Question

Let [\(\lambda\)] be the greatest integer less than or equal to \(\lambda\). The set of all values of \(\lambda\) for which the system of linear equations x+y+z=4, 3x+2y+5z=3, 9x + 4y + (28 + [\(\lambda\)])z = [\(\lambda\)] has a solution is :

• A. [-9, -8)
• B. (-\(\infty\), -9) \(\cup\) [-8, \(\infty\))
• C. R
• D. (-\(\infty\), -9) \(\cup\) (-9, \(\infty\))

Hint:

For a system of linear equations Ax=B, if det(A) \(\neq\) 0, there's always a unique solution. If det(A) = 0, you must check for consistency. A quick way is to check the relationship between the rows or columns. In this problem, notice that for the coefficient matrix when k=-9, \(R_3 = 3R_1 + 2R_2\) is not true, but \(R_3 = 6R_1 + 3R_2\) seems complex. The augmented matrix row operations are the most reliable method. If you get a row [0 0 0 | c] where c \(\neq\) 0, it's inconsistent. If you get [0 0 0 | 0], it's consistent.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We have a system of three linear equations where one coefficient and one constant term depend on the greatest integer function of \(\lambda\). We need to find all values of \(\lambda\) for which the system has a solution (is consistent).
Step 2: Simplify the System:
Let \(k = [\lambda]\), where k is an integer. The system becomes:
1) \(x + y + z = 4\)
2) \(3x + 2y + 5z = 3\)
3) \(9x + 4y + (28 + k)z = k\)
For a system to have a solution, it must be consistent. We can analyze this using the determinant of the coefficient matrix, A.
Step 3: Calculate the Determinant of the Coefficient Matrix:
\[ A = \begin{pmatrix} 1 & 1 & 1
3 & 2 & 5
9 & 4 & 28+k \end{pmatrix} \] \[ \text{det}(A) = 1(2(28+k) - 4(5)) - 1(3(28+k) - 9(5)) + 1(3(4) - 9(2)) \] \[ \text{det}(A) = (56 + 2k - 20) - (84 + 3k - 45) + (12 - 18) \] \[ \text{det}(A) = (36 + 2k) - (39 + 3k) - 6 \] \[ \text{det}(A) = 36 + 2k - 39 - 3k - 6 = -k - 9 \] Step 4: Analyze the Conditions for a Solution:
Case 1: Unique Solution
The system has a unique solution if \(\text{det}(A) \neq 0\).
\( -k - 9 \neq 0 \implies k \neq -9 \).
So, if \([\lambda] \neq -9\), the system is consistent and has a unique solution.
Case 2: Infinitely Many or No Solution
We must check for consistency when \(\text{det}(A) = 0\), which occurs when \(k = -9\).
This means we investigate the case where \([\lambda] = -9\). The system is:
1) \(x + y + z = 4\)
2) \(3x + 2y + 5z = 3\)
3) \(9x + 4y + 19z = -9\)
Let's use row operations to check for consistency. Consider the augmented matrix:
\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 4
3 & 2 & 5 & 3
9 & 4 & 19 & -9 \end{array} \right] \] Apply \(R_2 \rightarrow R_2 - 3R_1\) and \(R_3 \rightarrow R_3 - 9R_1\):
\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 4
0 & -1 & 2 & -9
0 & -5 & 10 & -45 \end{array} \right] \] Notice that the third row is exactly 5 times the second row (\(R_3 = 5R_2\)).
Apply \(R_3 \rightarrow R_3 - 5R_2\):
\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 4
0 & -1 & 2 & -9
0 & 0 & 0 & 0 \end{array} \right] \] Since we obtain a row of zeros (0 = 0), the system is consistent and has infinitely many solutions.
Step 5: Conclusion:
- If \([\lambda] \neq -9\), the system has a unique solution.
- If \([\lambda] = -9\), the system has infinitely many solutions.
In both cases, the system has a solution. Since \([\lambda]\) can be any integer, the system is consistent for all possible values of \([\lambda]\). Therefore, the system has a solution for all real values of \(\lambda\).
The set of all values is \(\mathbb{R}\).