Let S be the set of all column matrices (b1, b2, b3) such that b1, b2, b3 ∈ R and the system of equa

Q: Wait, why does $\Delta = 0$ force a condition on $b_1, b_2, b_3$?

Think of the system $AX = B$ as a transformation. When $\Delta = 0$, the matrix $A$ collapses the 3D space into a 2D plane or a 1D line. It loses information. If the vector $B$ does not lie within this 'collapsed' space (the column space), there is no way to reach it, and the system is inconsistent. The condition on $b_1, b_2, b_3$ is simply the mathematical requirement that $B$ must lie within the reachable space of $A$.

Q: Can't we just use Cramer's Rule to solve this?

Cramer's Rule is a powerful tool, but it has a fatal flaw: it requires $\Delta \neq 0$. When $\Delta = 0$, the formula involves division by zero, making it undefined. In such cases, Cramer's Rule cannot tell you if the system has infinitely many solutions or no solution at all. You must rely on the consistency conditions derived from row operations or linear combinations.

Q: Why did we check the determinant for the options instead of solving the system?

In JEE Advanced, time is your most precious currency. Solving a system of equations takes minutes; calculating a determinant takes seconds. If the determinant is non-zero, the system is guaranteed to have a unique solution for *any* $B$. Since the question asks for systems that work for *all* $B \in S$, and $S$ is a subset of $\mathbb{R}^3$, any system with $\Delta \neq 0$ is automatically a correct answer. It is the most elegant path to the solution.

Question

Let $S$ be the set of all column matrices $\begin{pmatrix} b_1 \ b_2 \ b_3 \end{pmatrix}$ such that $b_1, b_2, b_3 \in R$ and the system of equations (in real variables) $-x + 2y + 5z = b_1$ $2x - 4y + 3z = b_2$ $x - 2y + 2z = b_3$ has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each $\begin{pmatrix} b_1 \ b_2 \ b_3 \end{pmatrix} \in S$?

Options:

(A) $x+2y+3z = b_1, 4y + 5z = b_2$ and $x + 2y + 6z = b_3$

(B) $x + y + 3z = b_1, 5x + 2y + 6z = b_2$ and $-2x - y - 3z = b_3$

(C) $x + 2y - 5z = b_1, 2x – 4y + 10z = b_2$ and $x – 2y + 5z = b_3$

(D) $x + 2y + 5z = b_1, 2x + 3z = b_2$ and $x + 4y – 5z = b_3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In a three-dimensional system of linear equations, the determinant $\Delta$ of the coefficient matrix $A$ dictates the geometric behavior of the planes. When $\Delta = 0$, the planes are linearly dependent, meaning they do not intersect at a unique point. Instead, they may be parallel or intersect along a common line, leading to either infinite solutions or no solution at all.

We are given the system:
$$-x + 2y + 5z = b_1$$
$$2x - 4y + 3z = b_2$$
$$x - 2y + 2z = b_3$$

To verify the singularity, we calculate the determinant of the coefficient matrix $A$:
$$\Delta = -1(-8 + 6) - 2(4 - 3) + 5(-4 + 4) = 2 - 2 = 0$$
The system is indeed singular.

### Decoding the DNA of the System

Because $\Delta = 0$, the equations are linearly dependent. We seek a relationship of the form $P_2 + \lambda P_1 = \mu P_3$. By comparing the coefficients of $x, y,$ and $z$, we establish the following ratios:

$$\frac{2 - \lambda}{1} = \frac{-4 + 2\lambda}{-2} = \frac{3 + 5\lambda}{2} = \mu$$

Solving this system yields $\lambda = \frac{1}{7}$ and $\mu = \frac{13}{7}$. This relationship represents the "DNA" of the system.

For the system to be consistent, the constants $b_1, b_2, b_3$ must satisfy the same linear combination:
$$b_2 + \frac{1}{7}b_1 = \frac{13}{7}b_3$$

Multiplying by $7$, we obtain the condition for the set $S$:
$$b_1 + 7b_2 = 13b_3$$

### The Shortcut to Victory

To find systems that possess a solution for every vector $B \in S$, we look for systems where the coefficient matrix is invertible. If a matrix has a non-zero determinant ($\Delta 
eq 0$), it acts as an invertible map, ensuring a unique solution for any vector $B$ in $\mathbb{R}^3$.

If $\Delta 
eq 0$, the system is universally consistent. We do not need to check the consistency of each option against $S$ individually; we simply identify which matrices are non-singular.

### Testing the Candidates

**Testing Option A:**
The coefficient matrix is:
$$A_A = \begin{pmatrix} 1 & 2 & 3 \ 0 & 4 & 5 \ 1 & 2 & 6 \end{pmatrix}$$

Calculating the determinant:
$$\Delta_A = 1(24 - 10) - 0 + 1(10 - 12) = 14 - 2 = 12$$
Since $\Delta_A 
eq 0$, **Option A is a valid solution.**

**Testing Option B and C:**
In Option B, the third column is a multiple of the second, resulting in $\Delta = 0$. In Option C, the first equation is a multiple of the second, also resulting in $\Delta = 0$. These systems are singular and do not guarantee a solution for all $B \in S$.

**Testing Option D:**
The coefficient matrix is:
$$A_D = \begin{pmatrix} 1 & 2 & 5 \ 2 & 0 & 3 \ 1 & 4 & -5 \end{pmatrix}$$

Calculating the determinant:
$$\Delta_D = 1(0 - 12) - 2(-10 - 3) + 5(8 - 0) = -12 + 26 + 40 = 54$$
Since $\Delta_D 
eq 0$, **Option D is also a valid solution.**

Through the elegance of linear algebra, we conclude that the systems with non-zero determinants are **Option A and Option D**.

Select Answer:

Visualized Solution

System of Linear Equations

Determinant of Matrix $A$

Linear Dependence of Equations

Setting up the Relation

Solving for $λ$

Condition for Consistency (Set $S$ )

Testing the Options

Evaluating Option A

Evaluating Option B

Evaluating Option C

Evaluating Option D

Conclusion

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

Analyzing the Setup

Decoding the DNA of the System

The Shortcut to Victory

Testing the Candidates

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