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The system of linear equations x + y + z = 2, 2x + y – z = 3, 3x + 2y + kz = 4 has a unique solution if
- a)k = 0
- b)For all non zero k
- c)–1 < k < 1
- d)– 2 < k < 2
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The system of linear equations x + y + z = 2, 2x + y – z = 3, 3x...
Solution:
Given system of linear equations is
x + y + z = 2 ...(1)
2x + y + z = 3 ...(2)
3x + 2y + kz = 4 ...(3)
We can write equations (1) and (2) in matrix form as AX = B, where
A = \begin{pmatrix}
1 & 1 & 1\\
2 & 1 & 1\\
\end{pmatrix},
X = \begin{pmatrix}
x\\
y\\
z\\
\end{pmatrix},
B = \begin{pmatrix}
2\\
3\\
\end{pmatrix}
The solution to AX = B is given by X = A^-1B, where A^-1 is the inverse of matrix A.
Now, let's find the inverse of matrix A.
We have A = \begin{pmatrix}
1 & 1 & 1\\
2 & 1 & 1\\
\end{pmatrix}
To find A^-1, we need to find the determinant of matrix A. We have
|A| = \begin{vmatrix}
1 & 1 & 1\\
2 & 1 & 1\\
\end{vmatrix}
= 1(1-1) - 1(2-1) + 1(2-1)
= 0
Since |A| = 0, the matrix A is singular and does not have an inverse. This means that the system of equations (1) and (2) either has no solution or infinitely many solutions.
Now, let's consider equation (3).
If k = 0, then equation (3) becomes
3x + 2y = 4
This is a system of two linear equations with two unknowns, and it has a unique solution. Therefore, the given system of equations has a unique solution if k = 0.
If k is nonzero, then equation (3) is a system of three linear equations with three unknowns. Since equations (1) and (2) either have no solution or infinitely many solutions, we cannot determine whether the system of equations (1), (2), and (3) has a unique solution or not.
Therefore, option B is the correct answer.
Given system of linear equations is
x + y + z = 2 ...(1)
2x + y + z = 3 ...(2)
3x + 2y + kz = 4 ...(3)
We can write equations (1) and (2) in matrix form as AX = B, where
A = \begin{pmatrix}
1 & 1 & 1\\
2 & 1 & 1\\
\end{pmatrix},
X = \begin{pmatrix}
x\\
y\\
z\\
\end{pmatrix},
B = \begin{pmatrix}
2\\
3\\
\end{pmatrix}
The solution to AX = B is given by X = A^-1B, where A^-1 is the inverse of matrix A.
Now, let's find the inverse of matrix A.
We have A = \begin{pmatrix}
1 & 1 & 1\\
2 & 1 & 1\\
\end{pmatrix}
To find A^-1, we need to find the determinant of matrix A. We have
|A| = \begin{vmatrix}
1 & 1 & 1\\
2 & 1 & 1\\
\end{vmatrix}
= 1(1-1) - 1(2-1) + 1(2-1)
= 0
Since |A| = 0, the matrix A is singular and does not have an inverse. This means that the system of equations (1) and (2) either has no solution or infinitely many solutions.
Now, let's consider equation (3).
If k = 0, then equation (3) becomes
3x + 2y = 4
This is a system of two linear equations with two unknowns, and it has a unique solution. Therefore, the given system of equations has a unique solution if k = 0.
If k is nonzero, then equation (3) is a system of three linear equations with three unknowns. Since equations (1) and (2) either have no solution or infinitely many solutions, we cannot determine whether the system of equations (1), (2), and (3) has a unique solution or not.
Therefore, option B is the correct answer.
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