Q.If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$, then show that $|3A| = 27|A|$.

Concept: Scalar Multiplication — When a matrix is multiplied by a scalar, each element is multiplied by that scalar. For an $n \times n$ matrix, the determinant scales by $k^n$.

Here $A$ is $3 \times 3$, so $n = 3$.

Step 1: Compute $|A|$. Since $A$ is upper triangular, the determinant is the product of the diagonal entries:

Step 2: For a $3 \times 3$ matrix, $|kA| = k^3 |A|$. Thus: …

The property at work here is scalar multiplication of a determinant. Many students rush to compute $|3A|$ directly by first finding $3A$ and then evaluating its determinant. That works, but it misses the deeper pattern — and it's slower. Let's understand why the factor $3^3$ appears.

When you multiply a matrix $A$ by a scalar $k$, you multiply every entry of $A$ by $k$. Now, the determinant is a multilinear function of the rows (or columns). That means if you multiply a single row by $k$, the determinant gets multiplied by $k$. But here, all three rows are multiplied by $k$ — so the determinant gets multiplied by $k$ three times, once for each row.

For our $3 \times 3$ matrix, $n = 3$, so $|3A| = 3^3 |A| = 27|A|$. That's the entire logical skeleton. Now let's flesh it out step by step.

1. Write down $3A$ explicitly. Multiply each entry of $A$ by $3$:

2. Compute $|3A|$ directly (to verify). The matrix is upper triangular (all entries below the main diagonal are zero). For a triangular matrix, the determinant is simply the product of the diagonal entries:

3. Now compute $|A|$. $A$ is also upper triangular: …

Method: Verifying $|kA| = k^{\,n}|A|$ for a Triangular Matrix

Same scaling property as above, specialised to a case where the triangular structure lets you skip full cofactor expansion.

Steps

Step 1: Recognise the triangular shortcut

If $A$ is upper or lower triangular, $|A|$ is simply the product of its diagonal entries — no expansion needed.

Step 2: State the general scaling rule

For an $n\times n$ matrix, multiplying every entry by $k$ multiplies each of the $n$ rows by $k$, so

Step 3: Compute $|A|$ using the diagonal product

Multiply the diagonal entries of the original matrix.

Step 4: Compute $|kA|$ the same way …

Common Mistakes

Mistake 1: Using $|kA|=k|A|$ instead of $|kA|=k^n|A|$ for the correct order $n$

Why it's wrong: for this $3\times3$ matrix, the correct scaling factor is $k^3=27$, not $k=3$ — using the wrong exponent gives $12$ instead of the correct $108$. Correct approach: always identify $n$ (the matrix's own order) first and use $k^n$, not $k$.

Mistake 2: Needlessly performing a full $3\times3$ cofactor expansion on $3A$ instead of using its triangular structure …