Problem 53 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\mathrm{B}=\left[\begin{array}{ccc}5 & 2 \alpha & 1 \\ 0 & 2 & 1 \\\ \alpha & 3 & -1\end{array}\right]\) is the inverse of a \(3 \times 3\) matrix \(\mathrm{A}\), then the sum of all values of \(\alpha\) for which det \((\mathrm{A})+1=0\), is : [April 12, 2019 (I)] (a) 0 (b) \(-1\) (c) 1 (d) 2

Step-by-Step Solution

Answer

The sum of all values of \(\alpha\) is 1.

Step 1: Understanding the Problem

We are given a matrix \( B \) which is the inverse of another 3x3 matrix \( A \). It is asked to find the sum of all values of \( \alpha \) for which \( \text{det}(A) + 1 = 0 \).

Step 2: Using the Property of Inverse Matrices

Since \( B \) is the inverse of \( A \), we know that \( A \cdot B = I_3 \), where \( I_3 \) is the 3x3 identity matrix. The determinant of \( A \) can also be found using the property: \( \text{det}(A) = \frac{1}{\text{det}(B)} \).

Step 3: Setting up the Determinant Equation

We have the condition \( \text{det}(A) + 1 = 0 \). Hence, \( \text{det}(A) = -1 \). Thus, \( \frac{1}{\text{det}(B)} = -1 \), which simplifies to \( \text{det}(B) = -1 \).

Step 4: Calculating Det(B)

Calculate the determinant of matrix \( B \): \[ \text{det}(B) = 5 \times ((2) \times (-1) - (3) \times (1)) - 2\alpha \times (0 - \alpha) + 1 \times (0 \times 3 - 2 \times \alpha) \]This simplifies to:\[ \text{det}(B) = 5(-2 - 3) - 2\alpha(-\alpha) - 2\alpha \]\[ = 5(-5) + 2\alpha^2 - 2\alpha \]\[ = -25 + 2\alpha^2 - 2\alpha \].

Step 5: Solving the Equation for Alpha

Set the equation for the determinant to \(-25 + 2\alpha^2 - 2\alpha = -1\). Simplify to \(2\alpha^2 - 2\alpha - 24 = 0\). Divide the entire equation by 2 to simplify:\(\alpha^2 - \alpha - 12 = 0\). Now solve for \(\alpha\) using the quadratic formula \(\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -1\), \(c = -12\).

Step 6: Calculating the Discriminant and Roots

The discriminant is \(b^2 - 4ac = (-1)^2 - 4 \times 1 \times (-12) = 1 + 48 = 49\). So, the solutions are \(\alpha = \frac{-(-1) \pm \sqrt{49}}{2 \times 1}\). \(\alpha = \frac{1 \pm 7}{2}\). Thus, \(\alpha = 4\) and \(\alpha = -3\). Add these values to find the sum of all \(\alpha\): \(4 + (-3) = 1\).

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a scalar value that is a significant property in linear algebra. It provides insights into the matrix, such as whether it is invertible and certain geometric transformations it can represent. To compute the determinant of a 3x3 matrix, you can use the rule of Sarrus for simplicity.

• For instance, consider matrix \[B = \begin{bmatrix} 5 & 2\alpha & 1 \ 0 & 2 & 1 \ \alpha & 3 & -1 \end{bmatrix}.\]
• The determinant is computed as: \[det(B) = 5((2 \times (-1)) - (3 \times 1)) - 2\alpha(0 - \alpha) + 1(0 \times 3 - 2 \times \alpha).\]
• This simplifies to: \[det(B) = 5(-5) + 2\alpha^2 - 2\alpha.\]

A matrix is invertible if and only if its determinant is non-zero. Hence, calculating the determinant is crucial for determining matrix properties.

Inverse of a Matrix

The inverse of a matrix is a unique matrix that, when multiplied with the original matrix, results in the identity matrix. In simpler terms, it's the multiplicative inverse for matrices.

• For a matrix \( B \) that is the inverse of matrix \( A \), it follows that: \( A \cdot B = I \).
• The inverse exists only when the determinant of the matrix is not zero. If \( det(A) eq 0 \), the inverse \( A^{-1} \) can be calculated.
• For a matrix \( A \), relating its inverse to the determinant is given by: \( \text{det}(A \cdot A^{-1}) = \text{det}(I) = 1 \).

This reminds us that finding the inverse is both powerful and restricted by the absence of zero determinants.

Quadratic Equation

Quadratic equations are vital in various fields of math, representing equations of the form \( ax^2 + bx + c = 0 \). They can be solved using different methods, but one of the most common is the quadratic formula.

• For example, consider the quadratic \( \alpha^2 - \alpha - 12 = 0 \).
• This equation can be solved using the formula: \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Calculate the discriminant: \( b^2 - 4ac = (-1)^2 - 4 \times 1 \times (-12) = 49 \).
• The roots simplify to: \( \alpha = \frac{1 \pm 7}{2} \), giving roots \( \alpha = 4 \) and \( \alpha = -3 \).

Solving quadratic equations effectively finds critical values in many mathematical scenarios, including matrix operations where parameters are defined by quadratic relationships.