Which one is correct, the following system of linear equations 2x – 3y + 4z = 7, 3x – 4y + 5z = 8, 4x – 5y + 6z = 9 has?

The correct option is Infinitely many solutions.

To understand why this system of linear equations has infinitely many solutions, we can analyze the system step-by-step using matrix methods. Let us write down the given system of linear equations:

1) $2x-3y+4z=7$
2) $3x-4y+5z=8$
3) $4x-5y+6z=9$

We can represent this system as an augmented matrix $\left[A\right|B]$:

$\left[\begin{array}{}2& -3& 4& 7\end{array}\right]$ (Row 1, denoted as $R_1$)
$\left[\begin{array}{}3& -4& 5& 8\end{array}\right]$ (Row 2, denoted as $R_2$)
$\left[\begin{array}{}4& -5& 6& 9\end{array}\right]$ (Row 3, denoted as $R_3$)

Let us perform elementary row operations to reduce this matrix to row echelon form.

First, we can replace Row 3 ($R_3$) with $R_3-2R_1$:

$R_3\to R_3-2R_1$
For the x-coefficient: $4-2\left(2\right)=0$
For the y-coefficient: $-5-2(-3)=-5+6=1$
For the z-coefficient: $6-2\left(4\right)=6-8=-2$
For the constant term: $9-2\left(7\right)=9-14=-5$

Now, our intermediate augmented matrix has the new Row 3:

$\left[\begin{array}{}0& 1& -2& -5\end{array}\right]$

Next, we perform the row operation $R_2\to 2R_2-3R_1$ to eliminate x from Row 2:

For the x-coefficient: $2\left(3\right)-3\left(2\right)=6-6=0$
For the y-coefficient: $2(-4)-3(-3)=-8+9=1$
For the z-coefficient: $2\left(5\right)-3\left(4\right)=10-12=-2$
For the constant term: $2\left(8\right)-3\left(7\right)=16-21=-5$

Now, we compare the modified Row 2 and Row 3:

New Row 2:
New Row 3:

Since Row 2 and Row 3 are identical, subtracting Row 2 from Row 3 ($R_3\to R_3-R_2$) results in a row of all zeros:

$\left[\begin{array}{}0& 0& 0& 0\end{array}\right]$

This means the rank of both the coefficient matrix $A$ and the augmented matrix $\left[A\right|B]$ is equal to 2. Because the rank (2) is less than the number of variables (3: x, y, and z), the system is consistent and has at least one free variable, which leads to infinitely many solutions.