Answer:
Explanation:
Sure! To prove that A * A^(-1) = 1, we can calculate the inverse of matrix A and then multiply it with A. Let's do the calculations.
Matrix A = [[2, -2], [-1, 3]]
To find the inverse of A, we use the formula A^(-1) = (1 / det(A)) * adj(A), where det(A) is the determinant of A and adj(A) is the adjugate of A.
det(A) = (2 * 3) - (-2 * -1) = 6 - 2 = 4
adj(A) = [[3, 2], [1, 2]]
A^(-1) = (1 / 4) * [[3, 2], [1, 2]] = [[3/4, 1/2], [1/4, 1/2]]
Now, let's multiply A with A^(-1):
A * A^(-1) = [[2, -2], [-1, 3]] * [[3/4, 1/2], [1/4, 1/2]]
= [[(2 * 3/4) + (-2 * 1/4), (2 * 1/2) + (-2 * 1/2)], [(-1 * 3/4) + (3 * 1/4), (-1 * 1/2) + (3 * 1/2)]]
= [[6/4 - 2/4, 2/2 - 2/2], [-3/4 + 3/4, -1/2 + 3/2]]
= [[4/4, 0/2], [0/4, 2/2]]
= [[1, 0], [0, 1]]
As we can see, A * A^(-1) equals the identity matrix 1. Therefore, we have proved that A * A^(-1) = 1.
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Answer:
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Explanation:
Sure! To prove that A * A^(-1) = 1, we can calculate the inverse of matrix A and then multiply it with A. Let's do the calculations.
Matrix A = [[2, -2], [-1, 3]]
To find the inverse of A, we use the formula A^(-1) = (1 / det(A)) * adj(A), where det(A) is the determinant of A and adj(A) is the adjugate of A.
det(A) = (2 * 3) - (-2 * -1) = 6 - 2 = 4
adj(A) = [[3, 2], [1, 2]]
A^(-1) = (1 / 4) * [[3, 2], [1, 2]] = [[3/4, 1/2], [1/4, 1/2]]
Now, let's multiply A with A^(-1):
A * A^(-1) = [[2, -2], [-1, 3]] * [[3/4, 1/2], [1/4, 1/2]]
= [[(2 * 3/4) + (-2 * 1/4), (2 * 1/2) + (-2 * 1/2)], [(-1 * 3/4) + (3 * 1/4), (-1 * 1/2) + (3 * 1/2)]]
= [[6/4 - 2/4, 2/2 - 2/2], [-3/4 + 3/4, -1/2 + 3/2]]
= [[4/4, 0/2], [0/4, 2/2]]
= [[1, 0], [0, 1]]
As we can see, A * A^(-1) equals the identity matrix 1. Therefore, we have proved that A * A^(-1) = 1.