Determinants
System of equations determinant
nta_pyq_2025_apr
Grade 12
Question:
Let the system of equations:
$$2x + 3y + 5z = 9$$
$$7x + 3y - 2z = 8$$
$$12x + 3y - (4 + \lambda)z = 16 - \mu$$
have infinitely many solutions. Then the radius of the circle centred at $(\lambda, \mu)$ and touching the line $4x = 3y$ is
Step-by-Step Solution
Key Concept: Apply the matrix property for system of equations determinant and reduce it to determinant or parameter equations.
For infinitely many solutions, we need:
$$\begin{vmatrix}
2 & 3 & 5 \\
7 & 3 & -2 \\
12 & 3 & -(\lambda + 4)
\end{vmatrix} = 0$$
Expanding:
$$2(3(-(\lambda + 4)) + 6) - 3(7(-(\lambda + 4)) + 24) + 5(21 - 36) = 0$$
$$2(-3\lambda - 12 + 6) - 3(-7\lambda - 28 + 24) + 5(-15) = 0$$
$$2(-3\lambda - 6) - 3(-7\lambda - 4) - 75 = 0$$
$$-6\lambda - 12 + 21\lambda + 12 - 75 = 0$$
$$15\lambda - 75 = 0$$
$$\lambda = 5$$
Also, we need:
$$\begin{vmatrix}
9 & 3 & 5 \\
8 & 3 & -2 \\
16 - \mu & 3 & -(\lambda + 4)
\end{vmatrix} = 0$$
With $\lambda = 5$:
$$\begin{vmatrix}
9 & 3 & 5 \\
8 & 3 & -2 \\
16 - \mu & 3 & -9
\end{vmatrix} = 0$$
Expanding:
$$9(-27 + 6) - 3(-72 + 2(16 - \mu)) + 5(24 - 3(16 - \mu)) = 0$$
$$9(-21) - 3(-72 + 32 - 2\mu) + 5(24 - 48 + 3\mu) = 0$$
$$-189 - 3(-40 - 2\mu) + 5(-24 + 3\mu) = 0$$
$$-189 + 120 + 6\mu - 120 + 15\mu = 0$$
$$21\mu - 189 = 0$$
$$\mu = 9$$
Centre of circle: $(5, 9)$
Radius = distance from $(5, 9)$ to line $4x - 3y = 0$:
$$\text{radius} = \frac{|4(5) - 3(9)|}{\sqrt{16 + 9}} = \frac{|20 - 27|}{5} = \frac{7}{5}$$
Correct Answer: 2