Key Concept: Conditions for a System of Linear Equations to Have No Solution

For a system of linear equations in three variables, say $AX=B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix, we use determinants to determine the nature of solutions. Let $\Delta = \text{det}(A)$ be the determinant of the coefficient matrix. Let $\Delta_x, \Delta_y, \Delta_z$ be the determinants obtained by replacing the column of coefficients of $x, y, z$ respectively with the column of constant terms (the $B$ matrix).

A system of linear equations has no solution (inconsistent) if and only if:

1. The determinant of the coefficient matrix, $\Delta$, is equal to zero ($\Delta = 0$).
2. At least one of the determinants $\Delta_x, \Delta_y, \Delta_z$ is non-zero (i.e., $\Delta_x \neq 0$ or $\Delta_y \neq 0$ or $\Delta_z \neq 0$).

If $\Delta = 0$ and all of $\Delta_x, \Delta_y, \Delta_z$ are also zero, then the system has infinitely many solutions.

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Step-by-Step Derivation

1. Formulate the Coefficient Matrix (A) and Constant Matrix (B) The given system of linear equations is: $3x - 2y + z = b$ $5x - 8y + 9z = 3$ $2x + y + az = -1$

From these equations, we can write the coefficient matrix $A$ and the constant matrix $B$: $A = \begin{pmatrix} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & a \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} b \\ 3 \\ -1 \end{pmatrix}$ Explanation: We extract the coefficients of $x, y, z$ to form the matrix $A$ and the constant terms on the right-hand side to form the matrix $B$.

2. Calculate $\Delta$ and Determine the Value of 'a' for $\Delta = 0$ For the system to have no solution, the first condition is $\Delta = \text{det}(A) = 0$. Let's calculate $\Delta$: $\Delta = \begin{vmatrix} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & a \end{vmatrix}$ Expand the determinant along the first row: $\Delta = 3 \begin{vmatrix} -8 & 9 \\ 1 & a \end{vmatrix} - (-2) \begin{vmatrix} 5 & 9 \\ 2 & a \end{vmatrix} + 1 \begin{vmatrix} 5 & -8 \\ 2 & 1 \end{vmatrix}$ $\Delta = 3((-8)(a) - (9)(1)) + 2((5)(a) - (9)(2)) + 1((5)(1) - (-8)(2))$ $\Delta = 3(-8a - 9) + 2(5a - 18) + 1(5 + 16)$ $\Delta = -24a - 27 + 10a - 36 + 21$ $\Delta = (-24a + 10a) + (-27 - 36 + 21)$ $\Delta = -14a - 63 + 21$ $\Delta = -14a - 42$

Now, set $\Delta = 0$ to find the value of $a$: $-14a - 42 = 0$ $-14a = 42$ $a = \frac{42}{-14}$ $a = -3$ Explanation: Setting $\Delta=0$ is a necessary condition for both "no solution" and "infinitely many solutions". If $\Delta \ne 0$, there would be a unique solution, which contradicts the problem statement. So, $a$ must be $-3$.

3. Calculate $\Delta_x$ and Determine the Value(s) of 'b' for No Solution Now that we have $a = -3$, we need to ensure that at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero for the system to have no solution. Let's calculate $\Delta_x$. $\Delta_x$ is obtained by replacing the first column of $A$ with the constant matrix $B$: $\Delta_x = \begin{vmatrix} b & -2 & 1 \\ 3 & -8 & 9 \\ -1 & 1 & -3 \end{vmatrix}$ Expand the determinant along the first row: $\Delta_x = b \begin{vmatrix} -8 & 9 \\ 1 & -3 \end{vmatrix} - (-2) \begin{vmatrix} 3 & 9 \\ -1 & -3 \end{vmatrix} + 1 \begin{vmatrix} 3 & -8 \\ -1 & 1 \end{vmatrix}$ $\Delta_x = b((-8)(-3) - (9)(1)) + 2((3)(-3) - (9)(-1)) + 1((3)(1) - (-8)(-1))$ $\Delta_x = b(24 - 9) + 2(-9 + 9) + 1(3 - 8)$ $\Delta_x = b(15) + 2(0) + 1(-5)$ $\Delta_x = 15b - 5$

For the system to have no solution, we need $\Delta_x \neq 0$ (or $\Delta_y \neq 0$ or $\Delta_z \neq 0$). If $\Delta_x \neq 0$ while $\Delta=0$, the system is guaranteed to have no solution. $15b - 5 \neq 0$ $15b \neq 5$ $b \neq \frac{5}{15}$ $b \neq \frac{1}{3}$ Explanation: We need to find values of $b$ that make the system inconsistent. With $a=-3$ (making $\Delta=0$), if $\Delta_x$ is non-zero, then there's no solution. If $\Delta_x$ were zero, we would proceed to check $\Delta_y$ and $\Delta_z$.

