If A is nonsingular and (A - 2I) (A - 4I) = O, then (1/6)A + (4/3)A-1 is equal to kI. Then the value of k is .

Question

If A is nonsingular and (A - 2I) (A - 4I) = O, then (1/6)A + (4/3)A-1 is equal to kI. Then the value of k is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

We have, (A - 2I)(A - 4I) = 0 or A2 - 2A - 4A + 8I = O or A2 - 6A + 8I = O or A-1(A2 - 6A + 8I) = A-1O or A - 6I + 8A-1 = O or A + 8A-1 = 6I or (1/6)A + (4/3) A-1 = 1

Q. If $A$ is nonsingular and $(A - 2 I) (A - 4 I) = O$ , then $\frac{1}{6} A + \frac{4}{3} A^{- 1}$ is equal to $k I$ . Then the value of $k$ is ______.

We have,
$(A - 2 I) (A - 4 I) = 0$
or $A^{2} - 2 A - 4 A + 8 I = O$
or $A^{2} - 6 A + 8 I = O$
or $A^{- 1} (A^{2} - 6 A + 8 I) = A^{- 1} O$
or $A - 6 I + 8 A^{- 1} = O$
or $A + 8 A^{- 1} = 6 I$
or $\frac{1}{6} A + \frac{4}{3} A^{- 1} = 1$

Q. If A is nonsingular and (A−2I)(A−4I)=O, then 61​A+34​A−1 is equal to kI. Then the value of k is ______.

We have, (A−2I)(A−4I)=0 or A2−2A−4A+8I=O or A2−6A+8I=O or A−1(A2−6A+8I)=A−1O or A−6I+8A−1=O or A+8A−1=6I or 61​A+34​A−1=1

Q. If $A$ is nonsingular and $(A - 2 I) (A - 4 I) = O$ , then $\frac{1}{6} A + \frac{4}{3} A^{- 1}$ is equal to $k I$ . Then the value of $k$ is ______.

We have,
$(A - 2 I) (A - 4 I) = 0$
or $A^{2} - 2 A - 4 A + 8 I = O$
or $A^{2} - 6 A + 8 I = O$
or $A^{- 1} (A^{2} - 6 A + 8 I) = A^{- 1} O$
or $A - 6 I + 8 A^{- 1} = O$
or $A + 8 A^{- 1} = 6 I$
or $\frac{1}{6} A + \frac{4}{3} A^{- 1} = 1$