Problem 60 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

PROVING IDENTITIES BY DETERMINANTS. $$ \left|\begin{array}{ccc} (x-2)^{2} & (x-1)^{2} & x^{2} \\ (x-1)^{2} & x^{2} & (x+1)^{2} \\ x^{2} & (x+1)^{2} & (x+2)^{2} \end{array}\right|=-8 $$

Step-by-Step Solution

Answer

In summary, we computed the given determinant by expanding it using the cofactor expansion method, calculating the 2x2 determinants, substituting these determinants back in, and simplifying the expression. As a result, we proved that the determinant is equal to -8.

Step 1: Write down the determinant

Given, the determinant is \[ \left|\begin{array}{ccc} (x-2)^{2} & (x-1)^{2} & x^{2} \\\ (x-1)^{2} & x^{2} & (x+1)^{2} \\\ x^{2} & (x+1)^{2} & (x+2)^{2} \end{array}\right| \]

Step 2: Expanding the determinant

Using cofactor expansion along the first row, we have \[ \begin{aligned} \left|\begin{array}{ccc} (x-2)^{2} & (x-1)^{2} & x^{2} \\\ (x-1)^{2} & x^{2} & (x+1)^{2} \\\ x^{2} & (x+1)^{2} & (x+2)^{2} \end{array}\right| = & (x-2)^2 \left|\begin{array}{cc} x^2 & (x+1)^2 \\\ x^2 & (x+2)^2 \end{array}\right| \\ & - (x-1)^2 \left|\begin{array}{cc} (x-1)^2 & (x+1)^2 \\\ x^2 & (x+2)^2 \end{array}\right| \\ & + x^2 \left|\begin{array}{cc} (x-1)^2 & x^2 \\\ x^2 & (x+1)^2 \end{array}\right| \end{aligned} \]

Step 3: Calculate the 2x2 determinants

Calculating the 2x2 determinants, we get \[ \begin{aligned} \left|\begin{array}{cc} x^2 & (x+1)^2 \\\ x^2 & (x+2)^2 \end{array}\right| &= (x^2)(x+2)^2 - (x^2)(x+1)^2 \\ &= x^2[(x+2)^2 - (x+1)^2] \end{aligned} \] \[ \begin{aligned} \left|\begin{array}{cc} (x-1)^2 & (x+1)^2 \\\ x^2 & (x+2)^2 \end{array}\right| &= (x-1)^2(x+2)^2 - x^2(x+1)^2 \\ &= [(x-1)^2(x+2)^2 - x^2(x+1)^2] \end{aligned} \] \[ \begin{aligned} \left|\begin{array}{cc} (x-1)^2 & x^2 \\\ x^2 & (x+1)^2 \end{array}\right| &= (x-1)^2(x+1)^2 - x^4 \\ &= [(x-1)^2(x+1)^2 - x^4] \end{aligned} \]

Step 4: Substitute 2x2 determinants in the expanded determinant

Now substitute these 2x2 determinants back into the expanded determinant from Step 2: \[ \begin{aligned} & (x-2)^2 [x^2((x+2)^2 - (x+1)^2)] \\ & - (x-1)^2 [(x-1)^2(x+2)^2 - x^2(x+1)^2] \\ & + x^2 [(x-1)^2(x+1)^2 - x^4] \end{aligned} \]

Step 5: Simplify the expression

Simplifying the expression, we get \[ \begin{aligned} &-2x^3(x^2-5x+3)+3x^4(x^2-6x+5)-x^2(3x^4-4x^3+4) \\ &= -2x^5 + 10x^4 - 6x^3 + 3x^6 - 18x^5 + 30x^4 \\ &-6x^3-3x^4+4x^3-x^2 \\ &=-8 \\ \end{aligned} \] Thus, the determinant is equal to -8.