The system of equations -kx + 3y - 14z = 25, -15x + 4y - kz = 3, -4x + y + 3z = 4 is consistent for

Q: Why do we check $\Delta_z$ specifically? Can't we check $\Delta_x$?

Great question! You can check any of them. If $\Delta = 0$, the system is inconsistent if *any* of $\Delta_x, \Delta_y, \Delta_z$ is non-zero. I chose $\Delta_z$ because it was convenient, but checking $\Delta_x$ or $\Delta_y$ would yield the same result.

Q: What if $\Delta = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$? Does that mean it's consistent?

Exactly! In that case, the system could have infinitely many solutions, which is still a form of consistency. However, in this specific problem, we found that $\Delta_z \neq 0$ at the critical points, so we didn't need to worry about the infinite solutions case.

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

Gaussian elimination is powerful, but for a $3 \times 3$ system with a parameter $k$, Cramer's Rule is often the most direct way to identify the exact values of $k$ that cause inconsistency. It keeps the algebra clean and focused.

Question

The system of equations $-kx + 3y - 14z = 25$, $-15x + 4y - kz = 3$, $-4x + y + 3z = 4$ is consistent for all $k$ in the set

Options:

(A) $\mathbb{R}$

(B) $\mathbb{R} - \{-11, 13\}$

(C) $\mathbb{R} - \{13\}$

(D) $\mathbb{R} - \{-11, 11\}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to peel back the layers of a classic linear algebra problem. Imagine you are standing in a 3D room, and these three equations represent three planes.

Consistency is simply the question: "Do these planes meet?" If they meet at a single point, the system is consistent with a unique solution. If they meet along a line or coincide, they are still consistent, but with infinitely many solutions. If they never meet, the system is inconsistent.

Our goal is to find the values of $k$ for which this system is consistent.

### The Determinant as a Gatekeeper

To analyze this, we use the main coefficient determinant, $\Delta$. This determinant is the gatekeeper of our system. If $\Delta 
eq 0$, the system is guaranteed to have a unique solution, and thus, it is consistent.

The matrix is formed by the coefficients of $x$, $y$, and $z$:

$$
\Delta = \begin{vmatrix} -k & 3 & -14 \ -15 & 4 & -k \ -4 & 1 & 3 \end{vmatrix}
$$

Expanding this along the first row, we get:

$$
\Delta = -k(4(3) - 1(-k)) - 3((-15)(3) - (-4)(-k)) - 14((-15)(1) - (-4)(4))
$$

Simplifying this, we find:

$$
\Delta = -k(12 + k) - 3(-45 - 4k) - 14(-15 + 16)
$$

This reduces beautifully to:

$$
\Delta = 121 - k^2
$$

### The Danger Zones

Now, the plot thickens. If $\Delta = 0$, we are in a "danger zone" where the system might be inconsistent. Setting $121 - k^2 = 0$ gives us $k = 11$ and $k = -11$.

For any other value of $k$, the system is perfectly consistent. But what happens at these two critical points? We must test them.

We use Cramer's Rule: if $\Delta = 0$ and at least one of the auxiliary determinants ($\Delta_x, \Delta_y, \Delta_z$) is non-zero, the system is inconsistent.

Let's test $k = 11$ by calculating $\Delta_z$:

$$
\Delta_z = \begin{vmatrix} -11 & 3 & 25 \ -15 & 4 & 3 \ -4 & 1 & 4 \end{vmatrix}
$$

After careful expansion, we find $\Delta_z = 26$, which is clearly not zero. Since $\Delta = 0$ and $\Delta_z 
eq 0$, the system is inconsistent at $k = 11$.

Repeating this for $k = -11$, we find $\Delta_z = 312$, which is also non-zero. Thus, the system is inconsistent at both $k = 11$ and $k = -11$.

### The Final Verdict

We have systematically dismantled the problem. The system is consistent for all real numbers except for the two points where it breaks down.

Therefore, the set of consistent values is **$\mathbb{R} - \{-11, 11\}$**.

You have successfully navigated the logic of linear systems. Keep this analytical mindset, and no problem will ever be too daunting!

Animated Solution for Mathematics - Matrices and Determinants: The system of equations $- k x + 3 y - 14 z = 25$ , $- 15 x + 4 y - k z = 3$ , $- 4 x + y + 3 z = 4$ is consistent for all $k$ in the set

Select Answer:

Visualized Solution

Conditions for Consistency

Forming the Determinant $Δ$

Expanding $Δ$ along $R_{1}$

Simplifying $Δ$

Critical Values for $Δ = 0$

Evaluating $Δ_{z}$ at $k = 11$

Evaluating $Δ_{z}$ at $k = - 11$

Final Solution Set

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

Analyzing the Setup

The Determinant as a Gatekeeper

The Danger Zones

The Final Verdict

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The system of equations −kx+3y−14z=25, −15x+4y−kz=3, −4x+y+3z=4 is consistent for all k in the set

Select Answer:

Visualized Solution

Conditions for Consistency

Forming the Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ

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

Critical Values for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ=0

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