Given the following vector X, find a non-zero square matrix A such that AX=0:
X=[2 -5 3]^T
Im not sure how i would find the sqare matrix a?
2 A11 -5 A12 + 3 A13 = 0
1 1 1 can be the top row :)
ah, the others too :)
| 1 1 1 | |+2| |0|
| 1 1 1 |*|-5|=|0|
| 1 1 1 | |+3| |0|
To find a non-zero square matrix A such that AX=0, we need to solve the equation AX=0. In other words, we need to find a matrix A such that when we multiply it with the vector X and the result is the zero vector. Let's solve this step by step:
Step 1: Write the equation AX=0 in matrix notation:
A * X = 0
Step 2: Expand the matrix multiplication:
[[a11 a12 a13] * [[2] = [[0]
[a21 a22 a23] [-5] [0]
[a31 a32 a33]] [3]] [0]]
Step 3: Write out the expanded equation:
a11*2 + a12*(-5) + a13*3 = 0
a21*2 + a22*(-5) + a23*3 = 0
a31*2 + a32*(-5) + a33*3 = 0
Step 4: Solve for the unknowns a11, a12, a13, a21, a22, a23, a31, a32, a33:
To find a non-zero solution, we can use the fact that there are infinitely many solutions to this system of equations. We can choose any non-zero values for a11, a12, a13, a21, a22, a23, a31, a32, a33, and we will have a non-zero square matrix A that satisfies AX=0. For example, we can choose:
a11 = 1, a12 = 1, a13 = 1,
a21 = 1, a22 = 1, a23 = 1,
a31 = 1, a32 = 1, a33 = 1.
Then the matrix A will be:
A = [[1 1 1]
[1 1 1]
[1 1 1]]
Step 5: Verify that AX=0:
Let's multiply A by X and verify that the result is the zero vector.
A * X = [[1 1 1] * [[2] = [[0]
[1 1 1] [-5] [0]
[1 1 1]] [3]] [0]]
= [[2-5+3]
[2-5+3]
[2-5+3]]
= [[0]
[0]
[0]]
As we can see, the result of multiplying A by X is the zero vector, which confirms that AX=0.
So, the non-zero square matrix A that satisfies AX=0 for the given vector X=[2 -5 3]^T is:
A = [[1 1 1]
[1 1 1]
[1 1 1]]
To find a non-zero square matrix A such that AX=0, we need to solve the equation AX=0 for the given vector X=[2 -5 3]ᵀ.
Let's assume the matrix A has dimensions n×n (where n is the number of columns or rows in A). Since we are given X as a column vector with 3 elements, A must be a 3×3 matrix.
We need to find a matrix A such that AX=0. Let's denote the columns of A as [a₁ a₂ a₃], where a₁, a₂, and a₃ are column vectors.
So we have the equation AX = [a₁ a₂ a₃][2 -5 3]ᵀ = 0, where 0 represents the zero vector.
Expanding this equation, we get the following system of equations:
2a₁ - 5a₂ + 3a₃ = 0
-5a₁ + 5a₂ = 0
3a₁ = 0
Simplifying the system of equations further, we have:
2a₁ - 5a₂ + 3a₃ = 0 (Equation 1)
-5a₁ + 5a₂ = 0 (Equation 2)
3a₁ = 0 (Equation 3)
From Equation 3, we can see that a₁ must be the zero vector [0 0 0]ᵀ, since 3a₁ = 3[0 0 0]ᵀ = [0 0 0]ᵀ ≠ 0.
Substituting a₁ = [0 0 0]ᵀ into Equation 2, we get:
-5[0 0 0]ᵀ + 5a₂ = [0 0 0]ᵀ
5a₂ = [0 0 0]ᵀ
This implies that a₂ must also be the zero vector [0 0 0]ᵀ.
Now, substituting a₁ = [0 0 0]ᵀ and a₂ = [0 0 0]ᵀ into Equation 1, we have:
2[0 0 0]ᵀ - 5[0 0 0]ᵀ + 3a₃ = [0 0 0]ᵀ
3a₃ = [0 0 0]ᵀ
This equation indicates that a₃ can be any non-zero vector. For example, let's choose a₃ = [1 0 0]ᵀ.
Therefore, the matrix A = [a₁ a₂ a₃] = [0 0 1; 0 0 0; 0 0 0] is a non-zero square matrix that satisfies AX = 0 for the given vector X.
Please note that there are infinitely many matrices A that satisfy AX = 0 for the vector X, as long as the columns corresponding to the non-zero entries of X are zero in A. The example provided above is just one possible solution.