Final answer:
To prove I+5A is invertible, assume it is not and derive a contradiction. Show that A has eigenvalues 5 and -5. Write A as a combination of identity matrices. Substitute into I+5A. Conclude that I+5A is invertible.
Step-by-step explanation:
To prove that I+5A is invertible, let's assume that I+5A is not invertible. This means that there exists a nonzero vector x such that (I+5A)x = 0. Multiplying both sides by 5 and rearranging, we get 5x = -5Ax. Factoring out x, we have (5I+A)x = 0.
Now, consider the matrices 2I-A, 3I-A, and 4I-A. We are given that none of these are invertible. In other words, there exists a nonzero vector y such that (2I-A)y = 0, (3I-A)y = 0, and (4I-A)y = 0. Applying the same technique as before, we can write these equations as 2y = Ay, 3y = Ay, and 4y = Ay.
Let's consider the equation (2+3+4)(y) = A(2y+3y+4y). Simplifying, we have 9y = Ay+Ay+Ay = 3Ay. Dividing by 3, we get 3y = Ay. Comparing this equation to the earlier equations, we can conclude that 3y = Ay = 5y. This implies that 3y = 5y = Ay = 5Ay, which means that y is an eigenvector of A with eigenvalues 5 and -5.
Since A has eigenvalues 5 and -5, we can write A as A = (5I_1 + (-5)I_2) where I_1 and I_2 are the identity matrices corresponding to the eigenvalues 5 and -5 respectively. Now, let's consider I+5A. Substituting our expression for A, we have I+5A = I + 5(5I_1 + (-5)I_2) = I + 25I_1 + (-25)I_2 = 26I_1 + (-25)I_2.
Since I_1 and I_2 are identity matrices, I_1x = x and I_2x = x for any vector x. Therefore, (26I_1 + (-25)I_2)x = 26I_1x + (-25)I_2x = 26x + (-25)x = x. This shows that I+5A is invertible because it sends any nonzero vector to itself. Hence, we have proven that I+5A is invertible.