Q.If $A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$, then show that $|2A| = 4|A|$.

Concept: Scalar Multiplication — when a matrix is multiplied by a scalar, every element is multiplied by that scalar. The determinant of the resulting matrix is the scalar raised to the order of the matrix times the original determinant.

Step 1: Compute $|A|$.

$|A| = (1)(2) - (2)(4) = 2 - 8 = -6$.

Step 2: Compute $2A$:

$2A = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$.

Step 3: Compute $|2A|$:

$|2A| = (2)(4) - (4)(8) = 8 - 32 = -24$.

Step 4: Compare: $4|A| = 4(-6) = -24$, which equals $|2A|$.

The key idea is scalar multiplication of a determinant. When you multiply a matrix by a scalar $k$, every single entry gets multiplied by $k$. But a determinant is a multilinear function — it scales linearly with each row (or column). So if you have an $n \times n$ matrix, multiplying the whole matrix by $k$ multiplies each of the $n$ rows by $k$, and the determinant gets multiplied by $k$ a total of $n$ times. That is, $|kA| = k^n |A|$.

Here $n = 2$, so $|2A| = 2^2 |A| = 4|A|$. Let's verify this concretely.

1. Compute $|A|$ first. For $A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$,

2. Form $2A$ by multiplying every entry by 2.

3. Compute $|2A|$.

4. Compare $|2A|$ with $4|A|$.

They match exactly: $|2A| = -24 = 4|A|$.

Method: Verifying the Scalar-Multiplication Property $|kA| = k^{\,n}|A|$

Use this whenever asked to verify (or apply) how a determinant scales when the whole matrix is multiplied by a scalar.

Steps

Step 1: State the general property and why it holds

A determinant is multilinear — it scales by $k$ for every row that gets multiplied by $k$. Multiplying the whole $n\times n$ matrix by $k$ scales every row by $k$, so the determinant picks up a factor of $k$ once per row:

Step 2: Compute $|A|$ directly

Evaluate the original determinant using the standard $ad-bc$ (or cofactor) method.

Step 3: Form $kA$ by scaling every entry, then compute $|kA|$ directly

Multiply each entry of $A$ by $k$ (not just one row), then evaluate the new determinant the same way.

Step 4: Confirm $|kA| = k^{\,n}|A|$

Compare the two computed numbers against $k^n$ times the original determinant to confirm they match.

Common Mistakes

Mistake 1: Assuming $|kA| = k|A|$ for any matrix

Why it's wrong: this treats the matrix as if only a single row were scaled, but multiplying the whole matrix by $k$ scales every row by $k$ — for an $n\times n$ matrix the determinant picks up a factor of $k$ once per row, giving $k^n$, not $k$. Correct approach: always state $|kA|=k^n|A|$ and identify $n$ from the actual order of the matrix before applying it.

Mistake 2: Only scaling one row or column when forming $kA$

Why it's wrong: $kA$ means multiplying every entry of the matrix by $k$, not just one row — partial scaling gives a wrong intermediate matrix and a wrong determinant. Correct approach: multiply every single entry of $A$ by $k$ before computing $|kA|$ directly, as a check against the $k^n|A|$ formula.