If a is a 3-order square matrix, α is a 3-dimensional column vector If a is a 3-order square matrix, α is a 3-dimensional column vector, and the group of constant vectors α, a α, a & # 178; α is linearly independent, and a & # 179; α = 5A α - 3A & # 178; α, we prove that the matrix B = (α, a α, a ^ 4 α) is invertible My idea is: we can transform the topic into α, a α, a & # 178; α linearly independent, a α, a & # 178; α, a & # 179; α linearly related, and prove α, a α, a ^ 4 α linearly independent. But we still haven't done it

If a is a 3-order square matrix, α is a 3-dimensional column vector If a is a 3-order square matrix, α is a 3-dimensional column vector, and the group of constant vectors α, a α, a & # 178; α is linearly independent, and a & # 179; α = 5A α - 3A & # 178; α, we prove that the matrix B = (α, a α, a ^ 4 α) is invertible My idea is: we can transform the topic into α, a α, a & # 178; α linearly independent, a α, a & # 178; α, a & # 179; α linearly related, and prove α, a α, a ^ 4 α linearly independent. But we still haven't done it

A ^ 4 α = a (a ^ 3 α) = a (5a α - 3A ^ 2 α) = 5A ^ 2 α - 3A ^ 3 α = 5A ^ 2 α - 3 (5a α - 3A ^ 2 α) = 5A ^ 2 α - 15A α + 9A ^ 2 α = 14a ^ 2 α - 15A α B = (α, a α, a ^ 4 α) = (α, a α, 14a ^ 2 α - 15A α) = (α, a α, a ^ 2 α) k where k =