Let's go through each question step by step based on the solutions derived.
### Question 17: Elementary Operations on the Augmented Matrix
Given the options:
(a) [tex]\(\left(\begin{array}{ccc|c}
1 & 1 & 1 & m+1 \\
0 & -3 & m & -1 \\
0 & 0 & (m-3)(m+4) & 2(m+4)
\end{array}\right)\)[/tex]
(b) [tex]\(\left(\begin{array}{ccc|c}
1 & 1 & 1 & m+1 \\
0 & -3 & m & -1 \\
0 & 0 & (m+3)(m-4) & -2(m-4)
\end{array}\right)\)[/tex]
(c) [tex]\(\left(\begin{array}{ccc|c}
1 & 1 & 1 & m+1 \\
0 & -3 & m & -1 \\
0 & 0 & (m-3)(m+4) & m-3
\end{array}\right)\)[/tex]
From the given answer, the correct transformation of the matrix is:
[tex]\[
\left(\begin{array}{ccc|c}
1 & 1 & 1 & m+1 \\
0 & -3 & m & -1 \\
0 & 0 & (m-3)(m+4) & 2(m+4)
\end{array}\right)
\][/tex]
Thus, the correct choice is:
(a)
### Question 18: Whether the System (S) Admits a Unique Solution
A system admits a unique solution if the determinant of matrix [tex]\(A\)[/tex] is not zero. The conditions for this to happen are given as:
(a) [tex]\(m \neq -3\)[/tex] and [tex]\(m \neq 4\)[/tex]
(b) [tex]\(m \neq 3\)[/tex] and [tex]\(m \neq -4\)[/tex]
(c) [tex]\(m \neq -3\)[/tex] and [tex]\(m \neq -4\)[/tex]
From the given answer, the unique solution conditions are:
[tex]\(m \neq -4\)[/tex] and [tex]\(m \neq 3\)[/tex]
Thus, the correct choice is:
(b)
### Question 19: When the System (S) Does Not Admit Solutions
A system does not admit solutions if a row reduces to [tex]\(0 = k\)[/tex] where [tex]\(k\)[/tex] is non-zero. Based on the solutions:
(a) [tex]\(m = 3\)[/tex]
(b) [tex]\(m = 4\)[/tex]
(c) [tex]\(m = -4\)[/tex]
From the given answer, the conditions when the system does not admit solutions are:
[tex]\(m = 3\)[/tex]
Thus, the correct choice is:
(a)
### Question 20: When the System (S) Admits an Infinity of Solutions
A system admits infinity of solutions when there is a free variable, typically due to a row reducing to all zeros in [tex]\(A\)[/tex] and corresponding entry in [tex]\(b\)[/tex] also being zero. Based on the solutions:
(a) [tex]\(m = 4\)[/tex] and [tex]\(y = \frac{1 - 4z}{3}\)[/tex] and [tex]\(x = -\frac{z - 12}{3}\)[/tex]
(b) [tex]\(m = -4\)[/tex] and [tex]\(y = \frac{1 - 4z}{3}\)[/tex] and [tex]\(x = \frac{z - 10}{3}\)[/tex]
(c) [tex]\(m = 3\)[/tex] and [tex]\(y = \frac{1 - 4z}{3}\)[/tex] and [tex]\(x = \frac{z - 12}{3}\)[/tex]
From the given answer, the conditions for an infinity of solutions are:
[tex]\(m = 4\)[/tex] and [tex]\(y = \frac{1 - 4z}{3}\)[/tex] and [tex]\(x = -\frac{z - 12}{3}\)[/tex]
Thus, the correct choice is:
(a)
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In summary, the answers are:
- Q17: (a)
- Q18: (b)
- Q19: (a)
- Q20: (a)
