Let `A=[a_("ij")]` be `3xx3` matrix and `B=[b_("ij")]` be `3xx3` matrix such that `b_("ij")` is the sum of the elements of `i^(th)` row of A except `a_("ij")`. If det, `(A)=19`, then the value of det. (B) is ________ .

Explanation

To solve the problem, we need to find the determinant of matrix B, which is defined based on the elements of matrix A. Let's go through the steps systematically.

### Step 1: Define the Matrices
Let \( A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \)
and we know that \( \text{det}(A) = 19 \).

### Step 2: Define the Elements of Matrix B
Matrix B is defined such that each element \( b_{ij} \) is the sum of the elements of the \( i^{th} \) row of A, excluding \( a_{ij} \).

Thus, we can express the elements of B as follows:
- \( b_{11} = a_{12} + a_{13} \)
- \( b_{12} = a_{11} + a_{13} \)
- \( b_{13} = a_{11} + a_{12} \)
- \( b_{21} = a_{22} + a_{23} \)
- \( b_{22} = a_{21} + a_{23} \)
- \( b_{23} = a_{21} + a_{22} \)
- \( b_{31} = a_{32} + a_{33} \)
- \( b_{32} = a_{31} + a_{33} \)
- \( b_{33} = a_{31} + a_{32} \)

### Step 3: Write Matrix B
Thus, we can write matrix B as:
\[ B = \begin{bmatrix} a_{12} + a_{13} & a_{11} + a_{13} & a_{11} + a_{12} \\ a_{22} + a_{23} & a_{21} + a_{23} & a_{21} + a_{22} \\ a_{32} + a_{33} & a_{31} + a_{33} & a_{31} + a_{32} \end{bmatrix} \]

### Step 4: Express B in Terms of A
We can express B as a product of two matrices. Let \( C \) be the matrix formed by the elements of A:
\[ C = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \]

We can express B as:
\[ B = C \cdot D \]
where \( D \) is a transformation matrix that modifies the structure of A to obtain B.

### Step 5: Define the Transformation Matrix D
The transformation matrix \( D \) can be defined as:
\[ D = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \]

### Step 6: Calculate the Determinant of D
Now, we need to calculate \( \text{det}(D) \):
\[ \text{det}(D) = 0(0 \cdot 0 - 1 \cdot 1) - 1(1 \cdot 0 - 1 \cdot 1) + 1(1 \cdot 1 - 0 \cdot 1) \]
\[ = 0 - 1(-1) + 1(1) = 1 + 1 = 2 \]

### Step 7: Use the Determinant Property
Using the property of determinants:
\[ \text{det}(B) = \text{det}(C) \cdot \text{det}(D) \]
\[ \text{det}(B) = \text{det}(A) \cdot \text{det}(D) = 19 \cdot 2 = 38 \]

### Final Answer
Thus, the value of \( \text{det}(B) \) is \( 38 \).