Let`A= [[1,0,0],[2,1,0],[3,2,1]]` be a square matrix and ` C_(1), C_(2), C_(3)` be three
column matrices satisfying `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).`
The ratio of the trace of the matrix B to the matrix C, is

Options

A. `-9/5`
B. `-5/9`
C. `-2/3`
D. `-3/2`

Explanation

To solve the problem step by step, we will first find the matrix \( B \) using the conditions given for \( C_1, C_2, \) and \( C_3 \), and then we will calculate the matrix \( C \) and find the required ratio of the traces.

### Step 1: Find \( C_1 \)

Given \( AC_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \):

Let \( C_1 = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \).

Then, we have:

\[ A \cdot C_1 = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \]

This gives us the following equations:

1. \( 1a + 0b + 0c = 1 \) → \( a = 1 \)
2. \( 2a + 1b + 0c = 0 \) → \( 2(1) + b = 0 \) → \( b = -2 \)
3. \( 3a + 2b + 1c = 0 \) → \( 3(1) + 2(-2) + c = 0 \) → \( 3 - 4 + c = 0 \) → \( c = 1 \)

Thus, \( C_1 = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \).

### Step 2: Find \( C_2 \)

Given \( AC_2 = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} \):

Let \( C_2 = \begin{bmatrix} d \\ e \\ f \end{bmatrix} \).

Then, we have:

\[ A \cdot C_2 = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} d \\ e \\ f \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} \]

This gives us the following equations:

1. \( 1d + 0e + 0f = 2 \) → \( d = 2 \)
2. \( 2d + 1e + 0f = 3 \) → \( 2(2) + e = 3 \) → \( 4 + e = 3 \) → \( e = -1 \)
3. \( 3d + 2e + 1f = 0 \) → \( 3(2) + 2(-1) + f = 0 \) → \( 6 - 2 + f = 0 \) → \( f = -4 \)

Thus, \( C_2 = \begin{bmatrix} 2 \\ -1 \\ -4 \end{bmatrix} \).

### Step 3: Find \( C_3 \)

Given \( AC_3 = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} \):

Let \( C_3 = \begin{bmatrix} g \\ h \\ i \end{bmatrix} \).

Then, we have:

\[ A \cdot C_3 = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} g \\ h \\ i \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} \]

This gives us the following equations:

1. \( 1g + 0h + 0i = 2 \) → \( g = 2 \)
2. \( 2g + 1h + 0i = 3 \) → \( 2(2) + h = 3 \) → \( 4 + h = 3 \) → \( h = -1 \)
3. \( 3g + 2h + 1i = 1 \) → \( 3(2) + 2(-1) + i = 1 \) → \( 6 - 2 + i = 1 \) → \( i = -3 \)

Thus, \( C_3 = \begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix} \).

### Step 4: Construct Matrix \( B \)

Now, we can construct matrix \( B \) using \( C_1, C_2, C_3 \):

\[ B = \begin{bmatrix} C_1 & C_2 & C_3 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{bmatrix} \]

### Step 5: Calculate the Trace of Matrix \( B \)

The trace of a matrix is the sum of its diagonal elements:

\[ \text{Trace}(B) = 1 + (-1) + (-3) = -3 \]

### Step 6: Calculate Matrix \( C \)

Given \( C = \frac{1}{3} A \cdot B \):

\[ C = \frac{1}{3} \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{bmatrix} \]

Calculating \( A \cdot B \):

1. First row: \( \begin{bmatrix} 1 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} = 1 \), \( \begin{bmatrix} 1 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} 2 \\ -1 \\ -4 \end{bmatrix} = 2 \), \( \begin{bmatrix} 1 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix} = 2 \)
2. Second row: \( \begin{bmatrix} 2 & 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} = 2 - 2 = 0 \), \( \begin{bmatrix} 2 & 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 2 \\ -1 \\ -4 \end{bmatrix} = 4 - 1 = 3 \), \( \begin{bmatrix} 2 & 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix} = 4 - 1 = 3 \)
3. Third row: \( \begin{bmatrix} 3 & 2 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} = 3 - 4 + 1 = 0 \), \( \begin{bmatrix} 3 & 2 & 1 \end{bmatrix} \cdot \begin{bmatrix} 2 \\ -1 \\ -4 \end{bmatrix} = 6 - 2 - 4 = 0 \), \( \begin{bmatrix} 3 & 2 & 1 \end{bmatrix} \cdot \begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix} = 6 - 2 - 3 = 1 \)

Thus,

\[ A \cdot B = \begin{bmatrix} 1 & 2 & 2 \\ 0 & 3 & 3 \\ 0 & 0 & 1 \end{bmatrix} \]

Now,

\[ C = \frac{1}{3} \begin{bmatrix} 1 & 2 & 2 \\ 0 & 3 & 3 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ 0 & 1 & 1 \\ 0 & 0 & \frac{1}{3} \end{bmatrix} \]

### Step 7: Calculate the Trace of Matrix \( C \)

\[ \text{Trace}(C) = \frac{1}{3} + 1 + \frac{1}{3} = \frac{5}{3} \]

### Step 8: Find the Ratio of the Traces

The ratio of the trace of matrix \( B \) to the trace of matrix \( C \):

\[ \text{Ratio} = \frac{\text{Trace}(B)}{\text{Trace}(C)} = \frac{-3}{\frac{5}{3}} = -3 \cdot \frac{3}{5} = -\frac{9}{5} \]

### Final Answer

The ratio of the trace of the matrix \( B \) to the matrix \( C \) is \( -\frac{9}{5} \).

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