If the system of equations `ax+y=1,x+2y=3,2x+3y=5` are consistent, then a is given by

Options

A. 0
B. 1
C. 2
D. None of these

Answer

To determine the value of \( a \) for which the system of equations is consistent, we start with the given equations:

1. \( ax + y = 1 \)
2. \( x + 2y = 3 \)
3. \( 2x + 3y = 5 \)

### Step 1: Rewrite the equations in standard form
We can rewrite the equations in standard form:

1. \( ax + y - 1 = 0 \)
2. \( x + 2y - 3 = 0 \)
3. \( 2x + 3y - 5 = 0 \)

### Step 2: Form the coefficient matrix
The coefficients of the variables \( x \) and \( y \) can be arranged in a determinant form:

\[ \begin{vmatrix} a & 1 & -1 \\ 1 & 2 & -3 \\ 2 & 3 & -5 \end{vmatrix} \]

### Step 3: Calculate the determinant
To find the value of \( a \) such that the system is consistent, we need to set the determinant equal to zero:

\[ D = \begin{vmatrix} a & 1 & -1 \\ 1 & 2 & -3 \\ 2 & 3 & -5 \end{vmatrix} \]

Calculating the determinant using the formula for a 3x3 matrix:

\[ D = a \begin{vmatrix} 2 & -3 \\ 3 & -5 \end{vmatrix} - 1 \begin{vmatrix} 1 & -3 \\ 2 & -5 \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 2 & -3 \\ 3 & -5 \end{vmatrix} = (2)(-5) - (-3)(3) = -10 + 9 = -1 \)

2. \( \begin{vmatrix} 1 & -3 \\ 2 & -5 \end{vmatrix} = (1)(-5) - (-3)(2) = -5 + 6 = 1 \)

3. \( \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = (1)(3) - (2)(2) = 3 - 4 = -1 \)

Now substituting these back into the determinant equation:

\[ D = a(-1) - 1(1) - 1(-1) = -a - 1 + 1 = -a \]

### Step 4: Set the determinant to zero
For the system to be consistent, we set the determinant \( D \) equal to zero:

\[ -a = 0 \]

### Step 5: Solve for \( a \)
Thus, we find:

\[ a = 0 \]

### Conclusion
The value of \( a \) for which the given system of equations is consistent is \( a = 0 \).

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