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

We are given a 3x3 matrix \( A \) with \( \det(A) = -4 \). We are also given the matrix \( A + I \) and an equation involving the determinant of a related matrix. We need to find the value of \( m + n \).

Concept Used:

This problem uses several properties of determinants and adjugate matrices for a square matrix \( M \) of order \( p \):

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

Step-by-Step Solution:

Step 1: Find the value of the constant \( a \).

We are given the matrix \( A + I \):

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

We can find the matrix \( A \) by subtracting the identity matrix \( I \):

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

Now, we compute the determinant of \( A \) and set it equal to the given value, -4.

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

Expanding along the second row (which contains two zeros) simplifies the calculation:

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

We are given that \( \det(A) = -4 \), so:

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

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

Let's denote the expression inside the determinant as \( M \):

\[ M = (a + 1) \text{adj}((a - 1)A) \]

This is a scalar \( (a+1) \) multiplied by a matrix. Using the property \( \det(kX) = k^p \det(X) \) with \( p=3 \):

\[ \det(M) = (a + 1)^3 \det(\text{adj}((a - 1)A)) \]

Next, we use the property \( \det(\text{adj}(X)) = (\det(X))^{p-1} \) with \( p=3 \):

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

Now, we use the property \( \det(kX) = k^p \det(X) \) again:

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

Combining these results:

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

Step 3: Substitute the known values into the simplified expression.

We found \( a = 3 \) and we are given \( \det(A) = -4 \).

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

Final Computation & Result:

We are given that the value of the determinant is \( 2^m 3^n \).

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

By comparing the powers of the prime factors, we get:

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

The problem asks for the value of \( m + n \).

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

The value of \( m+n \) is 16.