Question

If $A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 5 & 4 \\ 1 & 0 & 5 \end{bmatrix}$, then the determinant of $\left(A^{2026} - 11A^{2025} - 9A^{2023}\right)$ is equal to:}

• A. \(9^{2026}\)
• B. \((-31)^3 3^{2025}\)
• C. \((-31)^3 3^{4048}\)
• D. \((31)^4 3^{4048}\)

Hint:

For any characteristic equation of a $3 \times 3$ matrix given by $\lambda^3 - S_1\lambda^2 + S_2\lambda - |A| = 0$, $S_1$ represents the trace ($\text{sum of diagonal entries}$) and $|A|$ represents the determinant. Identifying these invariants directly from the matrix saves valuable time during matrix polynomial reduction.

Answer 1

The Correct Option is C

Concept: According to the Cayley-Hamilton Theorem, every square matrix satisfies its own characteristic equation. If the characteristic polynomial of an $n \times n$ matrix $A$ is given by $P(\lambda) = \det(A - \lambda I) = 0$, then substituting $\lambda$ with the matrix $A$ yields the matrix equation $P(A) = O$, where $O$ is the zero matrix. Additionally, for any scalar $k$ and a square matrix $M$ of order $n \times n$, the determinant satisfies the scaling property: \[ \det(kM) = k^n \det(M) \] And for powers of a matrix, the determinant follows: \[ \det(M^p) = [\det(M)]^p \]

Step 1: Determine the characteristic equation of the matrix $A$.
The given matrix $A$ of order $3 \times 3$ is: \[ A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 5 & 4 \\ 1 & 0 & 5 \end{bmatrix} \] The characteristic equation is found by computing the determinant $\det(A - \lambda I) = 0$: \[ \begin{vmatrix} 1-\lambda & 1 & 2 2 & 5-\lambda & 4 1 & 0 & 5-\lambda \end{vmatrix} = 0 \] Expanding the determinant along the third row: \[ 1 \cdot \begin{vmatrix} 1 & 2 \\ 5-\lambda & 4 \end{vmatrix} - 0 \cdot \begin{vmatrix} 1-\lambda & 2 \\ 2 & 4 \end{vmatrix} + (5-\lambda) \cdot \begin{vmatrix} 1-\lambda & 1 \\ 2 & 5-\lambda \end{vmatrix} = 0 \] Evaluating the individual $2 \times 2$ determinants: \[ 1 \cdot [4 - 2(5-\lambda)] + (5-\lambda) \cdot [(1-\lambda)(5-\lambda) - 2] = 0 \] Simplifying the terms inside the brackets: \[ [4 - 10 + 2\lambda] + (5-\lambda) \cdot [\lambda^2 - 6\lambda + 5 - 2] = 0 \] \[ (2\lambda - 6) + (5-\lambda)(\lambda^2 - 6\lambda + 3) = 0 \] Expanding the second component completely: \[ 2\lambda - 6 + 5\lambda^2 - 30\lambda + 15 - \lambda^3 + 6\lambda^2 - 3\lambda = 0 \] Combining like terms to form the polynomial: \[ -\lambda^3 + 11\lambda^2 - 31\lambda + 9 = 0 \] Multiplying through by $-1$ gives the final characteristic equation: \[ \lambda^3 - 11\lambda^2 + 31\lambda - 9 = 0 \]

Step 2: Apply the Cayley-Hamilton Theorem.
Replacing the scalar variable $\lambda$ with the matrix $A$ and the constant term with the identity matrix $I$: \[ A^3 - 11A^2 + 31A - 9I = O \] Rearranging this relationship, we can isolate the highest-order terms: \[ A^3 - 11A^2 - 9I = -31A \quad \cdots (1) \]

Step 3: Simplify the expression inside the determinant.
We need to find the determinant of the matrix polynomial: \[ X = A^{2026} - 11A^{2025} - 9A^{2023} \] Factoring out the lowest power of $A$, which is $A^{2023}$, from the expression: \[ X = A^{2023} \left( A^3 - 11A^2 - 9I \right) \] Now, substituting the identity established in equation (1) into this expression: \[ X = A^{2023} (-31A) = -31 A^{2024} \]

Step 4: Compute the determinant of $X$.
Taking the determinant of both sides: \[ \det(X) = \det(-31 A^{2024}) \] Since $A$ is a $3 \times 3$ matrix, any scalar multiple pulled out of the determinant is raised to the power of $3$: \[ \det(X) = (-31)^3 \det(A^{2024}) = (-31)^3 [\det(A)]^{2024} \] From the characteristic equation $\lambda^3 - 11\lambda^2 + 31\lambda - 9 = 0$, the product of the eigenvalues (which equals the determinant of $A$) is given by the constant term: \[ \det(A) = 9 = 3^2 \] Substituting $\det(A) = 3^2$ back into our determinant expression: \[ \det(X) = (-31)^3 \cdot (3^2)^{2024} = (-31)^3 \cdot 3^{4048} \] This matches Option (C).