Problem 62 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\mathrm{A}\) be a \(3 \times 3\) matrix such that \(\mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}=0\). Statement \(-\mathrm{I}: \mathrm{A}^{-1}=\frac{1}{7}(5 \mathrm{I}-\mathrm{A})\) Statement II : the polynomial \(\mathrm{A}^{3}-2 \mathrm{~A}^{2}-3 \mathrm{~A}+\alpha\) can be reduced to \(5(\mathrm{~A}-4 \mathrm{I})\). [Online April 10, 2016] Then: (a) Both the statements are true. (b) Both the statements are false. (c) Statement-I is true, but Statement-II is false. (d) Statement I is false, but Statement-II is true.

Step-by-Step Solution

Answer

Both statements are true: option (a).

Step 1: Analyze the given matrix equation

The matrix equation given is \( \mathrm{A}^{2} - 5\mathrm{~A} + 7\mathrm{I} = 0 \). This is a quadratic equation in terms of the matrix \( \mathrm{A} \).

Step 2: Use the equation to find A^{-1}

Rearrange the matrix equation to express \( \mathrm{A}^{2} \) in terms of \( \mathrm{A} \) and \( \mathrm{I} \): \[ \mathrm{A}^{2} = 5\mathrm{~A} - 7\mathrm{I} \] Multiply both sides by \( \mathrm{A}^{-1} \) (provided it exists) to find \( \mathrm{A}^{-1} \): \[ \mathrm{A}\mathrm{A}^{-1} = (5\mathrm{~I} - 7\mathrm{A}^{-1})\mathrm{A}^{-1} \] This simplifies to: \[ \mathrm{I} = 5\mathrm{~I}\mathrm{A}^{-1} - 7 \mathrm{A}^{-1} \] Rearrange to get \( \mathrm{A}^{-1} = \frac{1}{7}(5\mathrm{~I} - \mathrm{A}) \), confirming Statement I is true.

Step 3: Evaluate the polynomial for Statement II

We need to check if the polynomial \( \mathrm{A}^{3} - 2\mathrm{~A}^{2} - 3\mathrm{~A} + \alpha \) is equal to \( 5(\mathrm{~A} - 4 \mathrm{I}) \). First, use \( \mathrm{A}^{2} = 5\mathrm{~A} - 7\mathrm{I} \) to compute \( \mathrm{A}^{3} \): \[ \mathrm{A}^{3} = \mathrm{A}(\mathrm{A}^{2}) = \mathrm{A}(5\mathrm{~A} - 7\mathrm{I}) = 5\mathrm{A}^{2} - 7\mathrm{A} \] Substitute \( \mathrm{A}^{2} = 5\mathrm{~A} - 7\mathrm{I} \) into \( \mathrm{A}^{3} \): \[ \mathrm{A}^{3} = 5(5\mathrm{~A} - 7\mathrm{I}) - 7\mathrm{A} = 25\mathrm{~A} - 35\mathrm{I} - 7\mathrm{A} = 18\mathrm{~A} - 35\mathrm{I} \] Plug \( \mathrm{A}^{3} \), \( \mathrm{A}^{2} \), and \( \mathrm{A} \) back into the polynomial: \[ 18\mathrm{~A} - 35\mathrm{I} - 2(5\mathrm{~A} - 7\mathrm{I}) - 3\mathrm{~A} + \alpha = 5(\mathrm{~A} - 4 \mathrm{I}) \] Simplify the equation to find \( \alpha \).

Step 4: Simplify and Check Statement II

Simplifying: \[ 18\mathrm{~A} - 35\mathrm{I} - 10\mathrm{~A} + 14\mathrm{I} - 3\mathrm{~A} + \alpha = 5\mathrm{~A} - 20\mathrm{I} \] \[ (18 - 10 - 3)\mathrm{~A} + (-35 + 14)\mathrm{I} + \alpha = 5\mathrm{~A} - 20\mathrm{I} \] \[ 5\mathrm{~A} - 21\mathrm{I} + \alpha = 5\mathrm{~A} - 20\mathrm{I} \] This requires \( \alpha = 1 \), making Statement II true as the two sides match for \( \alpha = 1 \).

Step 5: Conclusion: Evaluate truth of statements

Both statements are true: Statement I is confirmed true by algebraic manipulation of \( \mathrm{A} \) and its inverse, and Statement II is true as the polynomial simplifies to \( 5(\mathrm{~A} - 4\mathrm{I}) \) for \( \alpha = 1 \).

Key Concepts

Matrix Inverses

When dealing with matrices, the concept of a matrix inverse is similar to finding the reciprocal of a number. If a matrix \( \mathrm{A} \) is invertible, there exists another matrix, called \( \mathrm{A}^{-1} \), such that when multiplied together, they yield the identity matrix \( \mathrm{I} \). The identity matrix acts as the "1" in matrix algebra, meaning \( \mathrm{A} \cdot \mathrm{A}^{-1} = \mathrm{I} \). Not all matrices have inverses, only square matrices (matrices with the same number of rows and columns) might be invertible if they satisfy certain conditions, such as having a non-zero determinant.

The problem provided gives a matrix equation \( \mathrm{A}^{2} - 5\mathrm{~A} + 7\mathrm{I} = 0 \). From this, rearrangement allows us to derive \( \mathrm{A} \) in terms of \( \mathrm{I} \), leading to the inverse \( \mathrm{A}^{-1} = \frac{1}{7}(5\mathrm{~I} - \mathrm{A}) \). Verifying matrix inverses involves multiplying \( \mathrm{A} \) and \( \mathrm{A}^{-1} \) to check if they result in the identity matrix \( \mathrm{I} \), thus showing that the derived inverse expression is indeed valid.

Quadratic Matrix Equations

A quadratic matrix equation involves terms where a matrix is squared, similar to quadratic equations in algebra, but extended to two-dimensional arrays. Given in the problem is \( \mathrm{A}^{2} - 5 \mathrm{~A} + 7 \mathrm{I} = 0 \), where \( \mathrm{A}^{2} \) indicates \( \mathrm{A} \times \mathrm{A} \). This type of equation can often be solved or manipulated using matrix algebra techniques.

Quadratic matrix equations like the one given help solve for unknown matrices by enabling algebraic manipulations similar to those used with polynomial equations. In this case, rearranging for \( \mathrm{A}^{2} \) with respect to \( \mathrm{A} \) and \( \mathrm{I} \) allows us to approach inverse matrix computation and polynomial expressions effectively, as evidenced by further applications, such as in polynomial reduction.

Polynomial Reduction

Polynomial reduction involves expressing a polynomial, specifically in matrices, in simpler terms or an equivalent form. Polynomial expressions can sometimes be reduced using identities or established relationships between matrices. In the exercise, a polynomial expression \( \mathrm{A}^{3} - 2\mathrm{~A}^{2} - 3\mathrm{~A} + \alpha \) had to be checked against \( 5(\mathrm{~A} - 4 \mathrm{I}) \).

Using the relation \( \mathrm{A}^{2} = 5\mathrm{~A} - 7\mathrm{I} \), we compute \( \mathrm{A}^{3} \) and then proceed to substitute back into the polynomial. The simplification process evaluates if the polynomial can indeed be expressed in line with another given matrix expression through manipulations and substitutions. Here, polynomial reduction helps validate expressions and whether constants such as \( \alpha \) equate expressions realistically, like finding \( \alpha = 1 \) for statement consistency.