Question

Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).

If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to:

• A. 14
• B. 31
• C. 13
• D. 16

Hint:

The determinant of a matrix multiplied by its adjugate is the square of the determinant of the matrix.

Answer 1

The Correct Option is D

Given a 3x3 matrix \( A \) with \( \det(A) = -4 \), and the matrix \( A + I \), we need to determine the value of \( m + n \) based on an equation involving the determinant of a related matrix, where the determinant is expressed as \( 2^m 3^n \).

Concepts Utilized:

The solution relies on the following determinant and adjugate matrix properties for a square matrix \( M \) of order \( p \):

1. \( \det(kM) = k^p \det(M) \) for a scalar \( k \).
2. \( \text{adj}(kM) = k^{p-1} \text{adj}(M) \) for a scalar \( k \).
3. \( \det(\text{adj}(M)) = (\det(M))^{p-1} \).

Solution Breakdown:

Step 1: Determine the value of \( a \).

The matrix \( A + I \) is provided as:

\[A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}\]

Matrix \( A \) is found by subtracting the identity matrix \( I \):

\[A = (A + I) - I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{bmatrix}\]

The determinant of \( A \) is calculated and equated to -4:

\[\det(A) = \begin{vmatrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{vmatrix}\]

Expanding along the second row yields:

\[\det(A) = -2 \begin{vmatrix} a & 1 \\ 1 & 1 \end{vmatrix} = -2(a \cdot 1 - 1 \cdot 1) = -2(a - 1)\]

Given \( \det(A) = -4 \):

\[-2(a - 1) = -4 \implies a - 1 = 2 \implies a = 3\]

Step 2: Simplify \( \det((a + 1) \text{adj}((a - 1)A)) \).

Let \( M = (a + 1) \text{adj}((a - 1)A) \). Applying determinant properties:

\[\det(M) = (a + 1)^3 \det(\text{adj}((a - 1)A))\quad (\text{using } \det(kX) = k^p \det(X) \text{ with } p=3)\]

Using \( \det(\text{adj}(X)) = (\det(X))^{p-1} \) with \( p=3 \):

\[\det(\text{adj}((a - 1)A)) = (\det((a - 1)A))^2\]

Applying \( \det(kX) = k^p \det(X) \) again:

\[\det((a - 1)A) = (a - 1)^3 \det(A)\]

Combining these gives:

\[\det(M) = (a + 1)^3 \left( (a - 1)^3 \det(A) \right)^2 = (a + 1)^3 (a - 1)^6 (\det(A))^2\]

Step 3: Substitute known values.

With \( a = 3 \) and \( \det(A) = -4 \):

\[\det(M) = (3 + 1)^3 (3 - 1)^6 (-4)^2\]\[= (4)^3 (2)^6 (-4)^2 = (2^2)^3 (2^6) (-(2^2))^2 = (2^6)(2^6)(2^4)\]\[= 2^{6+6+4} = 2^{16}\]

Final Calculation & Result:

The determinant is given as \( 2^m 3^n \).

\[2^{16} = 2^m 3^n\]

Comparing powers of prime factors:

\[m = 16 \quad \text{and} \quad n = 0\]

The value of \( m + n \) is calculated as:

\[m + n = 16 + 0 = 16\]

Therefore, \( m+n = \) 16.