If 2[130x]+[y012]=[5618]2\begin{bmatrix} 1 & 3 \ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \ 1 & 8 \end{bmatrix}2[103x]+[y102]=[5168],then the value of (x + y) is:
4
-4
6
8
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Given,
2[130x]+[y012]=[5618]2\begin{bmatrix} 1 & 3 \ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \ 1 & 8 \end{bmatrix}2[103x]+[y102]=[5168]
Solving:
⇒[2602x]+[y012]=[5618]⇒[2+y612x+2]=[5618].\Rightarrow \begin{bmatrix} 2 & 6 \ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \ 1 & 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 + y & 6 \ 1 & 2x + 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \ 1 & 8 \end{bmatrix}.⇒[2062x]+[y102]=[5168]⇒[2+y162x+2]=[5168].
∴ 2 + y = 5
⇒ y = 5 - 2
⇒ y = 3
∴ 2x + 2 = 8
⇒ 2x = 8 - 2
⇒ 2x = 6
⇒ x = 62\dfrac{6}{2}26 = 3.
∴ x + y = 3 + 3 = 6.
Hence, option 3 is the correct option.
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[1111]\begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}[1111]
[0110]\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}[0110]
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none of these
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10
-10
-6
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a = 4, b = 2
a = 2, b = 4
both (a) and (b)
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0
2A
–A′
A′