GATE 2023 | Maths | Linear Alegbra

Question

Let \( M = \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & 4 \\ 0 & 0 & 1 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \). If \( 6M^{-1} = M^2 - 6M + \alpha I \) for some \( \alpha \in \mathbb{R} \), then the value of \( \alpha \) is equal to \(\underline{\hspace{1cm}}\).

Answer 1

Find the characteristic equation:

The characteristic equation of a matrix \(M\) is given by \(\text{det}(M - \lambda I) = 0\), where \(\text{det}\) denotes the determinant and \(\lambda\) represents the eigenvalues.

For the given matrix \(M\), we have:
\[ \text{det}\left(\begin{bmatrix} 3-\lambda & -1 & -2 \\ 0 & 2-\lambda & 4 \\ 0 & 0 & 1-\lambda \end{bmatrix}\right) = 0 \]

Simplifying the determinant, we get:
\[ (3-\lambda)((2-\lambda)(1-\lambda)) = 0 \]
\[ \lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0 \]

Rearrange the given equation:

\[ 6M^{-1} = M^2 - 6M + \alpha I \]

Multiply both sides by \(M\):
\[ 6I = M^3 - 6M^2 + \alpha M \]

Subtract \(6I\) from both sides:
\[ 0 = M^3 - 6M^2 + \alpha M - 6I \]

Substitute the characteristic equation:

The characteristic equation tells us that \(\lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0\) for the eigenvalues of \(M\). Therefore, we can substitute \(M\) into this equation:
\[ 0 = M^3 - 6M^2 + 11M - 6I \]

Compare coefficients:

Comparing the coefficients of both sides, we get:
\[ \alpha = 11 \]

Therefore, the value of \(\alpha\) is 11, as confirmed using the characteristic equation of the matrix.