Consider the matrices : $A = \begin{bmatrix} 2 & -5 \\ 3 & m \end{bmatrix}, B = \begin{bmatrix} 20 \\ m \end{bmatrix}$ and $X = \begin{bmatrix} x \\ y \end{bmatrix}$. 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_________

### Analyzing the Setup We are given the system $AX = B$, where: $$A = \begin{bmatrix} 2 & -5 \\ 3 & m \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 20 \\ m \end{bmatrix}$$ Translating this into linear equations, we obtain: $$2x - 5y = 20$$ $$3x + my = m$$ The intersection point $(x, y)$ must lie strictly in the third quadrant, which imposes the constraints **$x < 0$** and **$y < 0$**. ### The Power of Cramer's Rule To solve for $x$ and $y$, we first calculate the determinant of the coefficient matrix $A$: $$|A| = \begin{vmatrix} 2 & -5 \\ 3 & m \end{vmatrix} = 2m - (-15) = 2m + 15$$ Next, we calculate the determinants $\Delta_x$ and $\Delta_y$ by replacing the columns of $A$ with the constant vector $B$: $$\Delta_x = \begin{vmatrix} 20 & -5 \\ m & m \end{vmatrix} = 20m - (-5m) = 25m$$ $$\Delta_y = \begin{vmatrix} 2 & 20 \\ 3 & m \end{vmatrix} = 2m - 60$$ Using Cramer's Rule, the coordinates are: $$x = \frac{25m}{2m + 15}, \quad y = \frac{2m - 60}{2m + 15}$$ ### The Inequality Dance For $x < 0$, we require: $$\frac{25m}{2m + 15} < 0$$ Using the wavy curve method with critical points at $m = 0$ and $m = -7.5$, we find $x < 0$ when **$m \in (-7.5, 0)$**. For $y < 0$, we require: $$\frac{2m - 60}{2m + 15} 0$ within the interval $(-7.5, 0)$, the absolute value is redundant. We compute the definite integral: $$8 \left[ m^2 + 15m \right]_{-7.5}^{0}$$ Substituting the limits: $$8 \times [ (0) - ((-7.5)^2 + 15(-7.5)) ]$$ $$8 \times [ 0 - (56.25 - 112.5) ] = 8 \times [ 56.25 ]$$ The final result is **450**.

JEE Main 2024 (09 Apr Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Matrices and Determinants: 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 $A X = B$ has a negative solution (i.e., $x < 0$ and $y < 0$ ), be the interval $(a, b)$ . Then $8 \int_{a}^{b} ∣ A ∣ d m$ is equal to_________

Q: Why can't we just use substitution to solve for x and y? It feels more intuitive.

Substitution is a valid algebraic path, but it becomes cumbersome when dealing with a parameter like $m$ in the denominator. Cramer's Rule is the 'surgical' tool here; it allows us to isolate $x$ and $y$ as clean rational functions of $m$ immediately, which is exactly what we need for the inequality analysis.

Q: Why do we need the intersection of the two intervals for m?

Great question! The problem requires the solution $(x, y)$ to be in the third quadrant. This means $x < 0$ AND $y < 0$ must be true simultaneously. If we only satisfied one, the intersection point could land in the second or fourth quadrant. We need the overlap where both conditions hold.

Q: Wait, why did we drop the absolute value sign in the integral?

Always check the sign of the expression inside the absolute value within your limits of integration. Our interval for $m$ is $(-7.5, 0)$. If you test a value in this range, say $m = -1$, then $|A| = 2(-1) + 15 = 13$, which is positive. Since the expression is always positive in our interval, $|A| = 2m + 15$.

Enter Numerical Value:

Visualized Solution

Geometric Interpretation

System of equations represents two intersecting lines.
Condition $x < 0$ and $y < 0$ means intersection is in the 3rd quadrant.

Extracting Linear Equations

Given: $A X = B$
$[23 - 5 m] [x y] = [20 m]$
Equation 1: $2 x - 5 y = 20$
Equation 2: $3 x + m y = m$

Determinant of Coefficient Matrix $∣ A ∣$

Using Cramer's Rule, first find $∣ A ∣$ .
$∣ A ∣ = 23 - 5 m$
$∣ A ∣ = 2 m - (- 15) = 2 m + 15$

Solving for $x$

$Δ_{x} = 20 m - 5 m = 20 m - (- 5 m) = 25 m$
$x = \frac{Δ _{x}}{∣ A ∣} = \frac{25 m}{2 m + 15}$

Solving for $y$

$Δ_{y} = 23 20 m = 2 m - 60$
$y = \frac{Δ _{y}}{∣ A ∣} = \frac{2 m - 60}{2 m + 15}$

Condition for $x < 0$

We need $x < 0 ⟹ \frac{25 m}{2 m + 15} < 0$
Critical points: $m = 0, m = - 7.5$
Interval for $x < 0$ : $m \in (- 7.5, 0)$

Condition for $y < 0$

We need $y < 0 ⟹ \frac{2 m - 60}{2 m + 15} < 0$
Critical points: $m = 30, m = - 7.5$
Interval for $y < 0$ : $m \in (- 7.5, 30)$

