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Matrices and Determinants question

2023 · Shift 1 · Q31
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Matrices and Determinants question

2023 · Shift 1 · Q31

JEE AdvancedMathematicsMatrices and DeterminantsMCQ+3 / −1
Let α,β\alpha, \betaα,β and γ\gammaγ be real numbers. Consider the following system of linear equations

x+2y+z=7x+αz=112x−3y+βz=γ\begin{aligned} & x+2 y+z=7 \\\\ & x+\alpha z=11 \\\\ & 2 x-3 y+\beta z=\gamma \end{aligned}​x+2y+z=7x+αz=112x−3y+βz=γ​

Match each entry in List-I to the correct entries in List-II.

List - I List - II
(P) If β=12(7α−3)\beta=\frac{1}{2}(7 \alpha-3)β=21​(7α−3) and γ=28\gamma=28γ=28, then the system has (1) a unique solution
(Q) If β=12(7α−3)\beta=\frac{1}{2}(7 \alpha-3)β=21​(7α−3) and γeq28\gamma eq 28γeq28, then the system has (2) no solution
(R) If βeq12(7α−3)\beta eq \frac{1}{2}(7 \alpha-3)βeq21​(7α−3) where α=1\alpha=1α=1 and γeq28\gamma eq 28γeq28, then the system has (3) infinitely many solutions
(S) If βeq12(7α−3)\beta eq \frac{1}{2}(7 \alpha-3)βeq21​(7α−3) where α=1\alpha=1α=1 and γ=28\gamma=28γ=28, then the system has (4) x=11,y=−2x=11, y=-2x=11,y=−2 and z=0z=0z=0 as a solution
(5) x=−15,y=4x=-15, y=4x=−15,y=4 and z=0z=0z=0 as a solution

The correct option is:
  1. A
    (P)→(3)  (Q)→(2)  (R)→(1)  (S)→(4)(P) \rightarrow(3) ~~ (Q) \rightarrow(2) ~~(R) \rightarrow(1)~~ (S) \rightarrow(4)(P)→(3)  (Q)→(2)  (R)→(1)  (S)→(4)
  2. B
    (P)→(3)  (Q)→(2)  (R)→(5)  (S)→(4)(P) \rightarrow(3) ~~(Q) \rightarrow(2) ~~(R) \rightarrow(5)~~ (S) \rightarrow(4)(P)→(3)  (Q)→(2)  (R)→(5)  (S)→(4)
  3. C
    (P)→(2)  (Q)→(1)  (R)→(4)  (S)→(5)(P) \rightarrow(2)~~ (Q) \rightarrow(1) ~~ (R) \rightarrow(4) ~~ (S) \rightarrow(5)(P)→(2)  (Q)→(1)  (R)→(4)  (S)→(5)
  4. D
    (P)→(2)  (Q)→(1)  (R)→(1)  (S)→(3)(P) \rightarrow(2) ~~ (Q) \rightarrow(1) ~~ (R) \rightarrow(1) ~~ (S) \rightarrow(3)(P)→(2)  (Q)→(1)  (R)→(1)  (S)→(3)
View written solutionFree

Correct answer: A

The given system of linear equations is:

  1. x+2y+z=7x+2 y+z=7x+2y+z=7
  2. x+αz=11x+\alpha z=11x+αz=11
  3. 2x−3y+βz=γ2 x-3 y+\beta z=\gamma2x−3y+βz=γ

This system can be represented in the matrix form AX=BAX = BAX=B, where: A=(12110α2−3β)A = \begin{pmatrix} 1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta \end{pmatrix}A=​112​20−3​1αβ​​, X=(xyz)X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}X=​xyz​​, B=(711γ)B = \begin{pmatrix} 7 \\ 11 \\ \gamma \end{pmatrix}B=​711γ​​

Step 1: Analyze the determinant of the coefficient matrix A

The nature of the solution of the system of equations depends on the determinant of the coefficient matrix, det(A)det(A)det(A).

