Given a system of equations \(x\; + \;2y\; + \;2z\; = \;{b_1},\;\;\;\;\;5x\; + \;y\; + \;3z\; = \;{b_2}\) Which of the following is true about solutions?
The system will have infinitely many solutions for any given \({b_1}\) and \({b_2}\)
System of Equations Analysis
We are given a system of two linear equations with three variables:
- Equation 1: \(x + 2y + 2z = b_1\)
- Equation 2: \(5x + y + 3z = b_2\)
The question asks about the nature of the solutions for this system, specifically whether it has a unique solution, infinitely many solutions, no solution, or if the existence of a solution depends on the values of \(b_1\) and \(b_2\). To determine this, we can use methods from linear algebra, such as analyzing the ranks of the coefficient matrix and the augmented matrix.
Matrix Representation of the System
The system can be represented in matrix form as \(AX = B\), where:
- \(A\) is the coefficient matrix:
- \(X\) is the variable matrix:
- \(B\) is the constant matrix:
The coefficient matrix \(A\) is:
| \(A = \begin{bmatrix} 1 & 2 & 2 \\ 5 & 1 & 3 \end{bmatrix}\) |
The variable matrix \(X\) is:
| \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\) |
The constant matrix \(B\) is:
| \(B = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}\) |
The augmented matrix \([A|B]\) combines the coefficient matrix and the constant matrix:
| \([A|B] = \begin{bmatrix} 1 & 2 & 2 & | & b_1 \\ 5 & 1 & 3 & | & b_2 \end{bmatrix}\) |
Understanding Solution Types via Rank
The number of solutions to a system of linear equations depends on the relationship between the rank of the coefficient matrix (\(rank(A)\)) and the rank of the augmented matrix (\(rank([A|B])\)), compared to the number of variables (\(n\)).
- If \(rank(A) = rank([A|B]) = n\), the system has a unique solution.
- If \(rank(A) = rank([A|B]) < n\), the system has infinitely many solutions.
- If \(rank(A) < rank([A|B])\), the system has no solution.
In our system, the number of variables is \(n=3\).
Step-by-Step Row Reduction
To find the ranks, we use Gaussian elimination to transform the augmented matrix into row echelon form.
Start with the augmented matrix:
| \(\begin{bmatrix} 1 & 2 & 2 & | & b_1 \\ 5 & 1 & 3 & | & b_2 \end{bmatrix}\) |
We perform the row operation \(R_2 \leftarrow R_2 - 5R_1\) to eliminate the first element in the second row:
- New Row 2 = Row 2 - 5 * (Row 1)
- \(5 - 5 \times 1 = 0\)
- \(1 - 5 \times 2 = 1 - 10 = -9\)
- \(3 - 5 \times 2 = 3 - 10 = -7\)
- \(b_2 - 5 \times b_1 = b_2 - 5b_1\)
The matrix after the operation becomes:
| \(\begin{bmatrix} 1 & 2 & 2 & | & b_1 \\ 0 & -9 & -7 & | & b_2 - 5b_1 \end{bmatrix}\) |
Final Conclusion on Solutions
Now we determine the ranks:
- Rank of Coefficient Matrix \(A\): After row reduction, the coefficient part of the matrix is \(\begin{bmatrix} 1 & 2 & 2 \\ 0 & -9 & -7 \end{bmatrix}\). There are two non-zero rows, so \(rank(A) = 2\).
- Rank of Augmented Matrix \([A|B]\): The augmented matrix is \(\begin{bmatrix} 1 & 2 & 2 & | & b_1 \\ 0 & -9 & -7 & | & b_2 - 5b_1 \end{bmatrix}\). Since the first row is non-zero and the second row is also non-zero (as \(-9\) and \(-7\) are non-zero coefficients, regardless of \(b_1, b_2\)), there are two non-zero rows. Thus, \(rank([A|B]) = 2\).
Comparing the ranks with the number of variables:
- \(rank(A) = 2\)
- \(rank([A|B]) = 2\)
- Number of variables \(n = 3\)
Since \(rank(A) = rank([A|B]) = 2\), and this value is less than the number of variables (\(2 < 3\)), the system has infinitely many solutions for any given values of \(b_1\) and \(b_2\).
Important Questions from System of Linear Equations
A system of equations is said to be inconsistent if
If a system of simultaneous equations has infinite solutions, then that system of equations is called:
Consider the system of simultaneous equation,
x + 2y + z = 6
2x + y + 2z = 6
x + y + z = 5
The system has,
The system of equations x + 2y = 13 and 3x + 6y = 9 has:
For what value of k, the system linear equation has no solution
(3k + 1)x + 3y - 2 = 0
(k2 + 1)x + (k - 2)y - 5 = 0
