Let`A= [[1,0,0],[2,1,0],[3,2,1]]` be a square matrix and ` C_(1), C_(2), C_(3)` be three comumn matrices satisgying `AC_(1) = [[1],[0],[0]], AC_(2) = [[2],[3],[0]] and AC_(3)= [[2],[3],[1]]` of matrix B. If the matrix `C= 1/3 (AcdotB).`find `sin^(-1) (detA) + tan^(-1) (9det C) `

Options

A. `pi/4`
B. `pi/2`
C. `(3pi)/4`
D. `pi`

Explanation

To solve the problem step by step, we will follow the instructions given in the question.

### Step 1: Identify the matrix A and calculate its determinant.
Given matrix \( A \):
\[ A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \]

To find \( \text{det}(A) \), we can use the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is represented as:
\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

For our matrix \( A \):
- \( a = 1, b = 0, c = 0 \)
- \( d = 2, e = 1, f = 0 \)
- \( g = 3, h = 2, i = 1 \)

Plugging in the values:
\[ \text{det}(A) = 1(1 \cdot 1 - 0 \cdot 2) - 0(2 \cdot 1 - 0 \cdot 3) + 0(2 \cdot 0 - 1 \cdot 3) = 1(1) = 1 \]

### Step 2: Calculate the determinant of matrix B.
From the problem, we know:
\[ AC_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad AC_2 = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}, \quad AC_3 = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} \]
Thus, we can express matrix \( B \) as:
\[ B = \begin{bmatrix} C_1 & C_2 & C_3 \end{bmatrix} \]

The multiplication \( A \cdot B \) gives us:
\[ A \cdot B = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \cdot \begin{bmatrix} C_1 & C_2 & C_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 3 & 0 \\ 2 & 3 & 1 \end{bmatrix} \]

### Step 3: Calculate the determinant of the resulting matrix D.
Let:
\[ D = A \cdot B = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 3 & 0 \\ 2 & 3 & 1 \end{bmatrix} \]

To find \( \text{det}(D) \):
\[ \text{det}(D) = 1(3 \cdot 1 - 0 \cdot 3) - 0(2 \cdot 1 - 0 \cdot 2) + 0(2 \cdot 0 - 3 \cdot 2) = 1(3) = 3 \]

### Step 4: Calculate the determinant of matrix C.
Given:
\[ C = \frac{1}{3} (A \cdot B) = \frac{1}{3} D \]
Using the property of determinants:
\[ \text{det}(C) = \left(\frac{1}{3}\right)^3 \cdot \text{det}(D) = \frac{1}{27} \cdot 3 = \frac{1}{9} \]

### Step 5: Calculate \( \sin^{-1}(\text{det}(A)) + \tan^{-1}(9 \cdot \text{det}(C)) \).
Now we have:
- \( \text{det}(A) = 1 \)
- \( \text{det}(C) = \frac{1}{9} \)

Calculating:
\[ \sin^{-1}(\text{det}(A)) = \sin^{-1}(1) = \frac{\pi}{2} \text{ (or 90 degrees)} \]
\[ \tan^{-1}(9 \cdot \text{det}(C)) = \tan^{-1}(9 \cdot \frac{1}{9}) = \tan^{-1}(1) = \frac{\pi}{4} \text{ (or 45 degrees)} \]

Adding these:
\[ \sin^{-1}(\text{det}(A)) + \tan^{-1}(9 \cdot \text{det}(C)) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{2\pi}{4} + \frac{\pi}{4} = \frac{3\pi}{4} \text{ (or 135 degrees)} \]

### Final Answer:
\[ \sin^{-1}(\text{det}(A)) + \tan^{-1}(9 \cdot \text{det}(C)) = 135^\circ \]