Let the system of equations x + 5y - z = 1, 4x + 3y - 3z = 7, 24x + y + λz = μ, λ, μ ∈ R, have infin

Q: Why do we need $\Delta_1 = 0$ as well? Isn't $\Delta = 0$ enough?

That is a fantastic question! $\Delta = 0$ only tells us that the system *could* have infinitely many solutions or no solutions at all (it is the condition for the system to be singular). To ensure the system is actually consistent—meaning the planes intersect along a line rather than being parallel and non-intersecting—we must ensure the auxiliary determinants are also zero. Without this, you might be solving for a system that has no solutions!

Q: How do I know which variable to set as the parameter $k$?

You can choose any variable! However, looking at the equations, expressing $y$ as $k$ often simplifies the arithmetic significantly. The key is to ensure that when you substitute back, you maintain the integrity of the integer constraints. As long as your parameterization is consistent with the original equations, the choice of variable is purely a matter of algebraic convenience.

Q: Why do we reject non-integer values of $k$?

The problem explicitly states that $x$, $y$, and $z$ are integers. If $k$ is not an integer, then the resulting $x$, $y$, and $z$ will not be integers. Therefore, we only count the integer values of $k$ that satisfy the inequality. It is a crucial constraint that filters our solution set!

Question

Let the system of equations $x + 5y - z = 1$, $4x + 3y - 3z = 7$, $24x + y + \lambda z = \mu$, $\lambda, \mu \in \mathbb{R}$, have infinitely many solutions. Then the number of the solutions of this system, If $x, y, z$ are integers and satisfy $7 \le x + y + z \le 77$, is

Options:

(A) 3

(B) 6

(C) 5

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional space, looking at three planes defined by the equations:
$x + 5y - z = 1$
$4x + 3y - 3z = 7$
$24x + y + \lambda z = \mu$

Usually, three planes intersect at a single point. However, this system possesses infinitely many solutions, meaning the planes intersect along a common line. To capture this, we must ensure the system is singular by setting the main determinant $\Delta$ to zero.

$$
\Delta = 
\begin{vmatrix} 
1 & 5 & -1 \ 
4 & 3 & -3 \ 
24 & 1 & \lambda 
\end{vmatrix} = 0
$$

Expanding along the first row, we obtain:
$$1(3\lambda + 3) - 5(4\lambda + 72) - 1(4 - 72) = 0$$

This simplifies to $-17\lambda - 289 = 0$, which yields the value **$\lambda = -17$**.

### The Consistency Check

Finding $\lambda$ is only the first step. A system can have $\Delta = 0$ and still be inconsistent if the planes are parallel and non-coincident. To guarantee the planes intersect, we must ensure the auxiliary determinant $\Delta_1$ is also zero.

We replace the first column with the constants $1, 7, \mu$:

$$
\Delta_1 = 
\begin{vmatrix} 
1 & 5 & -1 \ 
7 & 3 & -3 \ 
\mu & 1 & -17 
\end{vmatrix} = 0
$$

Expanding this determinant, we find **$\mu = 45$**. The system is now perfectly consistent.

### The Power of Parameterization

We now have a system with one degree of freedom. To find integer solutions, we express $x$ and $y$ in terms of $z$ using the first two equations:
$x + 5y = z + 1$
$4x + 3y = 3z + 7$

By multiplying the first equation by $4$ and subtracting the second, we eliminate $x$:
$$17y = z - 3 \Rightarrow y = \frac{z - 3}{17}$$

For $y$ to be an integer, $z - 3$ must be a multiple of $17$. Let $z - 3 = 17k$, where $k$ is an integer. This yields:
$z = 17k + 3$
$y = k$

Substituting these into the first equation, we find $x = 12k + 4$. We have successfully parameterized the entire solution set using the integer $k$.

### The Final Constraint

The problem requires solutions satisfying the condition $7 \le x + y + z \le 77$. We calculate the sum $S = x + y + z$ in terms of $k$:
$$S = (12k + 4) + (k) + (17k + 3) = 30k + 7$$

Applying the constraint:
$$7 \le 30k + 7 \le 77$$

Subtracting $7$ from all sides, we get $0 \le 30k \le 70$, which simplifies to $0 \le k \le 2.33$. Since $k$ must be an integer, the possible values are **$k \in \{0, 1, 2\}$**.

Each value of $k$ corresponds to a unique integer triplet $(x, y, z)$. Thus, there are exactly **$3$ solutions**.

Select Answer:

Visualized Solution

$Δ = 0$

$λ = - 17$

$Δ_{1} = 0$

$μ = 45$

$y = \frac{z - 3}{17}$

$z = 17 k + 3$

$x = 12 k + 4$

$S = 30 k + 7$

$0 \leq k \leq 2.33$

$k \in {0, 1, 2}$

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

Analyzing the Setup

The Consistency Check

The Power of Parameterization

The Final Constraint

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Select Answer:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=−17

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ1​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ=45

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=17z−3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=17k+3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=12k+4

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S=30k+7

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0≤k≤2.33

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k∈{0,1,2}