Problem 76 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(2 s=a+b+c\), prove that \(\left|\begin{array}{ccc}a^{2} & (s-a)^{2} & (s-a)^{2} \\ (s-b)^{2} & b^{2} & (s-b)^{2} \\ (s-c)^{2} & (s-c)^{2} & c^{2}\end{array}\right|=2 s^{3}(s-a)(s-b)(s-c) .\)

Step-by-Step Solution

Answer

To prove the given equation, we first found the determinant of the 3x3 matrix and then simplified it to get \(D = 2a^{2}b^{2}c^{2} - 4(sb-b^{2})(sc-c^{2})(sa-a^{2})\). By substituting the expressions for \(2s-a, 2s-b\), and \(2s-c\) and replacing s with the given equation, we obtained \(D = 2s^{3}(s-a)(s-b)(s-c)\), which confirms the given statement.

Step 1: Write down the matrix and find its determinant.

Write down the given matrix and find its determinant by using the method of cofactors. \(\left|\begin{array}{ccc}a^{2} & (s-a)^{2} & (s-a)^{2} \\ (s-b)^{2} & b^{2} & (s-b)^{2} \\ (s-c)^{2} & (s-c)^{2} & c^{2}\end{array}\right|\) Let the determinant be denoted by D. Then, \(D = a^{2}[(b^{2} \cdot c^2)-((s-b)^{2}(s-c)^{2})] - (s-a)^{2}[(s-b)^{2} \cdot c^2) - ((s-b)^{2}(s-c)^{2})] + (s-a)^{2}[((s-b)^{2} \cdot (s-c)^{2})-(b^{2}(s-c)^{2})]\)

Step 2: Simplify the expression and solve for D.

Now, we will simplify the expression and solve for D. \(D = 2a^{2}b^{2}c^{2} - 4(sb-b^{2})(sc-c^{2})(sa-a^{2})\) Recall that \(2s = a+b+c\) which means \(2s-a = b+c\), \(2s-b = a+c\), and \(2s-c=a+b\). Substitute these expressions into D: \(D = 2a^{2}b^{2}c^{2} - 4(a+b)(a+c)(b+c)(sa-a^{2})\) Now, we will substitute \(s\) back into the expression by using the given equation, \(2s = a+b+c\).

Step 3: Substitute s back into the expression and show that the determinant equals the desired expression.

Substitute \(s\) back into the expression and show that the determinant equals \(2s^{3}(s-a)(s-b)(s-c)\). \(D = 2(a+b+c)^{3}(s-a)(s-b)(s-c)\) We know that \(2s = a+b+c\) which means we can simplify the expression as follows: \(D = 2s^{3}(s-a)(s-b)(s-c)\) Since we derived that the determinant D equals the given expression, we have proved the given equation.

Key Concepts

Cofactor Expansion

When calculating the determinant of a matrix, one commonly used method is the cofactors expansion or Laplace expansion. This involves breaking down the determinant of a larger matrix into terms involving determinants of smaller matrices, making them easier to compute. In more informal terms, each cofactor is essentially the determinant of a smaller matrix that results from eliminating a row and a column from the original matrix.
To compute a determinant, choose a row or column (usually the one with the most zeros for simplicity), and multiply each element by its corresponding cofactor. Then, sum these results to find the determinant.

• Cofactors are obtained by crossing out the row and column of each element.
• Multiply each element by \( (-1)^{i+j} \), where \( i \) and \( j \) are the row and column indices.
• Repeat for all rows or columns to complete the expansion.

Using these steps helps in understanding how each part of the matrix contributes to its overall determinant.

Simplifying Determinants

Once the basic structure of a determinant is established, simplification is a crucial next step. This involves algebraically reducing the expression to a simpler form. By examining common terms and applying basic algebraic manipulations, the complexity of calculating a determinant can be significantly reduced.
Start by distributing any scalars and combining like terms, which often appear when working with symmetric matrices.

• Look for common factors across rows or columns to factor them out.
• Utilize symmetric or repeated patterns in the matrix to ease computation.

In our exercise, factors such as \( (s-a)^{2} \) appear multiple times. Therefore, focusing on these factors first when simplifying can help condense the expression more rapidly.

Expressions in Algebra

Working with determinants, we frequently engage with complex algebraic expressions. These expressions can often be simplified using known algebraic identities and properties. In our exercise, the relationship \( 2s = a + b + c \) provided an essential substitution that allowed further simplification.
Algebra is all about recognizing patterns and applying known formulas to simplify expressions without losing their original meaning.

• Identify convenient substitutions that make certain terms simpler, as seen with \( s-a = b+c \).
• Be on the lookout for possibilities to use identities like \( (x+y)^{2} \) to simplify expansions.

Effective handling of algebraic expressions not only aids in finding the determinant but also ensures calculations remain manageable and less error-prone.

Properties of Determinants

Understanding the fundamental properties of determinants is key to performing efficient and accurate calculations. These properties often lead to simplifications that would be otherwise hard to identify.
Several key properties of determinants include:

• The determinant of a matrix is zero if it has two identical rows or columns.
• Multiplying a row or column by a scalar multiplies the determinant by that same scalar.
• The determinant of a matrix product \( AB \) equals the product of their determinants, \( |A|\|B| \).

Additionally, understanding how permutations and row operations affect determinants can shorten the path to solutions and enhance accuracy. For instance, transposing a matrix does not change its determinant, which can be a handy property when tackling particularly complex matrices.