To find a solution x to the equation Ax = -3a + 5b, where A is a 3x2 matrix, and we know u = [4, -4] and v = [4, 2] satisfy the equations Au = a and Av = b, we can use the linearity property of matrix multiplication.
Let's express the matrix A as [c1, c2], where c1 and c2 are the columns of A.
Since Au = a and Av = b, we have:
A * u = a
A * v = b
Expanding these equations using the column representation of A:
c1 * u + c2 * u = a
c1 * v + c2 * v = b
We can rewrite these equations as:
c1u + c2u = a
c1v + c2v = b
Now, let's substitute the given values:
4c1 - 4c2 = a (equation 1)
4c1 + 2c2 = b (equation 2)
Next, we want to find a solution x to Ax = -3a + 5b. Rewriting this equation using the column representation of A:
c1 * x1 + c2 * x2 = -3a + 5b
Substituting the values of a and b:
c1 * x1 + c2 * x2 = -3(4c1 - 4c2) + 5(4c1 + 2c2)
Simplifying:
c1 * x1 + c2 * x2 = -12c1 + 12c2 + 20c1 + 10c2
Combining like terms:
c1 * x1 + c2 * x2 = 8c1 + 22c2
Now, we have the equation:
c1 * x1 + c2 * x2 = 8c1 + 22c2 (equation 3)
To find a solution x, we need to express x in terms of the columns of A (c1 and c2). We can see that equation 3 matches the form of equations 1 and 2. Therefore, we can set:
8c1 + 22c2 = a (equation 4)
x1 = a
Now we have x1 in terms of a. To find x2, we can set:
8c1 + 22c2 = b (equation 5)
x2 = b
Therefore, a solution x to Ax = -3a + 5b can be expressed as:
x = [x1, x2] = [a, b]
Using the given values of u and v:
x = [4, -4] for u = [4, -4]
x = [4, 2] for v = [4, 2]
Hence, a solution x to Ax = -3a + 5b is x = [4, -4] when a = [4, -4] and b = [4, 2].