Let `P=[(1,2,1),(0,1,-1),(3,1,1)]`. If the product PQ has inverse `R=[(-1,0,1),(1,1,3),(2,0,2)]` then `Q^(-1)` equals

Options

A. `[(3,2,9),(-1,1,1),(0,1,8)]`
B. `[(5,2,9),(-1,1,1),(0,1,7)]`
C. `[(2,-1,0),(10,6,3),(8,6,4)]`
D. none of these

Explanation

To find \( Q^{-1} \) given that \( R = (PQ)^{-1} \), we can follow these steps:

### Step 1: Understand the relationship between \( R \), \( P \), and \( Q \)
We know from matrix theory that:
\[ (PQ)^{-1} = Q^{-1}P^{-1} \]
Thus, we can express \( R \) as:
\[ R = Q^{-1}P^{-1} \]

### Step 2: Rearrange the equation to find \( Q^{-1} \)
Multiplying both sides of the equation by \( P \) gives:
\[ RP = Q^{-1} \]
This means that to find \( Q^{-1} \), we need to compute the product \( RP \).

### Step 3: Calculate \( P \)
Given:
\[ P = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & 1 & 1 \end{pmatrix} \]

### Step 4: Calculate \( R \)
Given:
\[ R = \begin{pmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{pmatrix} \]

### Step 5: Compute the product \( RP \)
Now, we will calculate \( RP \):
\[ RP = \begin{pmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & 1 & 1 \end{pmatrix} \]

Calculating each element of the resulting matrix:

1. First row:
- First element: \((-1 \cdot 1) + (0 \cdot 0) + (1 \cdot 3) = -1 + 0 + 3 = 2\)
- Second element: \((-1 \cdot 2) + (0 \cdot 1) + (1 \cdot 1) = -2 + 0 + 1 = -1\)
- Third element: \((-1 \cdot 1) + (0 \cdot -1) + (1 \cdot 1) = -1 + 0 + 1 = 0\)

2. Second row:
- First element: \((1 \cdot 1) + (1 \cdot 0) + (3 \cdot 3) = 1 + 0 + 9 = 10\)
- Second element: \((1 \cdot 2) + (1 \cdot 1) + (3 \cdot 1) = 2 + 1 + 3 = 6\)
- Third element: \((1 \cdot 1) + (1 \cdot -1) + (3 \cdot 1) = 1 - 1 + 3 = 3\)

3. Third row:
- First element: \((2 \cdot 1) + (0 \cdot 0) + (2 \cdot 3) = 2 + 0 + 6 = 8\)
- Second element: \((2 \cdot 2) + (0 \cdot 1) + (2 \cdot 1) = 4 + 0 + 2 = 6\)
- Third element: \((2 \cdot 1) + (0 \cdot -1) + (2 \cdot 1) = 2 + 0 + 2 = 4\)

Thus, the product \( RP \) is:
\[ RP = \begin{pmatrix} 2 & -1 & 0 \\ 10 & 6 & 3 \\ 8 & 6 & 4 \end{pmatrix} \]

### Step 6: Conclusion
Therefore, we have:
\[ Q^{-1} = RP = \begin{pmatrix} 2 & -1 & 0 \\ 10 & 6 & 3 \\ 8 & 6 & 4 \end{pmatrix} \]