Consider the matrices: \(\mathrm{A}=\left[\begin{array}{cc}2 & -5 \\ 3 & \mathrm{~m}\end{array}\right], \mathrm{B}=\left[\begin{array}{c}20 \\ \mathrm{~m}\end{array}\right]\) and \(\mathrm{X}=\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y}\end{array}\right]\). Let the set of all m, for which the system of equations \(\mathrm{AX}=\mathrm{B}\) has a negative solution (i.e., \(x<0\) and \(y<0\)), be the interval (\(a, b\)). Then \(8 \int_{\mathrm{a}}^{\mathrm{b}}|\mathrm{A}| \mathrm{dm}\) is equal to ____.

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Matrices

Consider the matrices:

A = [23 - 5 m], B = [20 m]

and

X = [x y]

. Let the set of all m, for which the system of equations

AX = B

has a negative solution (i.e.,

x < 0

and

y < 0

), be the interval (

a, b

).

Then

8 \int_{a}^{b} ∣ A ∣ dm

is equal to ____.

Answer: 450

A = (23 - 5 m)

B = (20 m)

: system

2 x - 5 y = 20

3 x + m y = m

.

Solving by Cramer's rule:

∣ A ∣ = 2 m + 15

x = \frac{20 m - ( - 5 ) ( m )}{2 m + 15} = \frac{25 m}{2 m + 15}, y = \frac{2 m - 60}{2 m + 15}

For

y < 0

\frac{2 m - 60}{2 m + 15} < 0 ⟹ m \in (- \frac{15}{2}, 30)

.

For

x < 0

\frac{25 m}{2 m + 15} < 0 ⟹ m \in (- \frac{15}{2}, 0)

.

Both conditions together:

m \in (- \frac{15}{2}, 0) = (a, b)

8 \int_{a}^{b} ∣ A ∣ d m = 8 \int_{- 15/2}^{0} (2 m + 15) d m = 8 [m^{2} + 15 m]_{- 15/2}^{0} = 8 (0 - (\frac{225}{4} - \frac{225}{2})) = 8 \cdot \frac{225}{4} = 450

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Integer answer