Match List-I with List-II
| List-I | List-II | ||
| Matrix/equations | Values | ||
| (A) | \(\begin{bmatrix}2x+1&3y\\0&y^2-5y\end{bmatrix}=\begin{bmatrix}x+3&y^2+2\\0&-6\end{bmatrix}\) | (I) | \(x = 2, y = -1\) |
| (B) | \(\begin{bmatrix}1&2&-1\\x&0&3\\y&3&4\end{bmatrix}\) is symmetric | (II) | \(x = 2, y = 2\) |
| (C) | \(\begin{bmatrix}x&1\end{bmatrix}\begin{bmatrix}1&0\\-2&-3\end{bmatrix}\begin{bmatrix}5&2\\0&y\end{bmatrix}=O\) | (III) | \(x=-2, y = 2\) |
| (D) | \(\begin{bmatrix}x&0\\1&1\end{bmatrix}\begin{bmatrix}x&0\\1&1\end{bmatrix}=\begin{bmatrix}4&0\\-1&y/2\end{bmatrix}\) | (IV) | \(x = 2, y = 0\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Equal matrices have equal corresponding entries; this decides every item.
(A) \(\begin{bmatrix}2x+1&3y\\0&y^2-5y\end{bmatrix}=\begin{bmatrix}x+3&y^2+2\\0&-6\end{bmatrix}\)
This matches (II) x = 2, y = 2:
\(2x + 1 = x + 3 \Rightarrow x = 2\)
\(3y = y^2 + 2 \Rightarrow (y - 1)(y - 2) = 0 \Rightarrow y = 1 \text{ or } 2\)
\(y^2 - 5y = -6 \Rightarrow (y - 2)(y - 3) = 0 \Rightarrow y = 2 \text{ or } 3\)
Both equations hold only for y = 2.
(B) \(\begin{bmatrix}1&2&-1\\x&0&3\\y&3&4\end{bmatrix}\) is symmetric
This matches (I) x = 2, y = −1. A symmetric matrix has \(a_{ij} = a_{ji}\):
\(x = a_{21} = a_{12} = 2\)
\(y = a_{31} = a_{13} = -1\)
(C) \(\begin{bmatrix}x&1\end{bmatrix}\begin{bmatrix}1&0\\-2&-3\end{bmatrix}\begin{bmatrix}5&2\\0&y\end{bmatrix}=O\)
This matches (IV) x = 2, y = 0. Here O is the zero matrix of order 1 × 2.
\( \displaystyle \begin{bmatrix}x&1\end{bmatrix}\begin{bmatrix}1&0\\-2&-3\end{bmatrix} = \begin{bmatrix}x-2&-3\end{bmatrix} \)
\( \displaystyle \begin{bmatrix}x-2&-3\end{bmatrix}\begin{bmatrix}5&2\\0&y\end{bmatrix} = \begin{bmatrix}5(x-2)&2(x-2)-3y\end{bmatrix} \)
\(5(x - 2) = 0 \Rightarrow x = 2\)
\(2(0) - 3y = 0 \Rightarrow y = 0\)
(D) \(\begin{bmatrix}x&0\\1&1\end{bmatrix}\begin{bmatrix}x&0\\1&1\end{bmatrix}=\begin{bmatrix}4&0\\-1&y/2\end{bmatrix}\)
This matches (III) x = −2, y = 2:
\( \displaystyle \begin{bmatrix}x&0\\1&1\end{bmatrix}\begin{bmatrix}x&0\\1&1\end{bmatrix} = \begin{bmatrix}x^2&0\\x+1&1\end{bmatrix} \)
\(x^2 = 4 \text{ and } x + 1 = -1 \Rightarrow x = -2\)
\(1 = y/2 \Rightarrow y = 2\)
So the matching is (A)-(II), (B)-(I), (C)-(IV), (D)-(III).