Maths Q20 – JEE-MAIN 2025

If the system of linear equations

3 x + y + β z = 3

2 x + α y - z = - 3

x + 2 y + z = 4

has infinitely many solutions, then the value of

22 β - 9 α

is :

Clue 1 of 3: Conditions for Infinitely Many Solutions

For a system of linear equations $A X = B$ to have infinitely many solutions, the determinant of the coefficient matrix $D$ must be zero, and all determinants $D_{x}, D_{y}, D_{z}$ (obtained by replacing a column of $A$ with $B$ ) must also be zero.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Set up the coefficient matrix and apply the condition for infinitely many solutions✦ Active

For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix ( $D$ ) must be zero, and all determinants $D_{x}, D_{y}, D_{z}$ (obtained by replacing a column of the coefficient matrix with the constant terms) must also be zero. The coefficient matrix $A$ and the constant vector $B$ are:

A = (\begin{matrix} 3 & 1 & β \\ 2 & α & - 1 \\ 1 & 2 & 1 \end{matrix}), B = (\begin{matrix} 3 \\ - 3 \\ 4 \end{matrix})

The condition $D = det (A) = 0$ gives:

3 (α - (- 2)) - 1 (2 - (- 1)) + β (4 - α) = 0

3 (α + 2) - 1 (3) + β (4 - α) = 0

3 α + 6 - 3 + 4 β - α β = 0

3 α + 3 + 4 β - α β = 0 (Equation 1)

Step 2: Calculate

D_{y}

and solve for

β

○ Expand

Let's calculate $D_{y}$ by replacing the second column of $A$ with $B$ :

D_{y} = | \begin{matrix} 3 & 3 & β \\ 2 & - 3 & - 1 \\ 1 & 4 & 1 \end{matrix} |

Setting $D_{y} = 0$ :

3 (- 3 - (- 4)) - 3 (2 - (- 1)) + β (8 - (- 3)) = 0

3 (1) - 3 (3) + β (11) = 0

3 - 9 + 11 β = 0

- 6 + 11 β = 0 ⟹ 11 β = 6 ⟹ β = \frac{6}{11}

💡 Teacher's Secret Hint

Choosing $D_{y}$ or $D_{z}$ first can sometimes simplify calculations if one of them leads to a direct value for $α$ or $β$ .

Step 3: Substitute

β

into Equation 1 to find

α

, then calculate the required expression○ Expand

Substitute $β = \frac{6}{11}$ into Equation 1:

3 α + 3 + 4 (\frac{6}{11}) - α (\frac{6}{11}) = 0

3 α + 3 + \frac{24}{11} - \frac{6 α}{11} = 0

Multiply by 11 to clear denominators:

33 α + 33 + 24 - 6 α = 0

27 α + 57 = 0

27 α = - 57 ⟹ α = - \frac{57}{27} = - \frac{19}{9}

Now, calculate the value of $22 β - 9 α$ :

22 (\frac{6}{11}) - 9 (- \frac{19}{9}) = (2 \times 6) - (- 19) = 12 + 19 = 31

💡 Teacher's Secret Hint

Always double-check your arithmetic, especially with fractions and negative signs.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 20 – JEE-MAIN 2025

Clue 1 of 3: Conditions for Infinitely Many Solutions

🧠 Full Solution Path