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Mathematics

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},then the value of (x + y) is:

  1. 4

  2. -4

  3. 6

  4. 8

Matrices

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Answer

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}

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}.

∴ 2 + y = 5

⇒ y = 5 - 2

⇒ y = 3

∴ 2x + 2 = 8

⇒ 2x = 8 - 2

⇒ 2x = 6

⇒ x = 62\dfrac{6}{2} = 3.

∴ x + y = 3 + 3 = 6.

Hence, option 3 is the correct option.

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