If the system of equations 2x + 7y + λz = 3, 3x + 2y + 5z = 4, x + μy + 32z = -1 has infinitely many

Q: Wait, why do we need $D_1 = D_2 = D_3 = 0$ as well? Isn't $D = 0$ enough for infinite solutions?

Great question! $D = 0$ only tells us that the system does not have a unique solution. It could mean the system has *no* solution (inconsistent) or *infinitely many* solutions (dependent). To guarantee infinite solutions, we need the planes to be consistent, which is ensured by $D_1 = D_2 = D_3 = 0$.

Q: Why did we choose $D_3$ to find $\mu$ first? Could we have started with $D$?

You *could* start with $D$, but it would be a trap! $D$ contains both $\lambda$ and $\mu$. By choosing $D_3$, we strategically eliminate $\lambda$ from the equation, allowing us to solve for $\mu$ directly. Always look for the path of least resistance in algebra.

Q: Is there a faster way than Cramer's Rule? Maybe Gaussian elimination?

Gaussian elimination is a powerful tool, and yes, it works perfectly here! However, for a $3 \times 3$ system with parameters, Cramer's Rule is often more systematic and less prone to row-operation errors if you are comfortable with determinant expansion. Both methods are valid; choose the one where you make fewer calculation mistakes!

Question

If the system of equations $2x + 7y + \lambda z = 3$, $3x + 2y + 5z = 4$, $x + \mu y + 32z = -1$ has infinitely many solutions, then $(\lambda - \mu)$ is equal to________

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are not just solving a system of equations; we are exploring the beautiful, rigid, yet flexible world of linear algebra.

Imagine you are standing in a room where three walls represent our equations. Usually, these walls meet at a single point—a unique solution.

But today, the problem tells us something special: the system has **infinitely many solutions**. This means our three planes do not just meet at a point; they intersect along a common line.

To capture this, we use the condition that the main determinant $D$ must be zero, and the auxiliary determinants $D_1, D_2, D_3$ must also be zero.

### The Strategic Hunt for $\mu$

We have two unknowns, $\lambda$ and $\mu$. If we jump straight into the main determinant $D$, we will be stuck with an equation containing both variables. That is a dead end.

Instead, let us be tactical. Look at the third column of our coefficient matrix. It contains $\lambda$.

If we construct $D_3$ by replacing this column with the constant terms $(3, 4, -1)$, the $\lambda$ terms vanish! We are left with:

$$D_3 = \begin{vmatrix} 2 & 7 & 3 \ 3 & 2 & 4 \ 1 & \mu & -1 \end{vmatrix} = 0$$

Expanding this along the first row, we get:

$$2((2)(-1) - (4)(\mu)) - 7((3)(-1) - (4)(1)) + 3((3)(\mu) - (2)(1)) = 0$$

Simplifying this, we find:

$$2(-2 - 4\mu) - 7(-3 - 4) + 3(3\mu - 2) = 0$$

This leads to:

$$-4 - 8\mu + 49 + 9\mu - 6 = 0$$

Combining like terms, we get $\mu + 39 = 0$, so **$\mu = -39$**. We have successfully isolated our first variable!

### The Final Piece of the Puzzle

Now that we know $\mu = -39$, the main determinant $D$ is no longer a mystery. We substitute $\mu$ back into the matrix:

$$D = \begin{vmatrix} 2 & 7 & \lambda \ 3 & 2 & 5 \ 1 & -39 & 32 \end{vmatrix} = 0$$

Expanding along the first row again, we have:

$$2((2)(32) - (5)(-39)) - 7((3)(32) - (5)(1)) + \lambda((3)(-39) - (2)(1)) = 0$$

This looks intimidating, but stay calm. The arithmetic is just a series of steps:

$$2(64 + 195) - 7(96 - 5) + \lambda(-117 - 2) = 0$$

This simplifies to:

$$2(259) - 7(91) - 119\lambda = 0$$

$$518 - 637 - 119\lambda = 0$$

Finally, $-119 - 119\lambda = 0$, giving us **$\lambda = -1$**.

With $\lambda = -1$ and $\mu = -39$, the final step is a simple subtraction:

$$\lambda - \mu = (-1) - (-39) = 38$$

And there you have it! The elegance of the cancellation is the reward for your patience. Keep practicing, and you will see these patterns everywhere. The final answer is **38**.

Animated Solution for Mathematics - Matrices and Determinants: If the system of equations $2 x + 7 y + λ z = 3$ , $3 x + 2 y + 5 z = 4$ , $x + μ y + 32 z = - 1$ has infinitely many solutions, then $(λ - μ)$ is equal to________

Enter Numerical Value:

Visualized Solution

Condition for Infinite Solutions

Constructing $D_{3}$

Expanding $D_{3}$ along Row 1

Solving for $μ$

Constructing $D$

Expanding $D$ along Row 1

Simplifying the Equation

Solving for $λ$

Finding $(λ - μ)$

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

Analyzing the Setup

The Strategic Hunt for $μ$

The Final Piece of the Puzzle

Similar Questions

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Enter Numerical Value:

Visualized Solution

Condition for Infinite Solutions

Constructing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D3​

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D3​ along Row 1

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

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

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D along Row 1

Simplifying the Equation

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

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (λ−μ)

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

Analyzing the Setup

The Strategic Hunt for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ

The Final Piece of the Puzzle

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