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Question: $\begin{vmatrix} 3a & -a+b & -a+c \\ -b+a & 3b & -b+c \\ -c+a & -c+b & 3c \end{vmatrix} = 3(a+b+c)(a...

∣3a−a+b−a+c−b+a3b−b+c−c+a−c+b3c∣=3(a+b+c)(ab+bc+ca)\begin{vmatrix} 3a & -a+b & -a+c \\ -b+a & 3b & -b+c \\ -c+a & -c+b & 3c \end{vmatrix} = 3(a+b+c)(ab+bc+ca)

Answer

3(a+b+c)(ab+bc+ca)

Explanation

Solution

To prove the given identity, we will evaluate the determinant on the left-hand side (LHS) using properties of determinants and show that it equals the expression on the right-hand side (RHS).

Let the given determinant be Δ\Delta: Δ=∣3a−a+b−a+c−b+a3b−b+c−c+a−c+b3c∣\Delta = \begin{vmatrix} 3a & -a+b & -a+c \\ -b+a & 3b & -b+c \\ -c+a & -c+b & 3c \end{vmatrix}

Step 1: Apply Column Operations

Apply the operation C1→C1+C2+C3C_1 \to C_1 + C_2 + C_3. This operation adds the elements of the second and third columns to the first column.
The new elements of the first column will be:

  • For the first row: 3a+(−a+b)+(−a+c)=3a−a+b−a+c=a+b+c3a + (-a+b) + (-a+c) = 3a - a + b - a + c = a+b+c
  • For the second row: (−b+a)+3b+(−b+c)=−b+a+3b−b+c=a+b+c(-b+a) + 3b + (-b+c) = -b+a+3b-b+c = a+b+c
  • For the third row: (−c+a)+(−c+b)+3c=−c+a−c+b+3c=a+b+c(-c+a) + (-c+b) + 3c = -c+a-c+b+3c = a+b+c

So, the determinant becomes: Δ=∣a+b+c−a+b−a+ca+b+c3b−b+ca+b+c−c+b3c∣\Delta = \begin{vmatrix} a+b+c & -a+b & -a+c \\ a+b+c & 3b & -b+c \\ a+b+c & -c+b & 3c \end{vmatrix}

Step 2: Factor out Common Term

Now, we can take out the common factor (a+b+c)(a+b+c) from the first column: Δ=(a+b+c)∣1−a+b−a+c13b−b+c1−c+b3c∣\Delta = (a+b+c) \begin{vmatrix} 1 & -a+b & -a+c \\ 1 & 3b & -b+c \\ 1 & -c+b & 3c \end{vmatrix}

Step 3: Apply Row Operations to Create Zeros

To simplify the determinant further, we can make two elements in the first column zero using row operations.
Apply R2→R2−R1R_2 \to R_2 - R_1 and R3→R3−R1R_3 \to R_3 - R_1.

  • For R2→R2−R1R_2 \to R_2 - R_1:

    • 1−1=01 - 1 = 0
    • 3b−(−a+b)=3b+a−b=a+2b3b - (-a+b) = 3b + a - b = a+2b
    • (−b+c)−(−a+c)=−b+c+a−c=a−b(-b+c) - (-a+c) = -b+c+a-c = a-b

  • For R3→R3−R1R_3 \to R_3 - R_1:

    • 1−1=01 - 1 = 0
    • (−c+b)−(−a+b)=−c+b+a−b=a−c(-c+b) - (-a+b) = -c+b+a-b = a-c
    • 3c−(−a+c)=3c+a−c=a+2c3c - (-a+c) = 3c+a-c = a+2c

The determinant now becomes: Δ=(a+b+c)∣1−a+b−a+c0a+2ba−b0a−ca+2c∣\Delta = (a+b+c) \begin{vmatrix} 1 & -a+b & -a+c \\ 0 & a+2b & a-b \\ 0 & a-c & a+2c \end{vmatrix}

Step 4: Expand the Determinant

Expand the determinant along the first column. Since the first column has two zeros, the expansion will be simple: Δ=(a+b+c)×1×∣a+2ba−ba−ca+2c∣\Delta = (a+b+c) \times 1 \times \begin{vmatrix} a+2b & a-b \\ a-c & a+2c \end{vmatrix} Now, evaluate the 2×22 \times 2 determinant: ∣a+2ba−ba−ca+2c∣=(a+2b)(a+2c)−(a−b)(a−c)\begin{vmatrix} a+2b & a-b \\ a-c & a+2c \end{vmatrix} = (a+2b)(a+2c) - (a-b)(a-c) Expand the products:

  • (a+2b)(a+2c)=a2+2ac+2ab+4bc(a+2b)(a+2c) = a^2 + 2ac + 2ab + 4bc
  • (a−b)(a−c)=a2−ac−ab+bc(a-b)(a-c) = a^2 - ac - ab + bc

Subtract the second expansion from the first: (a2+2ac+2ab+4bc)−(a2−ac−ab+bc)(a^2 + 2ac + 2ab + 4bc) - (a^2 - ac - ab + bc) =a2+2ac+2ab+4bc−a2+ac+ab−bc= a^2 + 2ac + 2ab + 4bc - a^2 + ac + ab - bc =(a2−a2)+(2ac+ac)+(2ab+ab)+(4bc−bc)= (a^2 - a^2) + (2ac + ac) + (2ab + ab) + (4bc - bc) =0+3ac+3ab+3bc= 0 + 3ac + 3ab + 3bc =3(ab+bc+ca)= 3(ab+bc+ca)

Step 5: Final Result

Substitute this back into the expression for Δ\Delta: Δ=(a+b+c)[3(ab+bc+ca)]\Delta = (a+b+c) [3(ab+bc+ca)] Δ=3(a+b+c)(ab+bc+ca)\Delta = 3(a+b+c)(ab+bc+ca) This matches the RHS of the given identity.

Therefore, the identity is proven.