Question

Let A be a matrix such that adj (adj A) = $\begin{bmatrix} 5 & 2 & 1 \\ 3 & 1 & 0 \\ 2 & 2 & 1 \end{bmatrix}$ then the value of det $(3A^{-1})$ is

• A. 1
• B. $3^{11/4}$
• C. $3^{3/4}$
• D. 9

Answer 1

Answer: (B)

$3^{11/4}$

Answer 2

The given matrix is $adj(adj A) = B = \begin{bmatrix} 5 & 2 & 1 \\ 3 & 1 & 0 \\ 2 & 2 & 1 \end{bmatrix}$. The order of the matrix A is determined by the order of $adj(adj A)$. Since B is a $3 \times 3$ matrix, A must also be a $3 \times 3$ matrix. Let the order of A be $n=3$.

We need to find the value of $det(3A^{-1})$. Using the property $det(kA) = k^n det(A)$, we have $det(3A^{-1}) = 3^n det(A^{-1})$. For $n=3$, $det(3A^{-1}) = 3^3 det(A^{-1})$. Using the property $det(A^{-1}) = \frac{1}{det(A)}$, we have $det(3A^{-1}) = 3^3 \frac{1}{det(A)}$. To find the value, we need to determine $det(A)$.

We are given $adj(adj A) = B$. For a square matrix A of order $n \ge 2$, we have the property $adj(adj A) = (det A)^{n-2} A$. For $n=3$, this property becomes $adj(adj A) = (det A)^{3-2} A = (det A) A$. So, we have $(det A) A = B$.

Taking the determinant of both sides: $det((det A) A) = det(B)$. Using the property $det(kA) = k^n det(A)$ with $k=det A$ and $n=3$: $(det A)^3 det(A) = det(B)$. $(det A)^4 = det(B)$.

Now, we calculate the determinant of matrix B: $B = \begin{bmatrix} 5 & 2 & 1 \\ 3 & 1 & 0 \\ 2 & 2 & 1 \end{bmatrix}$ $det(B) = 5 \cdot \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 3 & 0 \\ 2 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & 1 \\ 2 & 2 \end{vmatrix}$ $det(B) = 5(1 \times 1 - 0 \times 2) - 2(3 \times 1 - 0 \times 2) + 1(3 \times 2 - 1 \times 2)$ $det(B) = 5(1) - 2(3) + 1(6 - 2)$ $det(B) = 5 - 6 + 4$ $det(B) = 3$.

Substituting the value of $det(B)$ into the equation $(det A)^4 = det(B)$: $(det A)^4 = 3$. This equation implies $det(A) = \pm 3^{1/4}$.

We need to find $det(3A^{-1}) = 3^3 \frac{1}{det(A)}$. If $det(A) = 3^{1/4}$, then $det(3A^{-1}) = 3^3 \frac{1}{3^{1/4}} = 3^{3 - 1/4} = 3^{12/4 - 1/4} = 3^{11/4}$. If $det(A) = -3^{1/4}$, then $det(3A^{-1}) = 3^3 \frac{1}{-3^{1/4}} = -3^{11/4}$.

The given options are all positive values. This suggests that either the problem assumes $det(A) > 0$, or the context implies it. Assuming $det(A) > 0$, we take $det(A) = 3^{1/4}$.

Then, $det(3A^{-1}) = 3^{11/4}$.

Comparing this with the given options: (1) 1 (2) $3^{11/4}$ (3) $3^{3/4}$ (4) 9

The calculated value $3^{11/4}$ matches option (2).

The final answer is $\boxed{3^{11/4}}$.

Explanation of the solution:

1. Identify the order of matrix A from the given matrix $adj(adj A)$. Since $adj(adj A)$ is $3 \times 3$, A is $3 \times 3$ (order $n=3$).
2. Use the property $det(kA) = k^n det(A)$ to express $det(3A^{-1})$ in terms of $det(A)$: $det(3A^{-1}) = 3^3 det(A^{-1}) = 3^3/det(A)$.
3. Use the property $adj(adj A) = (det A)^{n-2} A$. For $n=3$, this is $adj(adj A) = (det A) A$.
4. Take the determinant of both sides of $(det A) A = adj(adj A)$. This gives $(det A)^n det(A) = det(adj(adj A))$, which simplifies to $(det A)^4 = det(adj(adj A))$.
5. Calculate the determinant of the given matrix $adj(adj A)$. The determinant is 3.
6. Solve $(det A)^4 = 3$ for $det(A)$. This gives $det(A) = \pm 3^{1/4}$.
7. Assuming $det(A) > 0$ (based on the positive options), take $det(A) = 3^{1/4}$.
8. Substitute the value of $det(A)$ into the expression for $det(3A^{-1})$: $det(3A^{-1}) = 3^3 / 3^{1/4} = 3^{11/4}$.
9. Match the result with the given options.

The final answer is $\boxed{3^{11/4}}$.