Let `Z=[(1,1,3),(5,1,2),(3,1,0)] and P=[(1,0,2),(2,1,0),(3,0,1)]`. If `Z=PQ^(-1)`, where Q is a square matrix of order 3, then the value of `Tr((adjQ)P)` is equal to (where `Tr(A)` represents the trace of a matrix A i.e. the sum of all the diagonal elements of the matrix A and adjB represents the adjoint matrix of matrix B)

Options

A. 3
B. `-1`
C. 4
D. `(6)/(5)`

Answer

To solve the problem step-by-step, we will follow the instructions provided in the video transcript and derive the required value of \( \text{Tr}((\text{adj} Q) P) \).

### Step 1: Define the matrices
Given:
\[ Z = \begin{pmatrix} 1 & 1 & 3 \\ 5 & 1 & 2 \\ 3 & 1 & 0 \end{pmatrix}, \quad P = \begin{pmatrix} 1 & 0 & 2 \\ 2 & 1 & 0 \\ 3 & 0 & 1 \end{pmatrix} \]

### Step 2: Calculate the determinant of matrix \( Z \)
To find \( \det(Z) \), we can use the formula for the determinant of a 3x3 matrix:
\[ \det(Z) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( Z = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).

For our matrix \( Z \):
- \( a = 1, b = 1, c = 3 \)
- \( d = 5, e = 1, f = 2 \)
- \( g = 3, h = 1, i = 0 \)

Calculating:
\[ \det(Z) = 1(1 \cdot 0 - 2 \cdot 1) - 1(5 \cdot 0 - 2 \cdot 3) + 3(5 \cdot 1 - 1 \cdot 3) \]
\[ = 1(0 - 2) - 1(0 - 6) + 3(5 - 3) \]
\[ = -2 + 6 + 6 = 10 \]

### Step 3: Calculate the determinant of matrix \( P \)
Using the same formula for \( P \):
\[ P = \begin{pmatrix} 1 & 0 & 2 \\ 2 & 1 & 0 \\ 3 & 0 & 1 \end{pmatrix} \]
Calculating:
\[ \det(P) = 1(1 \cdot 1 - 0 \cdot 0) - 0(2 \cdot 1 - 0 \cdot 3) + 2(2 \cdot 0 - 1 \cdot 3) \]
\[ = 1(1) - 0 + 2(0 - 3) = 1 - 6 = -5 \]

### Step 4: Relate \( Z \), \( P \), and \( Q \)
From the problem statement, we have:
\[ Z = P Q^{-1} \]
Taking determinants on both sides:
\[ \det(Z) = \det(P) \cdot \det(Q^{-1}) = \det(P) \cdot \frac{1}{\det(Q)} \]
Substituting the values:
\[ 10 = -5 \cdot \frac{1}{\det(Q)} \]
Thus,
\[ \det(Q) = -\frac{5}{10} = -\frac{1}{2} \]

### Step 5: Use the relationship to find \( \text{Tr}((\text{adj} Q) P) \)
We know:
\[ Z \cdot \det(Q) = P \cdot \text{adj}(Q) \]
Substituting \( \det(Q) \):
\[ Z \cdot \left(-\frac{1}{2}\right) = P \cdot \text{adj}(Q) \]
Thus,
\[ -\frac{1}{2} Z = P \cdot \text{adj}(Q) \]

### Step 6: Calculate \( \text{Tr}((\text{adj} Q) P) \)
Taking the trace on both sides:
\[ \text{Tr}\left(-\frac{1}{2} Z\right) = \text{Tr}(P \cdot \text{adj}(Q)) \]
Using the property of trace:
\[ \text{Tr}(AB) = \text{Tr}(BA) \]
We have:
\[ \text{Tr}(P \cdot \text{adj}(Q)) = \text{Tr}((\text{adj}(Q)) P) \]
Calculating \( \text{Tr}(Z) \):
\[ \text{Tr}(Z) = 1 + 1 + 0 = 2 \]
Thus:
\[ -\frac{1}{2} \cdot 2 = -1 \]

### Final Result
Therefore, the value of \( \text{Tr}((\text{adj} Q) P) \) is:
\[ \text{Tr}((\text{adj} Q) P) = -1 \]