Suppose A is any 3 × 3 non-singular matrix and (A - 3I)(A - 5I) = O, where I = I3 and O = O3. If α A + β A-1 = 4I, then α + β is equal to :

Question

Suppose A is any 3 × 3 non-singular matrix and (A - 3I)(A - 5I) = O, where I = I3 and O = O3. If α A + β A-1 = 4I, then α + β is equal to : (A) 8 (B) 7 (C) 13 (D) 12

Tardigrade Admin · Accepted Answer

Given (A-3 I)(A-5 I)=0 ⇒ A2-8 A+15 I=0 ⇒ A-8 I+15 A-1=0 ⇒ (1/2) A+(15/2) A-1=4 I Given α A+β A -1=4 I So, α=(1/2), β=(15/2) Therefore, α+β=(16/2)=8

Q. Suppose $A$ is any $3 \times 3$ non-singular matrix and $(A - 3 I) (A - 5 I) = O$ , where $I = I_{3}$ and $O = O_{3}$ . If $α A + β A^{- 1} = 4 I$ , then $α + β$ is equal to :

8

7

13

12

Given $(A - 3 I) (A - 5 I) = 0$
$\Rightarrow A^{2} - 8 A + 15 I = 0$
$\Rightarrow A - 8 I + 15 A^{- 1} = 0$
$\Rightarrow \frac{1}{2} A + \frac{15}{2} A^{- 1} = 4 I$
Given $α A + β A^{- 1} = 4 I$
So, $α = \frac{1}{2}, β = \frac{15}{2}$
Therefore, $α + β = \frac{16}{2} = 8$

Q. Suppose A is any 3×3 non-singular matrix and (A−3I)(A−5I)=O, where I=I3​ and O=O3​. If αA+βA−1=4I, then α+β is equal to :

8

7

13

12

Given (A−3I)(A−5I)=0 ⇒A2−8A+15I=0 ⇒A−8I+15A−1=0 ⇒21​A+215​A−1=4I Given αA+βA−1=4I So, α=21​,β=215​ Therefore, α+β=216​=8

Q. Suppose $A$ is any $3 \times 3$ non-singular matrix and $(A - 3 I) (A - 5 I) = O$ , where $I = I_{3}$ and $O = O_{3}$ . If $α A + β A^{- 1} = 4 I$ , then $α + β$ is equal to :

Given $(A - 3 I) (A - 5 I) = 0$
$\Rightarrow A^{2} - 8 A + 15 I = 0$
$\Rightarrow A - 8 I + 15 A^{- 1} = 0$
$\Rightarrow \frac{1}{2} A + \frac{15}{2} A^{- 1} = 4 I$
Given $α A + β A^{- 1} = 4 I$
So, $α = \frac{1}{2}, β = \frac{15}{2}$
Therefore, $α + β = \frac{16}{2} = 8$