Let A,B and I (identity matrix) be the matrices of order 4 such that `A=[a_(ij)]_(4xx4)`' where `a_(lj){{:(0 if i = j),(1 if i ne j):}and A =B -I`
`{:(,"Column-I",,"Column-II"),((A),|B|=,,(P)4),((B),|B^(2)-4B+I|= ,,(Q)0),((C),-4+|B^(3)+B^(2)-8B|=,,(R)-1),((D),-1+|A^(-1)+I|"is greater than",,(S)-4),(,,,(T)1):}`

Explanation

To solve the problem, we need to analyze the matrices \( A \), \( B \), and the identity matrix \( I \) of order 4. Let's break down the steps to find the required determinants and match the options.

### Step 1: Define Matrix \( A \)
Matrix \( A \) is defined such that:
- \( a_{ij} = 0 \) if \( i = j \) (diagonal elements)
- \( a_{ij} = 1 \) if \( i \neq j \) (off-diagonal elements)

Thus, the matrix \( A \) looks like this:
\[ A = \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{pmatrix} \]

### Step 2: Define Matrix \( B \)
Given that \( A = B - I \), we can express \( B \) as:
\[ B = A + I \]
Where \( I \) is the identity matrix:
\[ I = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \]

Adding \( A \) and \( I \):
\[ B = \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \]

### Step 3: Calculate \( |B| \)
The determinant of matrix \( B \) can be calculated. Since all rows are identical, the determinant is:
\[ |B| = 0 \]
This matches with option \( Q \).

### Step 4: Calculate \( |B^2 - 4B + I| \)
First, we calculate \( B^2 \):
\[ B^2 = B \cdot B = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \end{pmatrix} \]

Now calculate \( |B^2 - 4B + I| \):
\[ B^2 - 4B + I = \begin{pmatrix} 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \end{pmatrix} - \begin{pmatrix} 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \]
This results in:
\[ = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = I \]
Thus, \( |B^2 - 4B + I| = 1 \).

### Step 5: Calculate \( |-4 + |B^3 + B^2 - 8B|| \)
Next, we calculate \( B^3 \):
\[ B^3 = B^2 \cdot B = \begin{pmatrix} 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 16 & 16 & 16 & 16 \\ 16 & 16 & 16 & 16 \\ 16 & 16 & 16 & 16 \\ 16 & 16 & 16 & 16 \end{pmatrix} \]

Now calculate \( B^3 + B^2 - 8B \):
\[ B^3 + B^2 - 8B = \begin{pmatrix} 16 & 16 & 16 & 16 \\ 16 & 16 & 16 & 16 \\ 16 & 16 & 16 & 16 \\ 16 & 16 & 16 & 16 \end{pmatrix} + \begin{pmatrix} 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \end{pmatrix} - \begin{pmatrix} 8 & 8 & 8 & 8 \\ 8 & 8 & 8 & 8 \\ 8 & 8 & 8 & 8 \\ 8 & 8 & 8 & 8 \end{pmatrix} \]
This results in:
\[ = \begin{pmatrix} 12 & 12 & 12 & 12 \\ 12 & 12 & 12 & 12 \\ 12 & 12 & 12 & 12 \\ 12 & 12 & 12 & 12 \end{pmatrix} \]
Thus, the determinant \( |B^3 + B^2 - 8B| = 0 \).

### Step 6: Calculate \( |-1 + |A^{-1} + I| > -4 \)
To find \( A^{-1} \), we note that \( A \) is singular (its determinant is 0), hence \( A^{-1} \) does not exist. Therefore, \( |A^{-1} + I| \) cannot be computed.

### Conclusion
From the calculations:
- \( |B| = 0 \) matches with \( Q \).
- \( |B^2 - 4B + I| = 1 \) matches with \( T \).
- \( |B^3 + B^2 - 8B| = 0 \) matches with \( S \).
- \( A^{-1} \) does not exist, so \( D \) cannot be matched.