det(A) &= \begin{vmatrix} 1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta \end{vmatrix} \\ &= 1(0 \cdot \beta - (-3) \cdot \alpha) - 2(1 \cdot \beta - 2 \cdot \alpha) + 1(1 \cdot (-3) - 2 \cdot 0) \\ &= 1(3\alpha) - 2(\beta - 2\alpha) - 3 \\ &= 3\alpha - 2\beta + 4\alpha - 3 \\ &= 7\alpha - 2\beta - 3 \end{aligned}$$ ### Step 2: Establish conditions for different types of solutions * **Unique Solution:** The system has a unique solution if and only if $det(A) \neq 0$. $7\alpha - 2\beta - 3 \neq 0 \implies 2\beta \neq 7\alpha - 3 \implies \beta \neq \frac{1}{2}(7\alpha - 3)$. * **No Solution or Infinitely Many Solutions:** The system has either no solution or infinitely many solutions if and only if $det(A) = 0$. $7\alpha - 2\beta - 3 = 0 \implies \beta = \frac{1}{2}(7\alpha - 3)$. To distinguish between no solution and infinitely many solutions when $det(A) = 0$, we analyze the augmented matrix $[A|B]$ using row operations. $[A|B] = \left[ \begin{array}{ccc|c} 1 & 2 & 1 & 7 \\ 1 & 0 & \alpha & 11 \\ 2 & -3 & \beta & \gamma \end{array} \right]$ Apply row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - 2R_1$: $\left[ \begin{array}{ccc|c} 1 & 2 & 1 & 7 \\ 0 & -2 & \alpha-1 & 4 \\ 0 & -7 & \beta-2 & \gamma-14 \end{array} \right]$ Apply $R_3 \to R_3 - \frac{7}{2}R_2$: $\left[ \begin{array}{ccc|c} 1 & 2 & 1 & 7 \\ 0 & -2 & \alpha-1 & 4 \\ 0 & 0 & (\beta-2) - \frac{7}{2}(\alpha-1) & (\gamma-14) - \frac{7}{2}(4) \end{array} \right]$ Simplifying the last row: $C_{33} = \beta-2 - \frac{7}{2}\alpha + \frac{7}{2} = \beta - \frac{7}{2}\alpha + \frac{3}{2} = \frac{1}{2}(2\beta - 7\alpha + 3) = -\frac{1}{2}(7\alpha - 2\beta - 3) = -\frac{1}{2}det(A)$ $D_3 = \gamma - 14 - 14 = \gamma - 28$ The last row corresponds to the equation: $(-\frac{1}{2}det(A))z = \gamma - 28$. When $det(A) = 0$, the equation becomes $0 \cdot z = \gamma - 28$. * If $\gamma \neq 28$, we get $0 = \text{non-zero}$, which is impossible. The system is inconsistent and has **no solution**. * If $\gamma = 28$, we get $0 = 0$, which is always true. The system is consistent and has **infinitely many solutions** (since one variable can be chosen freely). ### Step 3: Match List-I with List-II **(P) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$** The first condition implies $det(A) = 7\alpha - 2(\frac{1}{2}(7\alpha-3)) - 3 = 7\alpha - (7\alpha - 3) - 3 = 0$. The conditions are $det(A) = 0$ and $\gamma=28$. This corresponds to **infinitely many solutions**. Thus, **(P) $\rightarrow$ (3)**. **(Q) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$** The first condition implies $det(A) = 0$. The second condition is $\gamma \neq 28$. This corresponds to **no solution**. Thus, **(Q) $\rightarrow$ (2)**. **(R) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$** The condition $\beta \neq \frac{1}{2}(7 \alpha-3)$ implies $det(A) \neq 0$. The system has a **unique solution**, regardless of the specific values of $\alpha$ and $\gamma$. Thus, **(R) $\rightarrow$ (1)**. **(S) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$** The condition $\beta \neq \frac{1}{2}(7 \alpha-3)$ implies $det(A) \neq 0$, so the system has a unique solution. Let's find this solution for the given parameters $\alpha=1$ and $\gamma=28$. The condition on $\beta$ becomes $\beta \neq \frac{1}{2}(7(1)-3) = \frac{4}{2}=2$, so $\beta \neq 2$. The system is: 1. $x+2y+z=7$ 2. $x+z=11$ 3. $2x-3y+\beta z=28$ From equation (2), $x+z=11$. Substitute this into equation (1): $(x+z) + 2y = 7 implies 11 + 2y = 7 implies 2y = -4 implies y = -2$. Substitute $x=11-z$ and $y=-2$ into equation (3): $2(11-z) - 3(-2) + \beta z = 28$ $22 - 2z + 6 + \beta z = 28$ $28 + (\beta-2)z = 28$ $(\beta-2)z = 0$ Since we are given that $\beta \neq 2$, the term $(\beta-2)$ is non-zero. Therefore, we must have $z=0$. Then, $x = 11 - z = 11 - 0 = 11$. The unique solution is $(x,y,z) = (11, -2, 0)$. This matches option (4) in List-II. Thus, **(S) $\rightarrow$ (4)**. *Note on option (5):* Let's check if $x=-15, y=4, z=0$ can be a solution. Substituting into the first equation: $-15 + 2(4) + 0 = -15 + 8 = -7$. This is not equal to 7, so this can never be a solution. ### Step 4: Final Conclusion The final matching is: * (P) $\rightarrow$ (3) * (Q) $\rightarrow$ (2) * (R) $\rightarrow$ (1) * (S) $\rightarrow$ (4) This corresponds to the option **A**.
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