Consider a system of linear equation in three variables x,y,z
`a_1x+b_1y+ c_1z = d_1 , a_2x+ b_2y+c_2z=d_2 , a_3x + b_3y + c_3z=d_3`
The systems can be expressed by matrix equation `[(a_1,b_1,c_1),(a_2,b_2,c_2),(c_1,c_2,c_3)][(x),(y),(z)]=[(d_1),(d_2),(d_3)]`
if A is non-singular matrix then the solution of above system can be found by X =`A^(-1)B`, the solution in this case is unique.
if A is a singular matrix i.e. then the system will have
no unique solution if no solution (i.e. it is inconsistent) if
Where Adj A is the adjoint of the matrix A, which is obtained by taking transpose of the matrix obtained by replacing each element of matrix A with corresponding cofactors.
Now consider the following matrix.
`A=[(a,1,0),(1,b,d),(1,b,c)], B=[(a,1,1),(0,d,c),(f,g,h)], U=[(f),(g),(h)], V=[(a^2),(0),(0)], X=[(x),(y),(z)]`
If AX=U has infinitely many solutions then the equation BX=U is consistent if

Options

A. a=0
B. d=0
C. f=0
D. adf `ne` 0

Answer

To solve the problem step by step, we need to analyze the conditions under which the system of equations represented by the matrices A and B has infinitely many solutions and is consistent.

### Step 1: Understanding the System of Equations
We have two systems of equations represented by matrices:
1. \( AX = U \)
2. \( BX = U \)

Where:
- \( A = \begin{pmatrix} a & 1 & 0 \\ 1 & b & d \\ 1 & b & c \end{pmatrix} \)
- \( B = \begin{pmatrix} a & 1 & 1 \\ 0 & d & c \\ f & g & h \end{pmatrix} \)
- \( U = \begin{pmatrix} f \\ g \\ h \end{pmatrix} \)

### Step 2: Condition for Infinitely Many Solutions
For the equation \( AX = U \) to have infinitely many solutions, the following condition must hold:
- The rank of the augmented matrix \( [A | U] \) must be equal to the rank of the coefficient matrix \( A \), and both must be less than the number of variables (which is 3).

This implies:
\[ \text{rank}(A) = \text{rank}([A | U]) < 3 \]

### Step 3: Analyzing the Augmented Matrix
The augmented matrix \( [A | U] \) can be expressed as:
\[ [A | U] = \begin{pmatrix} a & 1 & 0 & f \\ 1 & b & d & g \\ 1 & b & c & h \end{pmatrix} \]

For the rank condition to hold, we need:
1. The third row to be a linear combination of the first two rows.
2. Specifically, we can deduce that \( d \) must equal \( c \) and \( g \) must equal \( h \) for the third row to become dependent on the first two rows.

### Step 4: Condition for Consistency of \( BX = U \)
Now, we need to check the consistency of the equation \( BX = U \):
\[ BX = U \implies \begin{pmatrix} a & 1 & 1 \\ 0 & d & c \\ f & g & h \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} f \\ g \\ h \end{pmatrix} \]

For the system \( BX = U \) to be consistent, we need:
\[ \text{rank}(B) = \text{rank}([B | U]) \]

### Step 5: Forming the Augmented Matrix for \( B \)
The augmented matrix for \( B \) is:
\[ [B | U] = \begin{pmatrix} a & 1 & 1 & f \\ 0 & d & c & g \\ f & g & h & h \end{pmatrix} \]

### Step 6: Analyzing the Rank Condition
For \( BX = U \) to be consistent, we can analyze the third row:
- If \( f = 0 \), it simplifies the matrix and allows us to maintain the rank condition:
\[ \text{rank}(B) = \text{rank}([B | U]) \]

### Conclusion
Thus, the condition for the equation \( BX = U \) to be consistent, given that \( AX = U \) has infinitely many solutions, is:
\[ f = 0 \]