Finding Interval $(a, b)$

Both conditions must hold simultaneously.
Intersection: $m \in (- 7.5, 0) \cap (- 7.5, 30)$
Result: $m \in (- 7.5, 0)$
Thus, $a = - 7.5$ and $b = 0$

Formulating the Integral

Evaluate: $8 \int_{a}^{b} ∣ A ∣ d m$
Substitute $a = - 7.5, b = 0$
Substitute $∣ A ∣ = 2 m + 15$
Integral: $8 \int_{- 7.5}^{0} (2 m + 15) d m$

Evaluating the Antiderivative

$\int (2 m + 15) d m = \frac{2 m ^{2}}{2} + 15 m$
Antiderivative: $m^{2} + 15 m$
Expression: $8 [m^{2} + 15 m]_{- 7.5}^{0}$

Applying Integration Limits

Upper limit ( $0$ ): $0^{2} + 15 (0) = 0$
Lower limit ( $- 7.5$ ): $(- 7.5)^{2} + 15 (- 7.5)$
$= 56.25 - 112.5 = - 56.25$
Result: $8 [0 - (- 56.25)]$

Final Calculation

$8 \times 56.25 = 450$
The value of the integral is $450$ .

00:00 / 00:00

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Solution Diagram

Analyzing the Setup

We are given the system

A X = B

, where:

A = [23 - 5 m], X = [x y], B = [20 m]

Translating this into linear equations, we obtain:

2 x - 5 y = 20

3 x + m y = m

The intersection point

(x, y)

must lie strictly in the third quadrant, which imposes the constraints $x < 0$ and $y < 0$ .

The Power of Cramer's Rule

To solve for

x

and

y

, we first calculate the determinant of the coefficient matrix

A

∣ A ∣ = 23 - 5 m = 2 m - (- 15) = 2 m + 15

Next, we calculate the determinants

Δ_{x}

and

Δ_{y}

by replacing the columns of

A

with the constant vector

B

Δ_{x} = 20 m - 5 m = 20 m - (- 5 m) = 25 m

Δ_{y} = 23 20 m = 2 m - 60

Using Cramer's Rule, the coordinates are:

x = \frac{25 m}{2 m + 15}, y = \frac{2 m - 60}{2 m + 15}

The Inequality Dance

For

x < 0

, we require:

\frac{25 m}{2 m + 15} < 0

Using the wavy curve method with critical points at

m = 0

and

m = - 7.5

, we find

x < 0

when $m \in (- 7.5, 0)$ .

For

y < 0

, we require:

\frac{2 m - 60}{2 m + 15} < 0

With critical points at

m = 30

and

m = - 7.5

, we find

y < 0

when $m \in (- 7.5, 30)$ .

To satisfy both conditions, we take the intersection of these sets:

(- 7.5, 0) \cap (- 7.5, 30) = (- 7.5, 0)

Thus, our interval is

(a, b) = (- 7.5, 0)

Final Calculation

We evaluate the integral

8 \int_{a}^{b} ∣ A ∣ d m

8 \int_{- 7.5}^{0} (2 m + 15) d m

Since

(2 m + 15) > 0

within the interval

(- 7.5, 0)

, the absolute value is redundant. We compute the definite integral:

8 [m^{2} + 15 m]_{- 7.5}^{0}

Substituting the limits:

8 \times [(0) - ((- 7.5)^{2} + 15 (- 7.5))]

8 \times [0 - (56.25 - 112.5)] = 8 \times [56.25]

The final result is 450.

Common Traps & FAQs

Why can't we just use substitution to solve for x and y? It feels more intuitive.

Substitution is a valid algebraic path, but it becomes cumbersome when dealing with a parameter like

m

in the denominator. Cramer's Rule is the 'surgical' tool here; it allows us to isolate

x

and

y

as clean rational functions of

m

immediately, which is exactly what we need for the inequality analysis.

Why do we need the intersection of the two intervals for m?

Great question! The problem requires the solution

(x, y)

to be in the third quadrant. This means

x < 0

AND

y < 0

must be true simultaneously. If we only satisfied one, the intersection point could land in the second or fourth quadrant. We need the overlap where both conditions hold.

Wait, why did we drop the absolute value sign in the integral?

Always check the sign of the expression inside the absolute value within your limits of integration. Our interval for

m

(- 7.5, 0)

. If you test a value in this range, say

m = - 1

, then

∣ A ∣ = 2 (- 1) + 15 = 13

, which is positive. Since the expression is always positive in our interval,

∣ A ∣ = 2 m + 15

Enter Numerical Value:

Visualized Solution

Geometric Interpretation

Extracting Linear Equations

Determinant of Coefficient Matrix $∣ A ∣$

Solving for $x$

Solving for $y$

Condition for $x < 0$

Condition for $y < 0$

Finding Interval $(a, b)$

Formulating the Integral

Evaluating the Antiderivative

Applying Integration Limits

Final Calculation

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Power of Cramer's Rule

The Inequality Dance

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Geometric Interpretation

Extracting Linear Equations

Determinant of Coefficient Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣A∣

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0

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Finding Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a,b)

Formulating the Integral

Evaluating the Antiderivative

Applying Integration Limits

Final Calculation

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Power of Cramer's Rule

The Inequality Dance

Final Calculation

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