Step 1: Understanding the Question:
We are given a matrix equation $A^2 - 4A + 3I = 0$ and need to find the inverse of the matrix $(A + 3I)$. Step 2: Key Formula or Approach:
Let $B = A + 3I$. We need to find $B^{-1}$.
If we can express $B^2 + pB + qI = 0$, then $B^{-1} = -\frac{1}{q}(B + pI)$. Step 3: Detailed Explanation:
From $B = A + 3I$, we have $A = B - 3I$.
Substitute $A$ into the given equation:
\[ (B - 3I)^2 - 4(B - 3I) + 3I = 0 \]
\[ (B^2 - 6B + 9I) - (4B - 12I) + 3I = 0 \]
\[ B^2 - 10B + 24I = 0 \]
Multiplying the equation by $B^{-1}$:
\[ B^{-1}B^2 - 10B^{-1}B + 24B^{-1}I = 0 \]
\[ B - 10I + 24B^{-1} = 0 \]
\[ 24B^{-1} = 10I - B \]
Substitute $B = A + 3I$ back:
\[ 24B^{-1} = 10I - (A + 3I) = 7I - A \]
\[ B^{-1} = \frac{7I - A}{24} = \frac{7I}{24} - \frac{1}{24}A \] Step 4: Final Answer:
The inverse is $\frac{7I}{24} - \frac{1}{24}A$.