KCET · Maths · Determinants
If \( 2\left[\begin{array}{ll}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right] \) then the value of \( x \) and \( y \) are
- A \( x=3, y=3 \)
- B \( x=-3, y=3 \)
- C \( x=3, y=-3 \)
- D \( x=-3, y=-3 \)
Answer & Solution
Correct Answer
(A) \( x=3, y=3 \)
Step-by-step Solution
Detailed explanation
Given that,
\(2\left[\begin{array}{cc}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]\)
\(\left[\begin{array}{cc}2+y & 6 \\ 1 & 2 x+2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]\)
Equating first elements, we have
\(2+y=5 \Rightarrow y=3\)
Equating last elements, we have
\(2 x+2=8 \Rightarrow x=3\)
Therefore, \(x=3, y=3\).
\(2\left[\begin{array}{cc}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]\)
\(\left[\begin{array}{cc}2+y & 6 \\ 1 & 2 x+2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]\)
Equating first elements, we have
\(2+y=5 \Rightarrow y=3\)
Equating last elements, we have
\(2 x+2=8 \Rightarrow x=3\)
Therefore, \(x=3, y=3\).
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