Answer:
The correct answer is [tex]A^{-1}[/tex] = [tex]\frac{1}{4}[/tex] A + [tex]\frac{3}{4 }[/tex] I
Step-by-step explanation:
Since A is an n × n invertible non singular matrix, [tex]A^{-1}[/tex] exists.
Given equation is [tex]A^{2}[/tex] + 3A -4I = O where I and O are the n × n identity and zero matrix respectively.
⇒ [tex]A^{2}[/tex] + 3A -4I = O
⇒ [tex]A^{-1}[/tex] ( [tex]A^{2}[/tex] + 3A -4I ) = [tex]A^{-1}[/tex] × O
⇒ [tex]A^{-1}[/tex] AA + 3 [tex]A^{-1}[/tex]A - 4[tex]A^{-1}[/tex]I = O
⇒ A + 3I - 4[tex]A^{-1}[/tex] = O
⇒ A + 3I = 4 [tex]A^{-1}[/tex]
⇒ [tex]A^{-1}[/tex] = [tex]\frac{1}{4}[/tex] A + [tex]\frac{3}{4 }[/tex] I
Thus the value of [tex]A^{-1}[/tex] can be given by the above equation.
