Problem 41 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(A\) be an \(n \times n\) matrix such that \(A^{n}=\alpha A\), where \(\alpha\) is a real number different from 1 and \(-1\). Then, the matrix \(A+I_{n}\) is (A) singular (B) non-singular, i.e., invertible (C) scalar matrix (D) None of these

Step-by-Step Solution

Answer

(B) Non-singular, i.e., invertible.

Step 1: Understand the Given Condition

We have a matrix \(A\) such that \(A^{n} = \alpha A\), where \(\alpha\) is a real number not equal to 1 or -1. This implies that the matrix \(A\) satisfies a polynomial equation of degree \(n\).

Step 2: Check Matrix Properties

Since \(A^{n} = \alpha A\), we can factor this equation to \(A^{n} - \alpha A = 0\). This hints that \(A(A^{n-1} - \alpha I_{n}) = 0\). Thus, \(A\) is either singular or \((A^{n-1} - \alpha I_{n})\) is singular.

Step 3: Investigate Further

Consider if \(A = 0\): then \(0^n = \alpha \times 0\), which is true, making \(A\) satisfy the condition trivially, leading \(A+I_{n} = I_{n}\), a non-singular matrix. Otherwise, \(A\) has some non-zero eigenvalues \(\lambda\) since \(A^{n} = \alpha A\) implies \(\lambda^{n} = \alpha \lambda\).

Step 4: Analyze Eigenvalue Relations

From \(\lambda^{n} = \alpha \lambda\), we obtain \(\lambda(\lambda^{n-1} - \alpha) = 0\). Therefore, \(\lambda = 0\) or \(\lambda^{n-1} = \alpha\). Since \(\lambda eq 0\), \(\lambda^{n-1} = \alpha\) must hold. Hence, non-zero eigenvalues of \(A\) satisfy this equation.

Step 5: Consider \(A + I_{n}\)

The structure of eigenvalues suggests \(\lambda_i = 0\) or \(\lambda_i^{n-1} = \alpha\). If \(\lambda_i = -1\) exists, \(\lambda_i + 1 = 0\), making \(A + I_{n}\) potentially singular. However, since \(\alpha eq -1\) ensures \(\lambda_i eq -1\), there are no eigenvalues \(-1\). Hence, \(A + I_{n}\) is non-singular.

Key Concepts

Eigenvalues

Eigenvalues are numbers associated with matrices, indicating important properties about their structure and behavior. To compute them, solve the equation \(Ax = \lambda x\), where \(A\) is the matrix and \(\lambda\) represents the eigenvalue. Here, \(x\) is a non-zero vector often called an eigenvector.

• In our problem, the matrix \(A\) is such that \(A^n = \alpha A\), where \(\alpha\) is a non-zero, non-one real number. This brings up a crucial relationship for eigenvalues, i.e., \(\lambda^n = \alpha \lambda\).
• This equation simplifies to \(\lambda(\lambda^{n-1} - \alpha) = 0\), indicating that \(\lambda = 0\) or \(\lambda^{n-1} = \alpha\).
• Understanding this helps us see which values make \(A\) singular or non-singular, aiding in answering the exercise question about the nature of \(A + I_n\).

Singular Matrix

A singular matrix is a square matrix that does not have an inverse. This happens if its determinant is zero.

• For a matrix \(A\), being singular indicates that its eigenvalues contain zero. If \(\lambda = 0\) for \(A\) as one of its eigenvalues, the determinant of \(A\) could be zero.
• In the problem at hand, we have found that if \(A\) has a non-zero eigenvalue, \(\lambda^{n-1}\) equals \(\alpha\). Since \(\alpha\) is not \(-1\), \(\lambda\) cannot be \(-1\), ensuring \(A + I_n\) stays non-singular.

Recognizing whether a matrix is singular or not is key to understanding its properties, especially in determining if it's invertible.

Scalar Matrix

A scalar matrix is a special type of square matrix where every diagonal entry is the same number, and all off-diagonal entries are zero. This can be expressed as \(cI_n\), where \(c\) is a scalar, and \(I_n\) is the identity matrix.

• In the context of our problem, the condition involves exploring if \(A + I_n\) is a scalar matrix. However, as seen from the original solution, \(A + I_n\) is not constrained to any specific scalar form.
• The analysis of eigenvalues showed that the potential eigenvalue \(\lambda = -1\) is non-existent due to \(\alpha eq -1\), ensuring \(A + I_n\) does not simplify to this special case.
• Ultimately, recognizing scalar matrices helps in classes of matrices where behavior is uniform and predictable, although it's not directly applicable for \(A + I_n\) in this specific exercise.