A matrix A=(acbd)A=(abcd) has eigenvectors (10)(10) and (−11)(−11) with eigenvalues -3 and 1 respectively. By making use of an appropriate diagonalisation, or otherwise, find the matrix AA and enter its entries in the boxes below. a=a= b=b= c=c= d=d=
Question
A matrix A=(acbd)A=(abcd) has eigenvectors (10)(10) and (−11)(−11) with eigenvalues -3 and 1 respectively. By making use of an appropriate diagonalisation, or otherwise, find the matrix AA and enter its entries in the boxes below. a=a= b=b= c=c= d=d=
Solution
To find the matrix A, we can use the fact that a matrix is diagonalizable if and only if it has a complete set of eigenvectors. In this case, we have two eigenvectors, (1,0) and (-1,1), and their corresponding eigenvalues, -3 and 1.
Step 1: Form the matrix P from the eigenvectors. Each column of P is an eigenvector of A. So, we have:
P = [(1, -1), (0, 1)]
Step 2: Form the diagonal matrix D from the eigenvalues. The diagonal entries of D are the eigenvalues of A. So, we have:
D = [(-3, 0), (0, 1)]
Step 3: Compute the inverse of P. The inverse of a 2x2 matrix [(a, b), (c, d)] is given by 1/(ad-bc) * [(d, -b), (-c, a)]. So, we have:
P_inverse = [(1, 1), (0, 1)]
Step 4: Compute the matrix A using the formula A = PDP_inverse. We have:
A = P * D * P_inverse = [(1, -1), (0, 1)] * [(-3, 0), (0, 1)] * [(1, 1), (0, 1)] = [(-3, -1), (0, 1)] * [(1, 1), (0, 1)] = [(-3, -4), (0, 1)]
So, the matrix A is:
A = [(-3, -4), (0, 1)]
Therefore, a = -3, b = -4, c = 0, and d = 1.
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