Problem 7 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(M\) is a \(3 \times 3\) matrix, where \(M^{\prime} M=I\) and \(\operatorname{det}(M)=1\), then \(\operatorname{det}(M-I)=\) (A) 1 (B) 0 (C) \(-1\) (D) None of these

Step-by-Step Solution

Answer

The determinant \(\text{det}(M-I)\) is 0. (Option B)

Step 1: Understanding the Properties of Matrix M

We are given a matrix \(M\) such that \(M'M = I\). This indicates that \(M\) is an orthogonal matrix, since the transpose of \(M\) multiplied by \(M\) results in the identity matrix \(I\). Furthermore, we are told that \(\text{det}(M) = 1\). For orthogonal matrices, the determinant can be either \(1\) or \(-1\), but here it is specifically \(1\).

Step 2: Calculating the Determinant \(\text{det}(M-I)\)

To find the determinant of \(M-I\), consider the properties of orthogonal matrices. Since \(M\) is orthogonal with determinant \(1\), it behaves like a rotation matrix. For a \(3 \times 3\) orthogonal matrix with determinant \(1\), the eigenvalues are typically \(1, \cos \theta + i \sin \theta, \cos \theta - i \sin \theta\). At least one of these eigenvalues is equal to \(1\).

Step 3: Using Eigenvalues to Determine \(\text{det}(M-I)\)

Now, since one of the eigenvalues of \(M\) is \(1\), the eigenvalues of \(M - I\) will be \(1 - 1 = 0\), and the others will be \(\cos \theta + i \sin \theta - 1\) and \(\cos \theta - i \sin \theta - 1\). Because one eigenvalue of \(M - I\) is \(0\), this directly implies that the determinant of \(M-I\) is the product of the eigenvalues, which includes \(0\).

Step 4: Conclusion

Since one of the eigenvalues of \(M-I\) is \(0\), the determinant \(\text{det}(M-I)\) must be \(0\). None of the other potential non-zero eigenvalues can affect this, as their product with \(0\) results in \(0\).

Key Concepts

Matrix Determinant

When we talk about the determinant of a matrix, like our matrix \(M\), we're looking at a particular value that gives us insights into certain properties of the matrix.

• Interpretation: For a \(3 \times 3\) matrix, the determinant can indicate whether the matrix is invertible. If \(\operatorname{det}(M) eq 0\), the matrix is invertible. For orthogonal matrices such as \(M\), they're always invertible because their determinant is either 1 or -1.
• Geometric Meaning: In the context of transformations, the determinant tells us about volume scaling. Specifically, a determinant of 1, like in our case, means the transformation preserves volume.
• Calculation: Determining \(\operatorname{det}(M)\) involves specific formulas when using numerical entries of the matrix or leveraging properties like orthogonality, which are used extensively in our example given \(M'M = I\) and \(\operatorname{det}(M)=1\).

Eigenvalues

Eigenvalues play a crucial role in understanding the properties of matrices, including our orthogonal matrix \(M\). They are scalars that give us insights about the transformation effects of \(M\).

• Definition: Eigenvalues are values \(\lambda\) such that there exists a non-zero vector \(v\) where \(Mv = \lambda v\). Here, \(\lambda\) describes how the transformation scales the vector \(v\).
• Orthogonal Matrices: For orthogonal matrices with a determinant of 1, like \(M\), one eigenvalue is always \(1\), ensuring it preserves vector lengths. The other eigenvalues describe possible rotations, typically expressed in the form \(\cos \theta + i \sin \theta\).
• Determinant and Eigenvalues: The product of the eigenvalues of a matrix equals its determinant. Therefore, for \(M-I\), one of the eigenvalues being 0 immediately indicates that \(\operatorname{det}(M-I) = 0\).

Rotation Matrix

A rotation matrix is a special kind of orthogonal matrix predominantly used in transformations that involve rotations in space.

• Structure: For a \(3 \times 3\) rotation matrix, entries are often dependent on the angle of rotation. This structure ensures that the matrix can rotate a point around an axis without changing the point's distance from the origin.
• Properties: They have a determinant of 1, and their eigenvalues typically include 1, \(\cos \theta + i \sin \theta\), and \(\cos \theta - i \sin \theta\). These eigenvalues capture the essence of the rotation, especially for axes not aligned with the axis of rotation.
• Application in \(M\): In our matrix \(M\), the rotation matrix behavior contributes to understanding \(\operatorname{det}(M-I)\). Since \(M\) is orthogonal and behaves like a rotation matrix, at least one eigenvalue of \(M\) is 1, leading to the determination that \(\operatorname{det}(M-I) = 0\).