Question

If a matrix A satisfies the equation \( A^3 - 6A^2 + 11A - 6I = 0 \), then \( A^{-1} \) can be

• A. \( \frac{1}{6}I \)
• B. \( 4I \)
• C. \( 3I \)
• D. \( \frac{1}{6}(A^2 - 6A + 11I) \) (This seems to be the form implied by option 4 checkmark, but option in image is just \(\frac{1}{6}\) without matrix terms. Let's assume option (d) is the full expression needed.) The options in the image are: \(1.* \frac{1}{6}I\), \(2.* 4I\), \(3.* 3I\), \(4.\checkmark \frac{1}{6}\). The checkmark is on "\(\frac{1}{6}\)". This likely implies the determinant is related or \(A^{-1} = \frac{1}{6} \times (\text{something})\). The typical way to find \(A^{-1}\) from characteristic polynomial is to multiply by \(A^{-1}\). If the option marked correct is just a scalar \(1/6\), the question is ill-posed as \(A^{-1}\) is a matrix. Assuming option (d) meant the expression.

Hint:

By the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation.
If a matrix A satisfies \(p(A)=0\), where \(p(\lambda)\) is its characteristic polynomial, and if the constant term of \(p(\lambda)\) is non-zero, then A is invertible.
To find \(A^{-1}\) from \(a_n A^n + \dots + a_1 A + a_0 I = 0\), (where \(a_0 \neq 0\)), rearrange to \(a_0 I = -(a_n A^n + \dots + a_1 A)\), then multiply by \(A^{-1}/a_0\).
Or, as done above, multiply the whole equation by \(A^{-1}\) and isolate \(A^{-1}\).

Answer 1

The Correct Option is D

Given the matrix equation \( A^3 - 6A^2 + 11A - 6I = 0 \). This is the characteristic equation (Cayley-Hamilton theorem states every matrix satisfies its own characteristic equation). To find \(A^{-1}\), we can multiply the entire equation by \(A^{-1}\), assuming A is non-singular (i.e., \(A^{-1}\) exists). If A were singular, \(\det(A)=0\), which means \(\lambda=0\) would be an eigenvalue. If \(\lambda=0\) is an eigenvalue, then substituting \(\lambda=0\) into the characteristic polynomial \(p(\lambda) = \lambda^3 - 6\lambda^2 + 11\lambda - 6\) should give 0. \(p(0) = 0 - 0 + 0 - 6 = -6\). Since \(p(0) = -6 \neq 0\), \(\lambda=0\) is not an eigenvalue, so \(\det(A) \neq 0\), and \(A^{-1}\) exists. Multiply the equation by \(A^{-1}\): \(A^{-1}(A^3 - 6A^2 + 11A - 6I) = A^{-1}(0)\) \(A^{-1}A^3 - 6A^{-1}A^2 + 11A^{-1}A - 6A^{-1}I = 0\) \(A^2 - 6A + 11I - 6A^{-1} = 0\) Now, solve for \(A^{-1}\): \(6A^{-1} = A^2 - 6A + 11I\) \[ A^{-1} = \frac{1}{6}(A^2 - 6A + 11I) \] This matches the form commonly expected for such problems. If option (d) in a full version was this expression, it would be correct. The option shown in the image is just "\(\frac{1}{6}\)" which is incorrect as \(A^{-1}\) is a matrix. Assuming the option was intended to be \(A^{-1} = \frac{1}{6}(A^2 - 6A + 11I)\). \[ \boxed{\frac{1}{6}(A^2 - 6A + 11I)} \]