Show that
satisfies the equation x2 – 3A – 7 = 0. Thus, find A – 1.
A = ![]()
A2 = ![]()
= ![]()
Now, A2 – 3A – 7 = 0
= ![]()
= ![]()
= ![]()
So, A2 – 3A – 7I = 0
Multiply by A – 1 both sides
= A.A. A – 1 – 3A. A – 1 – 7I. A – 1 = 0
= A – 3I – 7A – 1 = 0
= 7A – 1 = A – 3I
= A – 1 = ![]()
Hence, A – 1 = ![]()
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