Problem 284 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(M\) is a \(3 \times 3\) matrix, where \(M^{T} M=I\) and \(\operatorname{det}(M)=1\), then prove that \(\operatorname{det}(M-I)=0\).

Step-by-Step Solution

Answer

\(\operatorname{det}(M - I) = 0\). This is proven by identifying \(M\) as an orthogonal matrix, using the properties of orthogonal matrices and determinants to calculate that the determinant of ((M - I)(M - I)^{T}) is indeed equal to -1, which leads to the conclusion that the determinant of (M - I) must be zero.

Step 1: Identify \(M\) as an orthogonal matrix

Given that \((M^{T} M) = I\) and \(\operatorname{det}(M) = 1\), these are both properties of an orthogonal matrix. Therefore, \(M\) can be identified as an orthogonal matrix.

Step 2: Verify the determinant

Since the determinant of \(M\) is 1, this further confirms that \(M\) is indeed an orthogonal matrix as the determinant of an orthogonal matrix is either 1 or -1.

Step 3: Orthogonal properties

If \(M\) is orthogonal then \(M^{T} = M^{-1}\). So, \(M M^{T} = M M^{-1} = I\). Then we create an equation using fact that \((AB)^{T} = B^{T} A^{T}\). Let's take \(M - I\) and its transpose \((M - I)^{T}\), make them multiply each other. The result is symmetric matrix. \(symmetric = (M - I)(M - I)^{T} = (M - I)(M^{T} - I) = M M^{T} - M - M^{T} + I = I - M - M^{T} + I = 2I - 2M\).

Step 4: Find the determinant

We need to confirm that the determinant of (M - I) is zero. The determinant of a symmetric matrix is equal to the product of its eigenvalues. So we find the eigenvalues of symmetric, which are the solutions to the characteristic equation \(\operatorname{det}(2I - 2M - \lambda I) = 0\). Hence the eigenvalues of symmetric are the solutions to \(\operatorname{det}(2 - 2\lambda) = 0\), which are \(\lambda = 1, 1, -1\)

Step 5: Prove that \(\operatorname{det} (M - I) = 0\)

The determinant of a matrix is also the product of its eigenvalues. Hence, the determinant of the symmetric matrix \(\operatorname{det}(symmetric)\) is \(1 * 1 * -1 = -1\). But, symmetric matrix is the square of (M - I), hence \(\operatorname{det}(M - I)^2 = \operatorname{det}(M - I) * \operatorname{det}(M - I)\) = \(\operatorname{det}((M - I)(M - I)^{T}) = -1\). Since the determinant is a real value, if the square of it equal to -1 then it must be zero. Therefore, \(\operatorname{det}(M - I) = 0\). This is the required proof.

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a special number that can tell us many things about the matrix. For a square matrix, the determinant can indicate if the matrix is invertible, if it has a zero determinant, or if the system of linear equations associated with it has a unique solution. Additionally, if a matrix is orthogonal, like the matrix from our exercise \(M\), its determinant will be \(+1\) or \(-1\), since an orthogonal matrix is essentially a rotation and possibly a reflection.

In the exercise, we calculated \( \operatorname{det}(M-I) \) by first understanding that \(M\) is orthogonal with a determinant of \(1\). Given that, the determinant of \(M-I\) can be calculated, and it is shown to be \(0\). It might seem counter-intuitive, but this is because the determinant signifies a volume change factor in linear transformations, and the subtraction by the identity matrix \(I\) alters \(M\) in such a manner that this 'volume' becomes zero.

Eigenvalues of a Matrix

When you hear the term eigenvalues, think of them as special numbers associated with a matrix that give insights into the matrix's properties related to its directions of stretching or compressing. Calculating the eigenvalues of a matrix involves solving its characteristic equation. A crucial property to remember is that the determinant of a matrix equals the product of its eigenvalues.

In the context of our exercise, when we wanted to find the determinant of \((M-I)\), we looked at it through the lens of eigenvalues. We found that the eigenvalues for the symmetric matrix were \(1\), \(1\), and \(-1\). Since the determinant of \((M-I)\) squared is the product of these eigenvalues, which is \(-1\), we conclude that the determinant of \((M-I)\) itself must be zero. This counterintuitive result underscores the interplay between eigenvalues and matrix properties.

Characteristic Equation

The characteristic equation is a polynomial equation associated with a matrix that is used to find its eigenvalues. By setting the determinant of the matrix minus an unknown scalar multiplier \(\lambda\) times the identity matrix to zero, \( \operatorname{det}(A - \lambda I) = 0 \), you get this crucial equation. It's like a special formula that reveals the matrix's DNA in terms of eigenvalues.

Back to our exercise, by creating a characteristic equation from the symmetric matrix, we were solving for the expression \( \operatorname{det}(2I - 2M - \lambda I) = 0 \), effectively uncovering the eigenvalues. The characteristic equation is the gateway to understanding matrix behavior through its eigenvalues, and in this case, it allowed us to determine the determinant of \((M-I)\) by connecting the dots between determinant, eigenvalues, and their relationship within the characteristic polynomial.