Problem 70 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(A\) and \(B\) be real matrices of the form \(\left[\begin{array}{ll}\alpha & 0 \\ 0 & \beta\end{array}\right]\) and \(\left[\begin{array}{ll}0 & \gamma \\\ \delta & 0\end{array}\right]\), respectively. \(\quad\) Online May 12, 2012] Statement 1: \(A B-B A\) is always an invertible matrix. Statement 2: \(A B-B A\) is never an identity matrix. (a) Statement 1 is true, Statement 2 is false. (b) Statement 1 is false, Statement 2 is true. (c) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1 . (d) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1 .

Step-by-Step Solution

Answer

(d) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.

Step 1: Determine the product AB

To find the product of matrices \(A\) and \(B\), compute \(AB\) as follows:Matrix \(A\): \[A = \begin{bmatrix} \alpha & 0 \ 0 & \beta \end{bmatrix}\]Matrix \(B\): \[B = \begin{bmatrix} 0 & \gamma \ \delta & 0 \end{bmatrix}\]The product is:\[AB = \begin{bmatrix} \alpha & 0 \ 0 & \beta \end{bmatrix} \begin{bmatrix} 0 & \gamma \ \delta & 0 \end{bmatrix} = \begin{bmatrix} \alpha \cdot 0 + 0 \cdot \delta & \alpha \cdot \gamma + 0 \cdot 0 \ 0 \cdot 0 + \beta \cdot \delta & 0 \cdot \gamma + \beta \cdot 0 \end{bmatrix} = \begin{bmatrix} 0 & \alpha \gamma \ \beta \delta & 0 \end{bmatrix}\]

Step 2: Determine the product BA

Now find the product of matrices \(B\) and \(A\), compute \(BA\) as follows:\[BA = \begin{bmatrix} 0 & \gamma \ \delta & 0 \end{bmatrix} \begin{bmatrix} \alpha & 0 \ 0 & \beta \end{bmatrix} = \begin{bmatrix} 0 \cdot \alpha + \gamma \cdot 0 & 0 \cdot 0 + \gamma \cdot \beta \ \delta \cdot \alpha + 0 \cdot 0 & \delta \cdot 0 + 0 \cdot \beta \end{bmatrix} = \begin{bmatrix} 0 & \gamma \beta \ \delta \alpha & 0 \end{bmatrix}\]

Step 3: Calculate AB - BA

Subtract the matrix \(BA\) from \(AB\) as follows:\[AB - BA = \begin{bmatrix} 0 & \alpha \gamma \ \beta \delta & 0 \end{bmatrix} - \begin{bmatrix} 0 & \gamma \beta \ \delta \alpha & 0 \end{bmatrix} = \begin{bmatrix} 0 - 0 & \alpha \gamma - \gamma \beta \ \beta \delta - \delta \alpha & 0 - 0 \end{bmatrix} = \begin{bmatrix} 0 & \gamma (\alpha - \beta) \ \delta (\beta - \alpha) & 0 \end{bmatrix}\]

Step 4: Assess invertibility of AB - BA

For the matrix to be invertible, \(|AB - BA| eq 0\). The determinant is calculated as:Given:\[AB-BA = \begin{bmatrix} 0 & \gamma (\alpha - \beta) \ \delta (\beta - \alpha) & 0 \end{bmatrix} \]The determinant is:\[|AB - BA| = (0 \cdot 0) - (\gamma (\alpha - \beta))(\delta (\beta - \alpha)) = -\gamma \delta (\alpha - \beta)^2\]If \(\gamma, \delta, \alphaeq\beta eq 0\), then the determinant is non-zero, confirming it's invertible.

Step 5: Check if AB - BA is ever an identity matrix

To be an identity matrix, \(AB - BA\) must be \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\). Matching elements to 1 and 0 is impossible because the form of \(AB - BA\) is \(\begin{bmatrix} 0 & \gamma (\alpha - \beta) \ \delta (\beta - \alpha) & 0 \end{bmatrix}\) where non-diagonal elements can never be zero given non-zero \(\gamma, \delta\) and \(\alpha eq \beta\). Thus, it can never be the identity matrix.

Step 6: Evaluate correctness of statements

Statement 1 asserts \(AB - BA\) is always invertible – this is true when \(\gamma, \delta, \alpha, \beta\) are non-zero and distinct. Statement 2 asserts that \(AB - BA\) is never the identity matrix – this is true, as shown previously. However, Statement 2’s explanation doesn’t justify why \(AB - BA\) is always invertible, since that involves determinant conditions.

Key Concepts

Matrix Multiplication

Matrix multiplication is a fundamental operation where we multiply two matrices to get a resulting matrix. To find the product of two matrices, such as matrices \(A\) and \(B\), you follow a specific method. Each element in the resulting matrix is derived from the dot product of a row in the first matrix and a column in the second matrix. This involves multiplying corresponding entries and then summing them.

Let's break this down into a simpler explanation with a 2x2 matrix example. Suppose you have two matrices:

• Matrix \(A = \left[\begin{array}{cc} \alpha & 0 \ 0 & \beta \end{array}\right]\)
• Matrix \(B = \left[\begin{array}{cc} 0 & \gamma \ \delta & 0 \end{array}\right]\)

To find \(AB\), you multiply the rows of \(A\) by the columns of \(B\). This gives you:\[AB = \begin{bmatrix} \alpha \cdot 0 + 0 \cdot \delta & \alpha \cdot \gamma + 0 \cdot 0 \ 0 \cdot 0 + \beta \cdot \delta & 0 \cdot \gamma + \beta \cdot 0 \end{bmatrix} \]The importance of matrix multiplication is that it helps in linear transformations and solving systems of equations. It also has unique properties like being non-commutative, meaning \(AB eq BA\) in most cases.

Determinant of Matrices

The determinant of a matrix is a unique value that can be calculated from its elements. It provides insight into the matrix's properties, such as whether the matrix is invertible. For a 2x2 matrix, the determinant is computed as:\[|A| = ad - bc\]where \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \).

In the context of our exercise with matrices \(AB - BA\), we calculated the determinant as:\[|AB - BA| = -\gamma \delta (\alpha - \beta)^2\]The determinant offers critical information. If it's non-zero, the matrix is invertible. This is key for Statement 1's assessment of \(AB - BA\) being invertible. If \(\gamma, \delta\), and \(\alpha eq \beta\) are non-zero, then \(|AB - BA|\) remains non-zero, confirming its invertibility.

Knowing how to calculate and interpret this value is essential in various fields, including engineering and physics, where matrix equations often appear.

Identity Matrix

An identity matrix is a special kind of square matrix where all the elements on the main diagonal are ones, and all other elements are zeros. It is denoted as \(I_n\) for an \(n \times n\) matrix. For a 2x2 matrix, the identity matrix looks like this:\[I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]

The identity matrix plays an important role as it acts like the number 1 in matrix multiplication. Any matrix multiplied by the identity matrix remains unchanged, i.e., \(AI = IA = A\).

In our exercise, the matrix \(AB - BA\) was checked to see if it could ever be an identity matrix. However, as shown in the solution, because the elements \(\gamma (\alpha - \beta)\) and \(\delta (\beta - \alpha)\) can never simultaneously form the necessary zero pattern under given conditions, it is impossible for \(AB - BA\) to be an identity matrix.

Understanding identity matrices is essential in solving systems of equations and exploring matrix properties further. They are the backbone of matrix theories and often used in proofs and transformations.