4. Combine Conditions and Check the Options From our calculations, for the system to have no solution, we must have:

1. $a = -3$
2. $b \neq \frac{1}{3}$

Let's evaluate the given options based on these conditions:

• (A) $\left( {3,{1 \over 3}} \right)$: Here $a=3$. If $a=3$, then $\Delta = -14(3) - 42 = -42 - 42 = -84$. Since $\Delta = -84 \neq 0$, the system has a unique solution, not no solution.
• (B) $\left( { - 3,{1 \over 3}} \right)$: Here $a=-3$ and $b=1/3$.
  • For $a=-3$, $\Delta = 0$.
  • For $b=1/3$, $\Delta_x = 15(1/3) - 5 = 5 - 5 = 0$. Since $\Delta=0$ and $\Delta_x=0$, we would need to check $\Delta_y$ and $\Delta_z$. $\Delta_y = \begin{vmatrix} 3 & 1/3 & 1 \\ 5 & 3 & 9 \\ 2 & -1 & -3 \end{vmatrix} = 3(-9+9) - (1/3)(-15+9) + 1(-5-6) = 0 - (1/3)(-33) - 11 = 11 - 11 = 0$. $\Delta_z = \begin{vmatrix} 3 & -2 & 1/3 \\ 5 & -8 & 3 \\ 2 & 1 & -1 \end{vmatrix} = 3(8-3) - (-2)(-5-6) + (1/3)(5+16) = 3(5) + 2(-11) + (1/3)(21) = 15 - 22 + 7 = 0$. Since $\Delta = \Delta_x = \Delta_y = \Delta_z = 0$, the system has infinitely many solutions, not no solution.
• (C) $\left( { - 3, - {1 \over 3}} \right)$: Here $a=-3$ and $b=-1/3$.
  • For $a=-3$, $\Delta = 0$.
  • For $b=-1/3$, $\Delta_x = 15(-1/3) - 5 = -5 - 5 = -10$. Since $\Delta = 0$ and $\Delta_x = -10 \neq 0$, the system has no solution. This matches our derived conditions.
• (D) $\left( {3, - {1 \over 3}} \right)$: Here $a=3$. As explained for option (A), if $a=3$, then $\Delta = -84 \neq 0$, meaning the system has a unique solution, not no solution.

Therefore, the ordered pair $(a, b)$ for which the system has no solution is $(-3, -1/3)$.

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Tips for Success & Common Mistakes

• Determinant Calculation: Be extremely careful with signs and arithmetic when calculating determinants. A single sign error can lead to an incorrect value for $a$ or $b$.
• Conditions for Solutions: Memorize and clearly understand the conditions for unique solution ($\Delta \neq 0$), no solution ($\Delta = 0$ and at least one $\Delta_i \neq 0$), and infinitely many solutions ($\Delta = 0$ and all $\Delta_i = 0$). This is a fundamental concept for linear systems.
• Systematic Approach: Always start by calculating $\Delta$ first. This narrows down the possibilities for 'a'. Then, substitute the value(s) of 'a' into $\Delta_x, \Delta_y, \Delta_z$ to find the conditions for 'b'.
• Check All Conditions: If $\Delta = 0$ and one $\Delta_i = 0$, you must check the other $\Delta_j$ to distinguish between no solution and infinitely many solutions.

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Summary and Key Takeaway

For a system of linear equations to have no solution, the determinant of the coefficient matrix ($\Delta$) must be zero, AND at least one of the determinants formed by replacing a column of coefficients with the constant terms ($\Delta_x, \Delta_y, \Delta_z$) must be non-zero.

In this problem:

1. We found that $\Delta = -14a - 42$. Setting $\Delta = 0$ yields $a = -3$.
2. With $a = -3$, we calculated $\Delta_x = 15b - 5$. For no solution, we require $\Delta_x \neq 0$, which means $b \neq 1/3$.
3. Combining these, the system has no solution if $a=-3$ and $b \neq 1/3$.
4. Among the given options, only option (C) $(-3, -1/3)$ satisfies these conditions. Options (A) and (D) lead to unique solutions because $\Delta \neq 0$. Option (B) leads to infinitely many solutions because all determinants ($\Delta, \Delta_x, \Delta_y, \Delta_z$) are zero.

The final answer is $\boxed{\text{(A)}}$.