k^n$ scaling settles the adjugate too: from $A \cdot \operatorname{adj} A = (\det A) I$ with $\det A = 3$ ,

\det(\operatorname{adj} A) = (\det A)^{n-1} = 3^{2} = 9,

not

3

and not

3^3

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Common questions

Why is $\det(kA) = k^n \det A$ and not $k\,\det A$ ?: Because $kA$ multiplies every row by $k$ , and the determinant is linear in each row on its own — each scaled row contributes one factor of $k$ . An $n\times n$ matrix has $n$ rows, so the factors stack to $k^n$ . For a $3\times3$ matrix that is $k^3$ ; for a $2\times2$ , $k^2$ — never a single $k$ .
Does the determinant ever distribute over a sum?: No. $\det(A + B) = \det A + \det B$ fails in general — take $A = B = I$ , where $\det(A+B) = \det(2I) = 2^n$ but $\det A + \det B = 2$ . The determinant turns matrix products into number products ( $\det(AB) = \det A \cdot \det B$ ), but it does nothing clean with matrix sums.
Is the adjugate the same as the transpose?: No. The adjugate is the transpose of the cofactor matrix, not the transpose of $A$ ; they coincide only in special cases. The defining identity is $A \cdot \operatorname{adj} A = (\det A)\,I$ , which gives $\det(\operatorname{adj} A) = (\det A)^{n-1}$ and $\operatorname{adj}(\operatorname{adj} A) = (\det A)^{n-2} A$ — power laws in $\det A$ , not a transpose.
How does this trap show up in counting questions?: JEE Advanced 2024 Paper 1 Q10 asks how many $3\times3$ matrices with entries in $\{0,1\}$ have $\det \in \{-1, 1\}$ . You cannot count by scaling a base case, because the determinant is not a linear function of the entries — you must classify by the actual value. The answer is 168 (84 with $\det = 1$ and 84 with $\det = -1$ , paired by a row swap that flips the sign).

Related misconceptions

A zero determinant does not mean no solutiondet A = 0 means the solution is not unique, not that there is none — a consistency check decides infinitely many versus none.
Matrix multiplication does not commuteAB ≠ BA in general, and the inverse and transpose rules reverse the order: (AB)⁻¹ = B⁻¹A⁻¹.

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