Question

If M is a square matirx of order 3 satisfying $M^T M=4I$ and $|M|=8$ then the value of $|M-2I|$ will be

• A. 2
• B. 4
• C. 16
• D. 0

Answer 1

Answer: (D)

0

Answer 2

Given a square matrix M of order 3 such that $M^T M=4I$ and $|M|=8$. We need to find the value of $|M-2I|$.

Let $\lambda$ be an eigenvalue of M with corresponding eigenvector $x$. Then $Mx = \lambda x$.

Substitute this into the given equation $M^T M = 4I$: $M^T (Mx) = 4Ix$ $M^T (\lambda x) = 4x$ $\lambda M^T x = 4x$

Since $x$ is a non-zero eigenvector, and from $|M|=8$ we know M is invertible (so $\lambda \neq 0$): $M^T x = \frac{4}{\lambda} x$

This implies that if $\lambda$ is an eigenvalue of M, then $4/\lambda$ is an eigenvalue of $M^T$. Since a matrix and its transpose have the same eigenvalues, $4/\lambda$ must also be an eigenvalue of M.

Let the eigenvalues of M be $\lambda_1, \lambda_2, \lambda_3$. The set of eigenvalues $\{\lambda_1, \lambda_2, \lambda_3\}$ must be the same as $\{4/\lambda_1, 4/\lambda_2, 4/\lambda_3\}$.

We are given $|M|=8$, which means $\lambda_1 \lambda_2 \lambda_3 = 8$.

Now, let's consider the implications of the eigenvalue relationship:

1. If an eigenvalue $\lambda_i$ satisfies $\lambda_i = 4/\lambda_i$: Then $\lambda_i^2 = 4 \implies \lambda_i = \pm 2$. If all eigenvalues are of this form, then the possible sets of eigenvalues for M (whose product is 8) are:

• $\{2, 2, 2\}$ (product $2 \times 2 \times 2 = 8$)

• $\{2, -2, -2\}$ (product $2 \times (-2) \times (-2) = 8$)

In both these valid cases, 2 is an eigenvalue of M.

2. If eigenvalues are permuted (e.g., $\lambda_1 = 4/\lambda_2$, $\lambda_2 = 4/\lambda_1$, and $\lambda_3 = 4/\lambda_3$): From $\lambda_3 = 4/\lambda_3$, we get $\lambda_3 = \pm 2$. The product of eigenvalues is $\lambda_1 \lambda_2 \lambda_3 = (4/\lambda_2) \lambda_2 \lambda_3 = 4 \lambda_3$. Since the product is 8, $4 \lambda_3 = 8 \implies \lambda_3 = 2$. Thus, in this case, 2 is also an eigenvalue of M.

In all possible scenarios, 2 must be an eigenvalue of M. Let $\lambda_1 = 2$.

We need to find $|M-2I|$. The eigenvalues of the matrix $M-2I$ are $(\lambda_1-2), (\lambda_2-2), (\lambda_3-2)$. The determinant $|M-2I|$ is the product of these eigenvalues: $|M-2I| = (\lambda_1-2)(\lambda_2-2)(\lambda_3-2)$.

Since $\lambda_1 = 2$, one of the terms in the product is $(2-2)=0$. Therefore, $|M-2I| = (2-2)(\lambda_2-2)(\lambda_3-2) = 0 \times (\lambda_2-2)(\lambda_3-2) = 0$.

The value of $|M-2I|$ is 0.