Q.If the determinant $\begin{vmatrix} x + a & p + u & l + f \\ y + b & q + v & m + g \\ z + c & r + w & n + h \end{vmatrix}$ splits into exactly $K$ determinants of order $3$, each element of which contains only one term, then the value of $K$ is $8$.

Concept: Determinant Evaluation Using Identities — specifically, the linearity property: a determinant is linear in each row (or column). Here, every element is a sum of two terms, so each row is a sum of two row-vectors.

Step 1: Write the determinant as

$\begin{vmatrix} (x, p, l) + (a, u, f) \\ (y, q, m) + (b, v, g) \\ (z, r, n) + (c, w, h) \end{vmatrix}$ …

The key idea here is linearity of the determinant in each column. A determinant is a multilinear function — it behaves linearly when you add two vectors in a single column, keeping the other columns fixed. This property lets us break a complicated-looking determinant into a sum of simpler ones.

Let’s see why this works. If you have a determinant where one column is the sum of two vectors, say $\begin{vmatrix} \mathbf{c}_1 + \mathbf{d}_1 & \mathbf{c}_2 & \mathbf{c}_3 \end{vmatrix}$, it equals $\begin{vmatrix} \mathbf{c}_1 & \mathbf{c}_2 & \mathbf{c}_3 \end{vmatrix} + \begin{vmatrix} \mathbf{d}_1 & \mathbf{c}_2 & \mathbf{c}_3 \end{vmatrix}$. The same holds for any column. This is not a trick — it follows directly from the definition of determinant as an alternating multilinear form.

Now apply this to your problem. Each column of the given determinant is itself a sum of two vectors:

• Column 1: $(x+a,\; y+b,\; z+c)^T$ = $(x,y,z)^T + (a,b,c)^T$
• Column 2: $(p+u,\; q+v,\; r+w)^T$ = $(p,q,r)^T + (u,v,w)^T$
• Column 3: $(l+f,\; m+g,\; n+h)^T$ = $(l,m,n)^T + (f,g,h)^T$

So we have three columns, each a sum of two terms. When we expand using linearity, we do it one column at a time.

1. Start with the first column. Split it into two determinants:

2. Now each of these two determinants has its second column as a sum. Split each one again:

So far we have $2 \times 2 = 4$ determinants.

3. Each of these four determinants still has its third column as a sum. Split each one one more time. For example: …

Method: Counting the Split-Off Determinants Using Multilinearity

Use this whenever a question asks "into how many determinants does this split" when several columns (or rows) are each written as a sum of terms.

Steps

Step 1: Recall determinant linearity in each column separately

If one column of a determinant is $\mathbf c=\mathbf c'+\mathbf c''$, then

with every other column unchanged. This is a per-column rule, not a whole-matrix rule.

Step 2: Apply it independently, column by column

If column 1 splits into $m_1$ pieces, splitting it produces $m_1$ determinants. Now split column 2 in EACH of those $m_1$ determinants, producing $m_1\times m_2$ determinants, and so on for every splitting column.

Step 3: Multiply the counts to get the total …

Common Mistakes

Mistake 1: Confusing column-wise splitting with the false rule $\det(A+B)=\det A+\det B$

Why it's wrong: a determinant is linear in each column separately, not linear as a function of the whole matrix — splitting works one column at a time, giving $2^n$ (or more generally $\prod m_i$) terms, not just $2$. Correct approach: split one column at a time, and multiply the counts across all splitting columns.

Mistake 2: Forgetting that a repeated-column term would vanish in a related "evaluate" version of this question …