if matrix A=[2 ,1, 1,2](2*2),what are the eigenvalues and eigenvectors of A
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We have a 2 by 2 matrix A with entries 2, 1 in the first row and 1, 2 in the second row. To find eigenvalues and eigenvectors, we start with the characteristic equation: determinant of A minus lambda I equals zero.
Now we calculate the determinant. The determinant of a 2 by 2 matrix is ad minus bc. So we get 2 minus lambda times 2 minus lambda, minus 1 times 1. This simplifies to 2 minus lambda squared minus 1 equals zero. Solving this gives us 2 minus lambda squared equals 1, so 2 minus lambda equals plus or minus 1. Therefore, our eigenvalues are lambda 1 equals 1 and lambda 2 equals 3.
Now we find the eigenvector for lambda 1 equals 1. We solve the equation A minus lambda I times v equals zero. Substituting lambda equals 1, we get A minus I. This gives us the matrix with entries 1, 1 in the first row and 1, 1 in the second row. Setting this matrix times the vector x, y equal to the zero vector, we get the equation x plus y equals zero, which means y equals negative x. So our first eigenvector is negative 1, 1.
Now we find the eigenvector for lambda 2 equals 3. We solve A minus 3I times v equals zero. Substituting lambda equals 3, we get A minus 3I. This gives us the matrix with entries negative 1, 1 in the first row and 1, negative 1 in the second row. Setting this matrix times the vector x, y equal to the zero vector, we get negative x plus y equals zero, which means y equals x. So our second eigenvector is 1, 1.
To summarize our complete solution: For the matrix A with entries 2, 1, 1, 2, we found two eigenvalues: lambda 1 equals 1 and lambda 2 equals 3. The corresponding eigenvectors are: v1 equals negative 1, 1 for eigenvalue 1, and v2 equals 1, 1 for eigenvalue 3. We can verify that A times v1 equals 1 times v1, and A times v2 equals 3 times v2, confirming our solution is correct.