Aspire Faculty ID #11951 · Topic: NIMCET 2025 · Just now
NIMCET 2025

Let A and B be two square matrices of same order satisfying $A^2+5A+5I =0$ and $B^2+3B+I=0$ repectively. Where I is the identity matrix. Then the inverse of the matrix $C= BA+2B+2A+4I$ is

Solution

Given: $A^2 + 5A + 5I = 0 \quad\Rightarrow\quad A^2 = -5A - 5I$ 
$B^2 + 3B + I = 0 \quad\Rightarrow\quad B^2 = -3B - I$ 

Matrix: $C = BA + 2B + 2A + 4I$ 

We want $C^{-1}$. 

Step 1: Try a linear expression as inverse Assume: $C^{-1} = \alpha AB + \beta A + \gamma B + \delta I$ 
Test using the fact that quadratic equations give inverses as linear polynomials. 

Check option (2): 
$AB + A + 3B + 3I$ Multiply with $C$: 
$C(AB + A + 3B + 3I)$ 
$= (BA + 2B + 2A + 4I)(AB + A + 3B + 3I)$ 
Using reductions: 
$A^2 = -5A - 5I$ $B^2 = -3B - I$ 
Everything simplifies to: $I$ 
Hence: $C^{-1} = AB + A + 3B + 3I$
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