If matrix $A = \left[ \begin{array}{ccc}-1 & 2025 & 2026 \\ 0 & 2 & 2027 \\ 0 & 0 & -1\end{array} \right]$, then the sum of all elements in $\text{adj}(A^{-1})$ is equal to...

Question

Let the foot of the perpendicular from the point $ (1, 2, 4) $ on the line $ \frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} $ be $ P $. Then the distance of $ P $ from the plane $ 3x + 4y + 12z + 23 = 0 $ is:

Answer 1

Step 1: Use the identity
$\text{adj}(X) = |X|X^{-1}$ with $X = A^{-1}$. We have $|A^{-1}| = \frac{1}{|A|}$ and $(A^{-1})^{-1} = A$.

Step 2: So
$\text{adj}(A^{-1}) = \frac{1}{2}A$, since $|A| = (-1)(2)(-1) = 2$.

Step 3: Add the entries
The nine entries of $A$ add up to $(-1 + 2 - 1) + (2025 + 2026 + 2027) = 0 + 6078 = 6078$.

Step 4: Halve
$6078 \div 2 = 3039$.

Final Answer:
The sum is 3039. This is option (C). \[ \boxed{\text{(C) }3039} \]

If matrix \(A = \left[ \begin{array}{ccc}-1 & 2025 & 2026 \\ 0 & 2 & 2027 \\ 0 & 0 & -1\end{array} \right]\), then the sum of all elements in \(\text{adj}(A^{-1})\) is equal to...

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The Correct Option is C

Solution and Explanation

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