Let $A=\begin{bmatrix} 1 & 1 & 1 \\ -1 & -1 & -1 \\ 0 & 1 & -1 \end{bmatrix}$ and $b = \begin{bmatrix} 1/3 \\ -1/3 \\ 0 \end{bmatrix}$, then the system of linear equations $Ax = b$ has
Determining the Number of Solutions for Ax = b
We are given the system of linear equations represented by $Ax = b$, where:
$A = \begin{bmatrix} 1 & 1 & 1 \\ -1 & -1 & -1 \\ 0 & 1 & -1 \end{bmatrix}$ and $b = \begin{bmatrix} 1/3 \\ -1/3 \\ 0 \end{bmatrix}$.
To find the number of solutions, we analyze the ranks of the coefficient matrix $A$ and the augmented matrix $[A|b]$.
Calculating the Rank of Matrix A
First, find the determinant of $A$ to check for a unique solution:
$ \det(A) = 1 \begin{vmatrix} -1 & -1 \\ 1 & -1 \end{vmatrix} - 1 \begin{vmatrix} -1 & -1 \\ 0 & -1 \end{vmatrix} + 1 \begin{vmatrix} -1 & -1 \\ 0 & 1 \end{vmatrix} $ $ \det(A) = 1((-1)(-1) - (-1)(1)) - 1((-1)(-1) - (-1)(0)) + 1((-1)(1) - (-1)(0)) $ $ \det(A) = 1(1 + 1) - 1(1 - 0) + 1(-1 - 0) $ $ \det(A) = 2 - 1 - 1 = 0 $Since $\det(A) = 0$, the system does not have a unique solution.
Now, let's find the rank of $A$ using row reduction:
$A = \begin{bmatrix} 1 & 1 & 1 \\ -1 & -1 & -1 \\ 0 & 1 & -1 \end{bmatrix} \xrightarrow{R_2 \leftarrow R_2 + R_1} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 1 & -1 \end{bmatrix} \xrightarrow{R_2 \leftrightarrow R_3} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}$The row echelon form has 2 non-zero rows. Thus, rank(A) = 2.
Calculating the Rank of the Augmented Matrix [A|b]
Form the augmented matrix $[A|b]$:
$ [A|b] = \begin{bmatrix} 1 & 1 & 1 & | & 1/3 \\ -1 & -1 & -1 & | & -1/3 \\ 0 & 1 & -1 & | & 0 \end{bmatrix} $Perform row reduction:
$ [A|b] \xrightarrow{R_2 \leftarrow R_2 + R_1} \begin{bmatrix} 1 & 1 & 1 & | & 1/3 \\ 0 & 0 & 0 & | & 0 \\ 0 & 1 & -1 & | & 0 \end{bmatrix} \xrightarrow{R_2 \leftrightarrow R_3} \begin{bmatrix} 1 & 1 & 1 & | & 1/3 \\ 0 & 1 & -1 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{bmatrix} $The row echelon form has 2 non-zero rows. Thus, rank([A|b]) = 2.
Conclusion on the Number of Solutions
Compare the ranks:
- rank(A) = 2
- rank([A|b]) = 2
The number of variables ($n$) in the system is 3.
Since rank(A) = rank([A|b]) < n (specifically, 2 < 3), the system has infinitely many solutions.
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
