Let $A = [\begin{array}{cc}-5 & -3 \\ 2 & 1\end{array}]$. The Row transformation $R_1\rightarrow R_1+3R_2$ will transform matrix A into

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: Do the operation.
New first row: $(-5 + 3 \cdot 2,\ -3 + 3 \cdot 1) = (1, 0)$.

Step 2: Write the matrix.
\[ \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \]

Step 3: Test each option.
Upper triangular needs the bottom-left entry to be 0, but it is 2, so (A) is out. Identity needs 0 there, so (C) is out. Determinant $= 1 - 0 = 1 \neq 0$, so (D) is out. The top-right entry is 0, so it is lower triangular.

Final Answer:
Option (B). \[ \boxed{\text{Lower triangular matrix}} \]

Let \(A = [\begin{array}{cc}-5 & -3 \\ 2 & 1\end{array}]\). The Row transformation \(R_1\rightarrow R_1+3R_2\) will transform matrix A into

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The Correct Option is B

Solution and Explanation

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