Determinants: Step-by-Step Worked Examples (8)

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Tags Determinants

Question

Without expanding, prove that ∣1ab+c1bc+a1ca+b∣=0\begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b \end{vmatrix} = 0.

Solution — Step by Step

C3→C3+C2:∣1aa+b+c1ba+b+c1ca+b+c∣C_3 \to C_3 + C_2: \quad \begin{vmatrix} 1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c \end{vmatrix}

Now C3C_3 has the constant a+b+ca+b+c in every row.

=(a+b+c)∣1a11b11c1∣= (a+b+c)\begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix}

The first and third columns are identical (both all 1s). A determinant with two identical columns equals 0.

=(a+b+c)⋅0=0= (a+b+c) \cdot 0 = 0

Final answer: 0 (proved using column operations)

Why This Works

Two key properties:

  1. Adding a multiple of one column to another doesn’t change the determinant.
  2. If two columns (or rows) are identical, the determinant is 0.

Combining them is the heart of “prove without expanding” problems. Find an operation that creates duplicate columns/rows or a row/column of zeros.

Alternative Method

Expand the determinant directly:

1[b(a+b)−c(c+a)]−a[(a+b)−(c+a)]+(b+c)[c−b]1[b(a+b) - c(c+a)] - a[(a+b) - (c+a)] + (b+c)[c - b]

=ab+b2−c2−ac−a(b−c)+(b+c)(c−b)= ab + b^2 - c^2 - ac - a(b-c) + (b+c)(c-b)

=ab+b2−c2−ac−ab+ac+(c2−b2)=0= ab + b^2 - c^2 - ac - ab + ac + (c^2 - b^2) = 0

Same result, but much more algebra.

Common Mistake

Students apply C3→C3−C2C_3 \to C_3 - C_2 instead of C3→C3+C2C_3 \to C_3 + C_2. Both are valid operations, but the first doesn’t simplify C3C_3 to a constant — you’d get b+c−a,c+a−b,a+b−cb+c-a, c+a-b, a+b-c, no common factor.

Choose the operation that creates a constant column/row or duplicates an existing one.

For “prove det = 0” questions, look for: two equal rows/columns, a row/column of zeros, or rows/columns that are linear combinations of others. One of these three almost always works after a single operation.

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