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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) |
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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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