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If 2 \( ​​​​\begin{pmatrix} 3 & 4 \\ 5 & x \\ \end{pmatrix} \) + \( ​​​​\begin{pmatrix} 1 & y \\ 0 & 1 \\ \end{pmatrix} \) = \( ​​​​\begin{pmatrix} 7 & 0 \\ 10 & 5 \\ \end{pmatrix} \)

A. (x = -2, y = 8) 

B. (x = 2, y = -8) 

C. (x = 3, y = -6) 

D. (x = -3, y = 6)

1 Answer

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Best answer

\( 2​​​​\begin{pmatrix} 3 & 4 \\ 5 & x \\ \end{pmatrix} \) + \( ​​​​\begin{pmatrix} 1 & y \\ 0 & 1 \\ \end{pmatrix} \) = \( ​​​​\begin{pmatrix} 7 & 0 \\ 10 & 5 \\ \end{pmatrix} \)

To solve this problem we will use the comparison that is we will use that all the elements of L.H.S are equal to R.H.S .

\( ​​​​\begin{pmatrix} 6 & 8 \\ 10 & 2x \\ \end{pmatrix} \) + \( ​​​​\begin{pmatrix} 1 & y \\ 0 & 1 \\ \end{pmatrix} \)

\( ​​​​\begin{pmatrix} 7 &8+ y \\ 10 & 2x+1 \\ \end{pmatrix} \)

Comparing with R.H.S 

8 + y = 0 

y = - 8 

2x + 1 = 5 

2x = 4 

x = 2

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