Determining Nontrivial Solutions for Ax = 0 in a 3×3 Matrix with Two Pivot Positions

If A is 3×3 with two pivot positions, then Ax = 0 has a nontrivial solution.

To determine if Ax = 0 has a nontrivial solution, we need to examine the number of pivot positions in the augmented matrix [A | 0]

To determine if Ax = 0 has a nontrivial solution, we need to examine the number of pivot positions in the augmented matrix [A | 0].

Since matrix A is given to be a 3×3 matrix with two pivot positions, it means that there are two leading 1’s in the row-reduced echelon form of A.

In order for Ax = 0 to have a nontrivial solution, the number of free variables or non-pivotal variables should be greater than zero. This is because the presence of free variables indicates that there are infinite solutions to the system of equations.

In a 3×3 matrix, if there are two pivot positions, it means there is only one free variable (3 – 2 = 1). This implies that there is a nontrivial solution to Ax = 0.

To find the nontrivial solution, we need to set up the system of equations based on the given matrix A:

Let A be represented as:
| a b c |
| d e f |
| g h i |

Since we know that there are two pivot positions, let’s assume that the first two rows of the row-reduced echelon form of A have leading 1’s. This implies that the pivots are in the first column and the second column.

Setting up the equations based on this assumption:

x + d/a * y + g/a * z = 0 —- (Equation 1)
y + e/b * x + h/b * z = 0 —- (Equation 2)
0 + f/c * x + i/c * z = 0 —- (Equation 3)

From these equations, we can solve for the variables x, y, and z.

However, since we are looking for a nontrivial solution, our aim is to find values for x, y, and z that are not all zero. This can be achieved by assigning a value of 1 to the free variable. In this case, let’s assume z = 1.

Now, we can solve the equations to find the values of x and y.

From Equation 1, we have:
x + d/a * y + g/a * 1 = 0

From Equation 2, we have:
y + e/b * x + h/b * 1 = 0

From Equation 3, we have:
0 + f/c * x + i/c * 1 = 0

Solving these three equations will give us the nontrivial solution for Ax = 0, which satisfies the given conditions of having a 3×3 matrix with two pivot positions.

Note: The specific values of the coefficients in A will determine the actual numerical values of x, y, and z.

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