If $A = [\begin{array}{cc}2 & -1 \\ 0 & 2\end{array}]$ and $A^2+xA+yI_2 = O_2$, where $I_2$ and $O_2$ are the identity matrix and null matrix of order 2 respectively, then:

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 Cayley-Hamilton
For a 2 by 2 matrix, $A^2 - (\text{tr}A)A + (\det A)I = O$.

Step 2: Values
$\text{tr}A = 4$ and $\det A = 4$. So $A^2 - 4A + 4I = O$.

Step 3: Compare
Hence $x=-4$ and $y=4$, option (B).

Final Answer:
Option B. \[ \boxed{\text{(B)}\ x=-4,\ y=4} \]

If \(A = [\begin{array}{cc}2 & -1 \\ 0 & 2\end{array}]\) and \(A^2+xA+yI_2 = O_2\), where \(I_2\) and \(O_2\) are the identity matrix and null matrix of order 2 respectively, then:

Show Hint

The Correct Option is B

Solution and Explanation

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