To solve the system of linear equations, we start by forming the coefficient matrix [tex]\(A\)[/tex] and the constant vector [tex]\(b\)[/tex]:
[tex]\[
A = \begin{pmatrix}
2 & 3 & 5 \\
-4 & -2 & 4 \\
-6 & -1 & 13
\end{pmatrix}
\][/tex]
[tex]\[
b = \begin{pmatrix}
-23 \\
9 \\
-5
\end{pmatrix}
\][/tex]
Next, we need to determine whether this system has a unique solution, no solution, or infinitely many solutions. This can be achieved by examining the rank of the coefficient matrix [tex]\(A\)[/tex] and the augmented matrix [tex]\([A|b]\)[/tex].
1. Calculate the rank of [tex]\(A\)[/tex]:
To find the rank, reduce [tex]\(A\)[/tex] to its row echelon form or reduced row echelon form. After performing row operations:
[tex]\[
A = \begin{pmatrix}
2 & 3 & 5 \\
-4 & -2 & 4 \\
-6 & -1 & 13
\end{pmatrix}
\rightarrow
\begin{pmatrix}
2 & 3 & 5 \\
0 & 4 & 14 \\
0 & 8 & 28
\end{pmatrix}
\rightarrow
\begin{pmatrix}
2 & 3 & 5 \\
0 & 4 & 14 \\
0 & 0 & 0
\end{pmatrix}
\][/tex]
The rank of [tex]\(A\)[/tex] is 2, as there are two non-zero rows.
2. Calculate the rank of the augmented matrix [tex]\([A|b]\)[/tex]:
Form the augmented matrix [tex]\([A|b]\)[/tex] and reduce it:
[tex]\[
[A|b] = \begin{pmatrix}
2 & 3 & 5 & -23 \\
-4 & -2 & 4 & 9 \\
-6 & -1 & 13 & -5
\end{pmatrix}
\rightarrow
\begin{pmatrix}
2 & 3 & 5 & -23 \\
0 & 4 & 14 & -37 \\
0 & 0 & 0 & 0
\end{pmatrix}
\][/tex]
The rank of [tex]\([A|b]\)[/tex] is also 2, since there are two non-zero rows in the row echelon form, and the last row, although consisting purely of zeros in the coefficients, does not violate consistency.
Since the rank of [tex]\(A\)[/tex] is equal to the rank of [tex]\([A|b]\)[/tex], but less than the number of unknowns (which is 3), the system has infinitely many solutions and the equations are dependent. This is because there are fewer independent equations than variables, allowing for free parameters.
Based on this information, the correct answer is:
b. Infinitely many solutions; dependent
