Here's a detailed, step-by-step solution designed to be clear, elaborate, and educational for a JEE aspirant.

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1. Understanding the System of Linear Equations from Matrix Form

The problem provides us with three matrices: $A=\left[\begin{array}{cc}2 & -5 \\ 3 & m\end{array}\right], B=\left[\begin{array}{l}20 \\ m\end{array}\right], \text{ and } X=\left[\begin{array}{l}x \\ y\end{array}\right]$ The system of equations is given by $AX=B$. To understand this, we first perform the matrix multiplication $AX$: $A X = \left[\begin{array}{cc}2 & -5 \\ 3 & m\end{array}\right] \left[\begin{array}{l}x \\ y\end{array}\right] = \left[\begin{array}{c}2x + (-5)y \\ 3x + my\end{array}\right] = \left[\begin{array}{c}2x - 5y \\ 3x + my\end{array}\right]$ Now, equating this to matrix $B$: $\left[\begin{array}{c}2x - 5y \\ 3x + my\end{array}\right] = \left[\begin{array}{l}20 \\ m\end{array}\right]$ This gives us the following system of two linear equations in two variables $x$ and $y$: $(1) \quad 2x - 5y = 20$ $(2) \quad 3x + my = m$

Key Concept: A matrix equation $AX=B$ is a compact way to represent a system of linear equations. For a $2 \times 2$ matrix $A$ and column vectors $X$ and $B$, it translates to two linear equations.

2. Solving for $x$ and $y$ in terms of $m$

Our goal is to find expressions for $x$ and $y$ solely in terms of $m$. We can use either the substitution method or Cramer's rule. For this specific system, the substitution method seems straightforward.

Step 2.1: Express $x$ in terms of $y$ from equation (1) From equation (1), $2x = 20 + 5y$, so $x = \frac{20+5y}{2}$.

Step 2.2: Substitute $x$ into equation (2) to find $y$ in terms of $m$ Substitute $x = \frac{20+5y}{2}$ into equation (2): $3\left(\frac{20+5y}{2}\right) + my = m$ Multiply by 2 to clear the denominator: $3(20+5y) + 2my = 2m$ $60 + 15y + 2my = 2m$ Now, group terms containing $y$: $y(15 + 2m) = 2m - 60$ $y = \frac{2m - 60}{15 + 2m}$ We can factor out 2 from the numerator: $y = \frac{2(m - 30)}{15 + 2m}$

Step 2.3: Express $y$ in terms of $x$ from equation (1) From equation (1), $5y = 2x - 20$, so $y = \frac{2x-20}{5}$.

Step 2.4: Substitute $y$ into equation (2) to find $x$ in terms of $m$ Substitute $y = \frac{2x-20}{5}$ into equation (2): $3x + m\left(\frac{2x-20}{5}\right) = m$ Multiply by 5 to clear the denominator: $15x + m(2x-20) = 5m$ $15x + 2mx - 20m = 5m$ Group terms containing $x$: $x(15 + 2m) = 5m + 20m$ $x(15 + 2m) = 25m$ $x = \frac{25m}{15 + 2m}$

Tip: When solving systems of equations, always be mindful of potential denominators becoming zero. Here, $15+2m$ appears in both denominators. If $15+2m=0$ (i.e., $m = -15/2$), the system might have no unique solution, or infinitely many. We will address this in the next step.

3. Applying the Negative Solution Condition ($x<0$ and $y<0$)

The problem states that the system has a negative solution, meaning $x<0$ and $y<0$. We will use the expressions for $x$ and $y$ derived in the previous step to find the range of $m$ that satisfies these conditions.

Step 3.1: Condition for $y<0$ $y = \frac{2(m - 30)}{15 + 2m} < 0$ To solve this rational inequality, we find the critical points by setting the numerator and denominator to zero:

• Numerator: $2(m-30) = 0 \Rightarrow m = 30$
• Denominator: $15+2m = 0 \Rightarrow m = -\frac{15}{2}$

Now, we use a sign table or test intervals on a number line:

• If $m < -\frac{15}{2}$: Let $m=-10$. $\frac{2(-10-30)}{15+2(-10)} = \frac{2(-40)}{15-20} = \frac{-80}{-5} = 16 > 0$.
• If $-\frac{15}{2} < m < 30$: Let $m=0$. $\frac{2(0-30)}{15+2(0)} = \frac{-60}{15} = -4 < 0$.
• If $m > 30$: Let $m=40$. $\frac{2(40-30)}{15+2(40)} = \frac{2(10)}{15+80} = \frac{20}{95} > 0$.

So, $y<0$ when $m \in \left(-\frac{15}{2}, 30\right)$.

Common Mistake: Never cross-multiply when solving inequalities involving variables, as the sign of the multiplied term might be negative, flipping the inequality direction. Always use critical points and sign analysis.

Step 3.2: Condition for $x<0$ $x = \frac{25m}{15 + 2m} < 0$ Again, find the critical points:

• Numerator: $25m = 0 \Rightarrow m = 0$
• Denominator: $15+2m = 0 \Rightarrow m = -\frac{15}{2}$

Using a sign table or test intervals:

• If $m < -\frac{15}{2}$: Let $m=-10$. $\frac{25(-10)}{15+2(-10)} = \frac{-250}{15-20} = \frac{-250}{-5} = 50 > 0$.
• If $-\frac{15}{2} < m < 0$: Let $m=-1$. $\frac{25(-1)}{15+2(-1)} = \frac{-25}{15-2} = \frac{-25}{13} < 0$.
• If $m > 0$: Let $m=1$. $\frac{25(1)}{15+2(1)} = \frac{25}{17} > 0$.

So, $x<0$ when $m \in \left(-\frac{15}{2}, 0\right)$.

Step 3.3: Finding the interval $(a,b)$ for $m$ For the system to have a negative solution, BOTH $x<0$ AND $y<0$ must be true. Therefore, we need to find the intersection of the two intervals for $m$: $m \in \left(-\frac{15}{2}, 30\right) \cap \left(-\frac{15}{2}, 0\right)$ The intersection is $m \in \left(-\frac{15}{2}, 0\right)$. Thus, the interval $(a, b)$ is $\left(-\frac{15}{2}, 0\right)$. So, $a = -\frac{15}{2}$ and $b = 0$.

Important Note: The values $m = -15/2$ are excluded because they make the denominators zero, which means $x$ and $y$ would be undefined. This corresponds to the case where the determinant of the coefficient matrix $A$ is zero, meaning the system either has no unique solution or no solution at all (parallel or coincident lines). In either case, it wouldn't be a "negative solution" as defined.

4. Calculating the Determinant of $A$

The determinant of a $2 \times 2$ matrix $\left[\begin{array}{cc}p & q \\ r & s\end{array}\right]$ is given by $ps - qr$. For matrix $A=\left[\begin{array}{cc}2 & -5 \\ 3 & m\end{array}\right]$: $|A| = (2)(m) - (-5)(3) = 2m - (-15) = 2m + 15$

Key Concept: The determinant of the coefficient matrix is crucial for analyzing the nature of solutions to a system of linear equations. If $|A| \ne 0$, a unique solution exists. If $|A| = 0$, the system