If the system of equations \beginaligned & 2 x+3 y-z=5 & x+\alpha y+3 z=-4 & 3 x-y+\beta z=7 \endaligned has infinitely many solutions, then 13 \alpha \beta is equal to :

Matrices and Determinants PYQ — JEE Main 2024 · 1 Feb · Shift 1 · Q42 with Solution

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Correct answer: B

For a system of 3 linear equations in 3 variables to have infinitely many solutions, we need:

$\operatorname{rank}(A)=\operatorname{rank}([A|B])<3$

So the coefficient matrix must be singular, and the augmented matrix must be consistent with the same dependence.

The system is:

2x+3y-z&=5 \\ x+\alpha y+3z&=-4 \\ 3x-y+\beta z&=7 \end{aligned}$$ Coefficient matrix: $$A=\begin{pmatrix} 2&3&-1\\ 1&\alpha&3\\ 3&-1&\beta \end{pmatrix}$$ Augmented matrix: $$[A|B]=\begin{pmatrix} 2&3&-1&5\\ 1&\alpha&3&-4\\ 3&-1&\beta&7 \end{pmatrix}$$ --- 2. Since infinitely many solutions exist, one equation must be a linear combination of the other two. So let $$R_3=pR_1+qR_2$$ where $R_1,R_2,R_3$ are the three equations including constants. Thus, comparing coefficients of $x,y,z$ and constants: $$3=2p+q \quad ...(1)$$ $$-1=3p+\alpha q \quad ...(2)$$ $$\beta=-p+3q \quad ...(3)$$ $$7=5p-4q \quad ...(4)$$ --- 3. Solve for $p,q$ using (1) and (4): From (1): $$q=3-2p$$ Substitute into (4): $$7=5p-4(3-2p)=5p-12+8p=13p-12$$ $$13p=19$$ $$p=\frac{19}{13}$$ Then $$q=3-2\cdot \frac{19}{13}=\frac{39-38}{13}=\frac{1}{13}$$ --- 4. Now find $\alpha$ from (2): $$-1=3p+\alpha q$$ $$-1=3\cdot \frac{19}{13}+\alpha\cdot \frac{1}{13}$$ $$-1=\frac{57+\alpha}{13}$$ $$-13=57+\alpha$$ $$\alpha=-70$$ --- 5. Find $\beta$ from (3): $$\beta=-p+3q=-\frac{19}{13}+3\cdot \frac{1}{13}=-\frac{16}{13}$$ So, $$\alpha\beta=(-70)\left(-\frac{16}{13}\right)=\frac{1120}{13}$$ Hence, $$13\alpha\beta=1120$$ --- 6. Therefore the correct option is: $$\boxed{1120}$$ So option **B** is correct.

View written solutionFree

Correct answer: B

For a system of 3 linear equations in 3 variables to have infinitely many solutions, we need:

$\operatorname{rank}(A)=\operatorname{rank}([A|B])<3$

So the coefficient matrix must be singular, and the augmented matrix must be consistent with the same dependence.

The system is:

2x+3y-z&=5 \\ x+\alpha y+3z&=-4 \\ 3x-y+\beta z&=7 \end{aligned}$$ Coefficient matrix: $$A=\begin{pmatrix} 2&3&-1\\ 1&\alpha&3\\ 3&-1&\beta \end{pmatrix}$$ Augmented matrix: $$[A|B]=\begin{pmatrix} 2&3&-1&5\\ 1&\alpha&3&-4\\ 3&-1&\beta&7 \end{pmatrix}$$ --- 2. Since infinitely many solutions exist, one equation must be a linear combination of the other two. So let $$R_3=pR_1+qR_2$$ where $R_1,R_2,R_3$ are the three equations including constants. Thus, comparing coefficients of $x,y,z$ and constants: $$3=2p+q \quad ...(1)$$ $$-1=3p+\alpha q \quad ...(2)$$ $$\beta=-p+3q \quad ...(3)$$ $$7=5p-4q \quad ...(4)$$ --- 3. Solve for $p,q$ using (1) and (4): From (1): $$q=3-2p$$ Substitute into (4): $$7=5p-4(3-2p)=5p-12+8p=13p-12$$ $$13p=19$$ $$p=\frac{19}{13}$$ Then $$q=3-2\cdot \frac{19}{13}=\frac{39-38}{13}=\frac{1}{13}$$ --- 4. Now find $\alpha$ from (2): $$-1=3p+\alpha q$$ $$-1=3\cdot \frac{19}{13}+\alpha\cdot \frac{1}{13}$$ $$-1=\frac{57+\alpha}{13}$$ $$-13=57+\alpha$$ $$\alpha=-70$$ --- 5. Find $\beta$ from (3): $$\beta=-p+3q=-\frac{19}{13}+3\cdot \frac{1}{13}=-\frac{16}{13}$$ So, $$\alpha\beta=(-70)\left(-\frac{16}{13}\right)=\frac{1120}{13}$$ Hence, $$13\alpha\beta=1120$$ --- 6. Therefore the correct option is: $$\boxed{1120}$$ So option **B** is correct.