3 - 1112 - 0 11 - 4lendbmatrix beginbmatrix x - xy - y lend bmatrix -beginbmatrix 7 - 7 N2 x - 2 x l6 - 6lendbmatrix - -text then -x- y- text is A- 1-2B- 2-1C- -2-1D- 1-2

The correct option is B (2,1) nbsp;[ 3 amp; 1; 2 amp; 0; 1 amp; 4 nbsp; ][ x amp; x; y amp; y; ]= [ 7 amp; 7; 2x ...

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Byju's Answer

Standard XII

Mathematics

Consistency of Linear System of Equations

3 & 1112 & 0 ...

Question

$If ⎡ ⎢ ⎣ \begin{matrix} 3 & 1 2 & 0 1 & 4 \end{matrix} ⎤ ⎥ ⎦ [\begin{matrix} x & x y & y \end{matrix}] = ⎡ ⎢ ⎣ \begin{matrix} 7 & 7 2 x & 2 x 6 & 6 \end{matrix} ⎤ ⎥ ⎦, then (x, y) is$

(1, 2)

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(2, 1)

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(- 2, 1)

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(1, - 2)

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Solution

The correct option is B $(2, 1)$
$⎡ ⎢ ⎣ \begin{matrix} 3 & 1 2 & 0 1 & 4 \end{matrix} ⎤ ⎥ ⎦ [\begin{matrix} x & x y & y \end{matrix}] = ⎡ ⎢ ⎣ \begin{matrix} 7 & 7 2 x & 2 x 6 & 6 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow ⎡ ⎢ ⎣ \begin{matrix} 3 x + y & 3 x + y 2 x & 2 x x + 4 y & x + 4 y \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 7 & 7 2 x & 2 x 6 & 6 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow 3 x + y = 7 . . . (1) \Rightarrow x + 4 y = 6 . . . (2) \Rightarrow x = 2, y = 1$

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, then

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is equal to

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If ⎡⎢⎣312014⎤⎥⎦[xxyy]=⎡⎢⎣772x2x66⎤⎥⎦,then (x,y) is

The correct option is B (2,1) ⎡⎢⎣312014⎤⎥⎦[xxyy]=⎡⎢⎣772x2x66⎤⎥⎦⇒⎡⎢⎣3x+y3x+y2x2xx+4yx+4y⎤⎥⎦=⎡⎢⎣772x2x66⎤⎥⎦⇒3x+y=7 ...(1)⇒x+4y=6 ...(2)⇒x=2, y=1

$If ⎡ ⎢ ⎣ \begin{matrix} 3 & 1 2 & 0 1 & 4 \end{matrix} ⎤ ⎥ ⎦ [\begin{matrix} x & x y & y \end{matrix}] = ⎡ ⎢ ⎣ \begin{matrix} 7 & 7 2 x & 2 x 6 & 6 \end{matrix} ⎤ ⎥ ⎦, then (x, y) is$