Let $A$ be a matrix of order $3\times3$ with $\lvert A\rvert=5$. If $\bigl\lvert 2\,\mathrm{adj}\bigl(3A\,\mathrm{adj}(2A)\bigr)\bigr\rvert=2^{\alpha}3^{\beta}5^{\gamma}$, where $\alpha,\beta,\gamma\in\mathbb{N}$, then $\alpha+\beta+\gamma$ is equal to:

Step 1: Recall Properties of Determinants and Adjoint

We begin by recalling some fundamental properties of determinants and adjoints for an $n \times n$ matrix $X$. These properties are crucial for simplifying the given expression. Specifically, we'll use how scalars interact with determinants and adjoints, and the determinant of an adjoint matrix.

Step 2: Simplify the innermost adjoint term

The matrix $A$ is of order $3 \times 3$, so $n=3$. We first simplify the determinant of the innermost term, $\mathrm{adj}(2A)$. Using the property $\lvert \mathrm{adj}(X) \rvert = \lvert X \rvert^{n-1}$ and $\lvert kX \rvert = k^n \lvert X \rvert$, we find that $\lvert \mathrm{adj}(2A) \rvert = \lvert 2A \rvert^2 = (2^3 \lvert A \rvert)^2$. Given $\lvert A \rvert = 5$, this simplifies to $(8 \cdot 5)^2 = 40^2 = 1600$.

Step 3: Simplify the expression inside the outermost adjoint

Next, we evaluate the determinant of the expression $3A \mathrm{adj}(2A)$. Using the property $\lvert XY \rvert = \lvert X \rvert \lvert Y \rvert$, we can write this as $\lvert 3A \rvert \lvert \mathrm{adj}(2A) \rvert$. We know $\lvert 3A \rvert = 3^3 \lvert A \rvert = 27 \cdot 5 = 135$, and we calculated $\lvert \mathrm{adj}(2A) \rvert = 1600$. Multiplying these values gives $135 \cdot 1600 = 216000$.

Step 4: Evaluate the full expression

Finally, we evaluate the determinant of the entire expression. Using the property $\lvert kX \rvert = k^n \lvert X \rvert$ and $\lvert \mathrm{adj}(X) \rvert = \lvert X \rvert^{n-1}$, we get $2^3 \lvert \mathrm{adj}(3A \mathrm{adj}(2A)) \rvert = 8 \cdot \lvert 3A \mathrm{adj}(2A) \rvert^2$. Substituting the value $216000$ for $\lvert 3A \mathrm{adj}(2A) \rvert$, we have $8 \cdot (216000)^2$. We then express $216000$ in its prime factorization: $216000 = 216 \cdot 1000 = 6^3 \cdot 10^3 = (2 \cdot 3)^3 \cdot (2 \cdot 5)^3 = 2^3 \cdot 3^3 \cdot 2^3 \cdot 5^3 = 2^6 \cdot 3^3 \cdot 5^3$. Squaring this gives $(2^6 \cdot 3^3 \cdot 5^3)^2 = 2^{12} \cdot 3^6 \cdot 5^6$. Multiplying by $2^3$ gives $2^3 \cdot 2^{12} \cdot 3^6 \cdot 5^6 = 2^{15} \cdot 3^6 \cdot 5^6$.

Step 5: Determine $\alpha, \beta, \gamma$ and their sum

By comparing the final expression $2^{15} \cdot 3^6 \cdot 5^6$ with the given form $2^{\alpha}3^{\beta}5^{\gamma}$, we can identify the values of $\alpha$, $\beta$, and $\gamma$. We find $\alpha = 15$, $\beta = 6$, and $\gamma = 6$. The problem asks for the sum $\alpha + \beta + \gamma$, which is $15 + 6 + 6 = 27$.

More from this chapter

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