Let $A = \(\begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix}\) If $\det (\text{adj (adj (3A))})= 2^m \cdot 3^n$, $m, n \in \mathbb{N}$, then $m + n$ is equal to
Matrix Simplification and Determinant Calculation
First, simplify the given matrix $A$ using row operations to find its determinant.
Given matrix:
$ A = \begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix} $Apply row operations:
- $R_2 \leftarrow R_2 - 2R_1$
- $R_3 \leftarrow R_3 - 3R_1$
The matrix transforms into:
$ \begin{bmatrix} 2 & 2+p & 2+p+q \\ 0 & 2 & 4+p \\ 0 & 6 & 14+3p \end{bmatrix} $Now, calculate the determinant of $A$ using the cofactor expansion along the first column:
$ \det(A) = 2 \times \begin{vmatrix} 2 & 4+p \\ 6 & 14+3p \end{vmatrix} $ $ \det(A) = 2 \times [2(14+3p) - 6(4+p)] $ $ \det(A) = 2 \times [28 + 6p - 24 - 6p] $ $ \det(A) = 2 \times [4] = 8 $Properties of Adjoint and Determinant
We need to evaluate $\det(\text{adj}(\text{adj}(3A)))$. Let the matrix size be $n=3$. We use the following properties for an $n \times n$ matrix $X$:
- $\det(kA) = k^n \det(A)$
- $\text{adj}(\text{adj}(X)) = (\det(X))^{n-2} X$ (for $n \ge 3$)
Let $X = 3A$. Using the property $\text{adj}(\text{adj}(X)) = (\det(X))^{n-2} X$, we get:
$ \text{adj}(\text{adj}(3A)) = (\det(3A))^{3-2} (3A) = (\det(3A)) (3A) $Now, find the determinant of this expression:
$ \det(\text{adj}(\text{adj}(3A))) = \det((\det(3A)) (3A)) $Using $\det(kA) = k^n \det(A)$, where $k = \det(3A)$ and the matrix is $3A$ (size $n=3$):
$ \det(\text{adj}(\text{adj}(3A))) = (\det(3A))^3 \det(3A) = (\det(3A))^4 $Calculating $\det(\text{adj}(\text{adj}(3A)))$
First, calculate $\det(3A)$ using $\det(kA) = k^n \det(A)$ with $k=3$ and $n=3$:
$ \det(3A) = 3^3 \det(A) = 27 \times 8 = 216 $Now substitute this value back into the expression for $\det(\text{adj}(\text{adj}(3A)))$:
$ \det(\text{adj}(\text{adj}(3A))) = (216)^4 $Express 216 in terms of its prime factors:
$ 216 = 6^3 = (2 \times 3)^3 = 2^3 \times 3^3 $Therefore:
$ (216)^4 = (2^3 \times 3^3)^4 = (2^3)^4 \times (3^3)^4 = 2^{12} \times 3^{12} $We are given that $\det(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n$. So:
$ 2^{12} \cdot 3^{12} = 2^m \cdot 3^n $Finding m + n
By comparing the powers of 2 and 3, we find:
- $m = 12$
- $n = 12$
Both $m$ and $n$ are natural numbers, as required.
The value of $m + n$ is:
$ m + n = 12 + 12 = 24 $Similar Questions
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