If A is 3×3 with two pivot positions, then Ax = b has a solution for every b in R3.
To determine if the equation Ax = b has a solution for every b in R3, we need to analyze the properties of matrix A
To determine if the equation Ax = b has a solution for every b in R3, we need to analyze the properties of matrix A.
In a 3×3 matrix, having two pivot positions means that only two rows in the matrix have leading non-zero entries (pivots), and the remaining row consists of all zeros.
When we write the augmented matrix [A | b], it would look something like:
| a11 a12 a13 | | x1 | | b1 |
| a21 a22 a23 | * | x2 | = | b2 |
| 0 0 0 | | x3 | | b3 |
Since the bottom row consists of all zeros, it does not contribute to the system of equations formed by Ax = b.
Therefore, we can rewrite the system as:
a11 * x1 + a12 * x2 + a13 * x3 = b1
a21 * x1 + a22 * x2 + a23 * x3 = b2
This system can be written as a matrix equation as:
| a11 a12 a13 | | x1 | | b1 |
| a21 a22 a23 | * | x2 | = | b2 |
In order for this system to have a solution for every b in R3, the coefficient matrix [A] must be invertible, or equivalently, have a non-zero determinant.
Let’s try to determine if the matrix A is invertible. Since A is a 3×3 matrix and has two pivot positions, we can conclude that it is not invertible because it lacks a third pivot, which means the determinant is zero.
Hence, we can conclude that the equation Ax = b does not have a solution for every b in R3 when matrix A is 3×3 with two pivot positions.
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