Matrices & Determinants
System of Linear Equations
Grade Class 12
Question:
Consider a system of linear equations $a_ix + b_iy + c_iz = d_i$ (where $a_i, b_i, c_i \neq 0$ and $i = 1,2,3$ ) & $(\alpha,\beta,\gamma)$ is its unique solution, then match list-I with list-II<br><table><tr><td><b>List-I</b></td><td><b>List-II</b></td></tr><tr><td>(I) If $a_i = k, d_i = k^2, (k \neq 0)$ and $\alpha + \beta + \gamma = 2$, then $k$ is</td><td>(P) 1</td></tr><tr><td>(II) If $a_i = d_i = k \neq 0$, then $\alpha + \beta + \gamma$ is</td><td>(Q) 2</td></tr><tr><td>(III) If $a_i = k > 0, d_i = k + 1$, then $\alpha + \beta + \gamma$ can be</td><td>(R) 0</td></tr><tr><td>(IV) If $a_i = k < 0, d_i = k + 1$, then $\alpha + \beta + \gamma$ can be</td><td>(S) 3</td></tr><tr><td></td><td>(T) -1</td></tr></table>
(A) I $\rightarrow$ P,Q; II $\rightarrow$ R; III $\rightarrow$ Q, S; IV $\rightarrow$ T
(B) I $\rightarrow$ P; II $\rightarrow$ Q; III $\rightarrow$ R, S; IV $\rightarrow$ T
(C) I $\rightarrow$ Q; II $\rightarrow$ P; III $\rightarrow$ S; IV $\rightarrow$ T
(D) I $\rightarrow$ Q; II $\rightarrow$ P; III $\rightarrow$ Q, S; IV $\rightarrow$ R, T
Step-by-Step Solution
Key Concept: For a system of linear equations with a unique solution, the determinant of the coefficient matrix must be non-zero. By substituting the given conditions into the system, we can solve for the variables and their sum.
The system is $a_ix + b_iy + c_iz = d_i$. Since it has a unique solution, the determinant of the coefficient matrix is non-zero. By analyzing each case, we find the values of $\alpha + \beta + \gamma$. For (I), $a_i = k, d_i = k^2$, the system becomes $k(x+y+z) = k^2$, so $x+y+z = k$. Given $\alpha+\beta+\gamma=2$, we get $k=2$. For (II), $a_i=d_i=k$, $x+y+z=1$, so $\alpha+\beta+\gamma=1$. For (III) and (IV), substituting $d_i=k+1$ and $a_i=k$ leads to $x+y+z = (k+1)/k = 1 + 1/k$.
Correct Answer: (D)