Problem 69 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(x_{1} \neq 0, x_{2} \neq 0, x_{3} \neq 0\), then the determinant \(\left|\begin{array}{ccc}x_{1}+a_{1} b_{1} & a_{1} b_{2} & a_{1} b_{3} \\\ a_{2} b_{1} & x_{2}+a_{2} b_{2} & a_{2} b_{3} \\ a_{3} b_{1} & a_{3} b_{2} & x_{3}+a_{3} b_{3}\end{array}\right|\) is equal to (A) \(x_{1} x_{2} x_{3}\left(1+\frac{a_{1} b_{1}}{x_{1}}+\frac{a_{2} b_{2}}{x_{2}}+\frac{a_{3} b_{3}}{x_{3}}\right)\) (B) \(-x_{1} x_{2} x_{3}\left(1+\frac{a_{1} b_{1}}{x_{1}}+\frac{a_{2} b_{2}}{x_{2}}+\frac{a_{3} b_{3}}{x_{3}}\right)\) (C) \(x_{1} x_{2} x_{3}\left(1-\frac{a_{1} b_{1}}{x_{1}}-\frac{a_{2} b_{2}}{x_{2}}-\frac{a_{3} b_{3}}{x_{3}}\right)\) (D) None of these

Step-by-Step Solution

Answer

(A) x_1 x_2 x_3 \left(1+\frac{a_{1}b_{1}}{x_{1}}+\frac{a_{2}b_{2}}{x_{2}}+\frac{a_{3}b_{3}}{x_{3}}\right)

Step 1: Identify Matrix Format

We have a 3x3 matrix with specific entries: the diagonal elements are of the form \(x_i + a_i b_i\) and the off-diagonal elements are \(a_i b_j\) for \(i eq j\). The matrix can be written as: \[ \begin{bmatrix} x_1 + a_1 b_1 & a_1 b_2 & a_1 b_3 \ a_2 b_1 & x_2 + a_2 b_2 & a_2 b_3 \ a_3 b_1 & a_3 b_2 & x_3 + a_3 b_3 \end{bmatrix} \]

Step 2: Simplify Using Properties of Determinants

A special form allows simplification: if \(M = X + A \, B^T\), where \(X\) is a diagonal matrix, \(A\) and \(B\) are column vectors, then \(\det(M) = \det(X) (1 + \mathrm{trace}((X^{-1})AB^T))\). For our matrix, \(X\) is \(\mathrm{diag}(x_1, x_2, x_3)\), and \((AB^T)_{ij} = a_i b_j\).

Step 3: Calculate the Determinant of the Diagonal Matrix

The determinant of the matrix \(X = \begin{bmatrix} x_1 & 0 & 0 \ 0 & x_2 & 0 \ 0 & 0 & x_3 \end{bmatrix} \) is \(x_1 x_2 x_3\). Because it's a diagonal matrix, the determinant is simply the product of its diagonal entries.

Step 4: Calculate Trace Contribution

For the trace contribution, since \(X^{-1} = \mathrm{diag}(1/x_1, 1/x_2, 1/x_3)\), we calculate the trace of \( (X^{-1})AB^T \) as follows:\[ \mathrm{trace}\left( \begin{bmatrix} a_1 \frac{b_1}{x_1} & a_2 \frac{b_2}{x_2} & a_3 \frac{b_3}{x_3} \end{bmatrix} \right) = \frac{a_1 b_1}{x_1} + \frac{a_2 b_2}{x_2} + \frac{a_3 b_3}{x_3}\]. So for the determinant, we get the expression \( 1 + \frac{a_1 b_1}{x_1} + \frac{a_2 b_2}{x_2} + \frac{a_3 b_3}{x_3} \).

Step 5: Combine Results

Combine the results of Steps 3 and 4. Thus, the determinant of the matrix is \( x_1 x_2 x_3 \left( 1 + \frac{a_1 b_1}{x_1} + \frac{a_2 b_2}{x_2} + \frac{a_3 b_3}{x_3} \right) \).

Step 6: Mathematically Match with Options

The obtained determinant expression matches exactly with option (A) \(x_1 x_2 x_3\left(1+\frac{a_{1} b_{1}}{x_{1}}+\frac{a_{2} b_{2}}{x_{2}}+\frac{a_{3} b_{3}}{x_{3}}\right)\). Therefore, the correct answer is (A).

Key Concepts

Properties of Determinants

Determinants have various properties that make them a powerful tool in linear algebra. One of these properties is related to the calculation of the determinant when you have a special form of a matrix. For instance, if you encounter a matrix that can be decomposed into the form \( M = X + AB^T \), where \( X \) is a diagonal matrix, and \( A \) and \( B \) are column vectors, this enables an easier path to finding the determinant.

In such scenarios, the determinant can be simplified with the formula:

• \( \det(M) = \det(X) \left( 1 + \mathrm{trace}((X^{-1})AB^T) \right) \)

Knowing this, the determinant of \( M \) is reduced to a simpler calculation, involving just the diagonal matrix \( X \) and its inverse. This property is particularly beneficial because multiplying and inverting a diagonal matrix is straightforward, and finding the trace involves just summing the diagonal elements.

Diagonal Matrix

A diagonal matrix is a type of matrix where all the elements outside the main diagonal are zero. Simply put, only the diagonals from the top left to the bottom right (or vice versa) have non-zero values, and every other element is zero.

Diagonal matrices have unique characteristics:

• Easy to manipulate due to the simplicity of operations like multiplication and inversion.
• The determinant is just the product of the diagonal entries \( x_1 \times x_2 \times x_3 \).

For any matrix involving complex entries, converting or identifying it as a diagonal matrix when applicable can save a tremendous amount of calculation effort. In our scenario, the matrix \( X \) is already a diagonal matrix, so calculating its determinant is just about multiplying its diagonal elements.

Matrix Simplification

Matrix simplification refers to the process of making complicated matrix operations more feasible by breaking down the matrix into simpler forms. Specifically in determinant calculation, if a matrix can be expressed as a combination of simpler matrices like a diagonal matrix and additional vectors, simplification becomes highly effective.

In practice, simplifying a matrix can involve:

• Identifying diagonal structures within a matrix.
• Using vector operations to describe matrix transformations, such as outer products.
• Applying determinant properties to reduce computational complexity.

In our initial matrix, this simplification entailed recognizing \( M = X + AB^T \) structure, allowing a focus on fundamental operations like matrix determinant multiplication and trace calculation. These ideas help develop a more efficient and clear approach to solving matrix-based problems.