Question

Let the system of linear equations $ x+y+ kz =2$ $ 2 x+3 y-z=1 $ $ 3 x+4 y +2 z= k$ have infinitely many solutions Then the system $ ( k +1) x+(2 k -1) y=7$ $ (2 k +1) x+( k +5) y=10$ has:

• A. infinitely many solutions
• B. unique solution satisfying $x-y=1$
• C. no solution
• D. unique solution satisfying $x+y=1$

Answer 1

The Correct Option is D

We are given the system of equations:

\[ x + y + kz = 2, \quad 2x + 3y - z = 1, \quad 3x + 4y + 2z = k. \]

To find the value of \(k\) for which the system has infinitely many solutions, we first find the determinant of the coefficient matrix.

Step 1: Coefficient Matrix

The coefficient matrix is:

\[ A = \begin{bmatrix} 1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2 \end{bmatrix}. \]

Step 2: Determinant of the Matrix

The determinant of \(A\) is calculated as:

\[ \text{Det}(A) = \begin{vmatrix} 1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2 \end{vmatrix}. \]

Expanding the determinant along the first row:

\[ \text{Det}(A) = 1 \cdot \begin{vmatrix} 3 & -1 \\ 4 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix} + k \cdot \begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix}. \]

Step 3: Calculate the 2x2 Determinants

Compute each 2x2 determinant:

• \[ \begin{vmatrix} 3 & -1 \\ 4 & 2 \end{vmatrix} = (3)(2) - (4)(-1) = 6 + 4 = 10. \]
• \[ \begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix} = (2)(2) - (3)(-1) = 4 + 3 = 7. \]
• \[ \begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix} = (2)(4) - (3)(3) = 8 - 9 = -1. \]

Step 4: Substitute Back

Substitute these values into the determinant expression:

\[ \text{Det}(A) = 1(10) - 1(7) + k(-1) = 10 - 7 - k = 3 - k. \]

Step 5: Infinitely Many Solutions

For the system to have infinitely many solutions, the determinant must be zero:

\[ 3 - k = 0 \implies k = 3. \]

Step 6: Reduced System

For \(k = 3\), the system becomes:

\[ 4x + 5y = 7 \quad \text{(1)}, \]

\[ 7x + 8y = 10 \quad \text{(2)}. \]

Subtract equation (1) from equation (2):

\[ (7x + 8y) - (4x + 5y) = 10 - 7. \]

This simplifies to:

\[ 3x + 3y = 3 \implies x + y = 1. \]

Conclusion

The value of \(k\) for which the system has infinitely many solutions is:

\[ \boxed{3}. \]

Furthermore, the system satisfies \(x + y = 1\).

Answer 2

$\left|\begin{array}{ccc}1& 1& k\\ 2& 3& -1\\ 3& 4& 2\end{array}\right|=0$
$\Rightarrow 1(10)-1(7)+k(-1)-0$
$\Rightarrow k=3$
For $k=3,2^{md}$ system is
$4x+5y=\mathrm{7....}$(1)
and $7x+8y=\mathrm{10....}$(2)
Clearly, they have a unique solution
(2) $-(1)\Rightarrow 3x+3y=3$
$\Rightarrow x+y=1$