If $A = [\begin{array}{cc}5a & -b \\ 3 & 2\end{array}]$ and $A\,(\text{adj }A) = AA^T$, Then $5a+b =$

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 determinants:
Take determinants of $A\,\operatorname{adj}A = AA^T$ is unnecessary; instead compare entries directly.

Step 2: Entries:
$|A| I$ has zero off-diagonals and equal diagonals. $AA^T$ diagonals: $25a^2+b^2$ and $13$; off-diagonal $15a-2b$.

Step 3: Solve the system:
$b = 7.5a$ and $10a + 3b = 13$ give $a = 0.4$, $b = 3$. Hence $5a + b = 2 + 3 = 5$, option (C).

Final Answer:
5a + b equals 5. \[ \boxed{\text{(C) }5} \]

If \(A = [\begin{array}{cc}5a & -b \\ 3 & 2\end{array}]\) and \(A\,(\text{adj }A) = AA^T\), Then \(5a+b =\)

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The Correct Option is C

Solution and Explanation

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