Let $a \in \mathbf{R}$ 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 $\det ((a+1)\text{adj}((a-1)A))$ is $2^m 3^n$, $m, n \in \{0, 1, 2, \dots, 20\}$, then $m+n$ is equal to :
16
Finding the Value of 'a' from Matrix Equation
We are given the matrix equation $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$, where $A$ is a $3 \times 3$ matrix and $I$ is the identity matrix.
First, find the matrix $A$:
$A = (A + I) - I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{bmatrix}$
We are given that $\det(A) = -4$. Calculate the determinant of the derived matrix $A$:
$\det(A) = 0 \begin{vmatrix} 0 & 0 \\ 1 & 1 \end{vmatrix} - a \begin{vmatrix} 2 & 0 \\ a & 1 \end{vmatrix} + 1 \begin{vmatrix} 2 & 0 \\ a & 1 \end{vmatrix}$
$\det(A) = 0 - a(2 \cdot 1 - 0 \cdot a) + 1(2 \cdot 1 - 0 \cdot a)$
$\det(A) = -a(2) + 1(2) = -2a + 2$
Equating this to the given determinant:
$-2a + 2 = -4$
$ -2a = -6$
$a = 3$
Calculating the Determinant of the Adjugate Matrix
Now, substitute $a=3$ into the expression $\det((a+1)\text{adj}((a-1)A))$.
The expression becomes $\det((3+1)\text{adj}((3-1)A)) = \det(4 \cdot \text{adj}(2A))$.
We need to use the properties of determinants and adjugate matrices:
- For an $n \times n$ matrix $M$, $\det(kM) = k^n \det(M)$.
- For an $n \times n$ matrix $M$, $\det(\text{adj}(M)) = (\det M)^{n-1}$.
Here, $A$ is a $3 \times 3$ matrix ($n=3$).
First, calculate $\det(2A)$:
$\det(2A) = 2^3 \det(A) = 8 \times (-4) = -32$.
Next, calculate $\det(\text{adj}(2A))$:
$\det(\text{adj}(2A)) = (\det(2A))^{3-1} = (-32)^2 = 1024$.
Finally, calculate $\det(4 \cdot \text{adj}(2A))$:
Let $M = \text{adj}(2A)$. Since $M$ is a $3 \times 3$ matrix, we have:
$\det(4 \cdot M) = 4^3 \det(M) = 4^3 \times \det(\text{adj}(2A))$
$\det(4 \cdot \text{adj}(2A)) = 64 \times 1024$
Determining 'm' and 'n' and Finding m+n
Express the result in powers of 2:
$64 = 2^6$
$1024 = 2^{10}$
So, $64 \times 1024 = 2^6 \times 2^{10} = 2^{16}$.
We are given that the determinant equals $2^m 3^n$. Therefore:
$2^{16} = 2^m 3^n$
Comparing the powers, we get:
$m = 16$
$n = 0$
Both $m=16$ and $n=0$ are within the allowed set $\{0, 1, 2, \dots, 20\}$.
The question asks for the value of $m+n$:
$m+n = 16 + 0 = 16$.
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