If the system of linear equations \( x + y + z = 1 \), \( 2x + 2y + 3z = 6 \), \( x + 4y + 9z = 3 \) has a unique solution \( x = \alpha, y = \beta, z = \gamma \), then the value of \( \beta \) is:
Show Hint
- 3
- 6
- \(-10 \)
- \(-4 \)
The Correct Option is C
Solution and Explanation
Concept:& nbsp;
For a system of linear equations \[ AX=B, \] if \[ |A|\neq0, \] then the system has a unique solution. Using Cramer's Rule, the required variable can be found as
• \[ y=\frac{\Delta_y}{\Delta}, \] where \(y=\beta\).
• \(\Delta\) is the determinant of the coefficient matrix.
• \(\Delta_y\) is the determinant obtained by replacing the second column of the coefficient matrix with the constant vector.
Step 1: Calculate \(\Delta\).
\[ \Delta= \begin{vmatrix} 1 & amp; 1 & amp; 1\\ 2 & amp; 2 & amp; 3\\ 1 & amp; 4 & amp; 9 \end{vmatrix}. \]
Expanding along the first row,
\[ \Delta = 1(18-12) -1(18-3) +1(8-2) = 6-15+6 = -3. \]
Step 2: Calculate \(\Delta_y\).
Replace the second column with the constants \[ \begin{bmatrix} 1\\ 6\\ 3 \end{bmatrix}. \]
\[ \Delta_y= \begin{vmatrix} 1 & amp; 1 & amp; 1\\ 2 & amp; 6 & amp; 3\\ 1 & amp; 3 & amp; 9 \end{vmatrix}. \]
Expanding along the first row,
\[ \Delta_y = 1(54-9) -1(18-3) +1(6-6) = 45-15+0 = 30. \]
Step 3: Find \(\beta\).
\[ \beta = \frac{\Delta_y}{\Delta} = \frac{30}{-3} = -10. \]
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