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MCQMediumJEE 2024Solving Linear Equations (Matrix Method)

JEE Mathematics 2024 Question with Solution

Consider the matrices A=[2−53m]A = \begin{bmatrix} 2 & -5 \\ 3 & m \end{bmatrix}, B=[20m]B = \begin{bmatrix} 20 \\ m \end{bmatrix}, and X=[xy]X = \begin{bmatrix} x \\ y \end{bmatrix}. Let the set of all mm, for which the system of equations AX=BAX = B has a negative solution (i.e., x<0x < 0 and y<0y < 0), be the interval (a,b)(a, b). Then 8∫ba∣A∣ dm8\int_b^a |A| \, dm is equal to:

  • A

    450450

  • B

    400400

  • C

    500500

  • D

    350350

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

A=[2−53m],B=[20m],X=[xy]A = \begin{bmatrix} 2 & -5 \\ 3 & m \end{bmatrix}, \quad B = \begin{bmatrix} 20 \\ m \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}

Find: The value of 8∫ba∣A∣ dm8\int_b^a |A| \, dm where AX=BAX = B has a negative solution.

From AX=BAX = B, the system is

2x−5y=202x - 5y = 20

and

3x+my=m3x + my = m

From the working,

y=2m−602m+15y = \frac{2m - 60}{2m + 15}

For y<0y < 0, we get

−152<m<30-\frac{15}{2} < m < 30

Also,

x=25m2m+15x = \frac{25m}{2m + 15}

For x<0x < 0, we get

−152<m<0-\frac{15}{2} < m < 0

Thus the required interval is

(a,b)=(−152,0)(a,b) = \left(-\frac{15}{2}, 0\right)

The determinant of matrix AA is

∣A∣=2m+15|A| = 2m + 15

Over (−152,0)\left(-\frac{15}{2}, 0\right), we have ∣2m+15∣=2m+15|2m+15| = 2m+15.

Therefore,

8∫−1520(2m+15) dm=8[m2+15m]−15208 \int_{-\frac{15}{2}}^{0} (2m + 15) \, dm = 8 \left[ m^2 + 15m \right]_{-\frac{15}{2}}^{0}

Now,

[m2+15m]−1520=0−(2254−2252)=2254\left[ m^2 + 15m \right]_{-\frac{15}{2}}^{0} = 0 - \left( \frac{225}{4} - \frac{225}{2} \right) = \frac{225}{4}

Hence,

8×2254=4508 \times \frac{225}{4} = 450

Therefore, the correct option is A.

Algebraic Derivation

Given:

2x−5y=20,3x+my=m2x - 5y = 20, \quad 3x + my = m

Find: The interval of mm for which x<0x < 0 and y<0y < 0, then evaluate the integral.

From

2x−5y=202x - 5y = 20

we get

2x=20+5y2x = 20 + 5y

and hence

x=10+52yx = 10 + \frac{5}{2}y

Substitute into the second equation:

3(10+52y)+my=m3\left(10 + \frac{5}{2}y\right) + my = m

So,

30+152y+my=m30 + \frac{15}{2}y + my = m

which gives

y(152+m)=m−30y\left(\frac{15}{2} + m\right) = m - 30

Thus,

y=m−30m+152=2m−602m+15y = \frac{m-30}{m+\frac{15}{2}} = \frac{2m-60}{2m+15}

For y<0y < 0, numerator and denominator must have opposite signs, so

−152<m<30-\frac{15}{2} < m < 30

Now use

x=10+52⋅m−30m+152x = 10 + \frac{5}{2} \cdot \frac{m-30}{m+\frac{15}{2}}

Simplifying,

x=25m2m+15x = \frac{25m}{2m+15}

For x<0x < 0, we get

−152<m<0-\frac{15}{2} < m < 0

Hence,

(a,b)=(−152,0)(a,b) = \left(-\frac{15}{2}, 0\right)

and

∣A∣=det⁡(A)=2m+15|A| = \det(A) = 2m + 15

Therefore,

8∫−1520∣2m+15∣ dm=8∫−1520(2m+15) dm=4508\int_{-\frac{15}{2}}^0 |2m+15| \, dm = 8\int_{-\frac{15}{2}}^0 (2m+15) \, dm = 450

So the final answer is 450450.

Common mistakes

  • Students often find the interval for only y<0y < 0 and stop at −152<m<30-\frac{15}{2} < m < 30. This is incomplete because the question requires both x<0x < 0 and y<0y < 0. Intersect both conditions before integrating.

  • A common mistake is using ∣A∣|A| as the matrix itself instead of the determinant. Here ∣A∣=det⁡(A)=2m+15|A| = \det(A) = 2m + 15, not the matrix AA.

  • Some students reverse the limits incorrectly from the interval (a,b)(a,b). Since a=−152a = -\frac{15}{2} and b=0b = 0, identify the correct bounds carefully before evaluating the integral.

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