Solve for x and y- -2 0- 3 1 - -1- 2 x -3- -2- 1 -2- y- 3 -

amp;[[ 2 amp; 0; 3 amp; 1 ]][[ 1; 2 x ]]+3[[ 2; 1 ]]=2[[ y; 3 ]] [0.5 Mark] amp;⇒[[ 2+0; 3+2 x ]]+[[ 6; 3 ]]=[[ 2 y; 6 ]] ...

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Byju's Answer

Standard XII

Physics

Solenoid

Solve for x a...

Question

Solve for $x$ and $y :$
$[\begin{matrix} - 2 & 0 3 & 1 \end{matrix}] [\begin{matrix} - 1 2 x \end{matrix}] + 3 [\begin{matrix} - 2 1 \end{matrix}] = 2 [\begin{matrix} y 3 \end{matrix}]$ .

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Solution

$\begin{matrix} [\begin{matrix} - 2 & 0 3 & 1 \end{matrix}] [\begin{matrix} - 1 2 x \end{matrix}] + 3 [\begin{matrix} - 2 1 \end{matrix}] = 2 [\begin{matrix} y 3 \end{matrix}] [0.5 Mark] \Rightarrow [\begin{matrix} 2 + 0 - 3 + 2 x \end{matrix}] + [\begin{matrix} - 6 3 \end{matrix}] = [\begin{matrix} 2 y 6 \end{matrix}] \Rightarrow [\begin{matrix} 2 - 3 + 2 x \end{matrix}] + [\begin{matrix} - 6 3 \end{matrix}] = [\begin{matrix} 2 y 6 \end{matrix}] [0.5 Mark] \Rightarrow [\begin{matrix} 2 - 6 - 3 + 2 x + 3 \end{matrix}] = [\begin{matrix} 2 y 6 \end{matrix}] \Rightarrow [\begin{matrix} - 4 2 x \end{matrix}] = [\begin{matrix} 2 y 6 \end{matrix}] \Rightarrow 2 y = - 4 and 2 x = 6 \Rightarrow y = - 2 and x = 3 \end{matrix}$
Thus, the values of $x$ and $y$ are: $3, - 2$ [1 Mark]

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Similar questions

Complete the magic square:

$\begin{matrix} 14 & 0 8 & 6 & 11 4 & 7 2 & 1 & 12 \end{matrix}$

If x varies inversely as y, then

$\begin{matrix} x & □ & 60 y & 2 & 10 \end{matrix}$

Q. lf

∣ ∣ ∣ ∣ \begin{matrix} 1 & a & b c 1 & b & c a 1 & c & a b \end{matrix} ∣ ∣ ∣ ∣ = λ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b^{2} & c^{2} a & b & c 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣

then

λ

is equal to

Q. (a)

If $H = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 2 & 3 & 1 3 & 1 & 2 \end{matrix} ⎤ ⎥ ⎦, G = ⎡ ⎢ ⎣ \begin{matrix} 2 & 6 & - 10 0 & - 4 & 8 2 & - 8 & 4 \end{matrix} ⎤ ⎥ ⎦$

Find $3 H - \frac{1}{2} G .$
$[3]$

(b) If $3 A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 2 2 & 1 & - 2 x & 2 & y \end{matrix} ⎤ ⎥ ⎦$ and $A \cdot A^{T} = I$ , then find the value of $x + y$ .
$[3]$

(c) If $A = [\begin{matrix} 2 & 5 1 & 3 \end{matrix}], B = [\begin{matrix} 4 & - 2 - 1 & 3 \end{matrix}]$ and $I$ is Identity matrix of same order and $A^{t}$ is the transpose of matrix $A$ find $A^{t} . B + B I$ .
$[4]$

Q. Let

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 & 1 2 & 1 & - 3 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦

and

10 B = ⎡ ⎢ ⎣ \begin{matrix} 4 & 2 & 2 - 5 & 0 & α 1 & - 2 & 3 \end{matrix} ⎤ ⎥ ⎦

. If

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is the inverse

o f A

, then

α

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Solve for x and y: [−2031][−12x]+3[−21]=2[y3].

[−2031][−12x]+3[−21]=2[y3] [0.5 Mark]⇒[2+0−3+2x]+[−63]=[2y6]⇒[2−3+2x]+[−63]=[2y6] [0.5 Mark]⇒[2−6−3+2x+3]=[2y6]⇒[−42x]=[2y6]⇒2y=−4 and 2x=6⇒y=−2 and x=3 Thus, the values of x and y are: 3,−2 [1 Mark]

Solve for $x$ and $y :$
$[\begin{matrix} - 2 & 0 3 & 1 \end{matrix}] [\begin{matrix} - 1 2 x \end{matrix}] + 3 [\begin{matrix} - 2 1 \end{matrix}] = 2 [\begin{matrix} y 3 \end{matrix}]$ .