cindy974
Answered

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Only the a) part please!!!!

Only The A Part Please class=

Sagot :

loneguy

[tex](i)\\\\2A = \begin{pmatrix}3x \times 2 & 1 \times 2 & 5 \times 2\\0\times 2 & 2x \times 2 & -3 \times 2\end{pmatrix} = \begin{pmatrix}6x & 2 & 10\\0 & 4x & -6\end{pmatrix}\\\\\\(ii)\\\\2A + B = C\\\\\implies \begin{pmatrix}6x & 2 & 10\\0 & 4x & -6\end{pmatrix} + \begin{pmatrix}y & 0 & -4\\1 & -3y & 2\end{pmatrix} = \begin{pmatrix}31& 2 & 6\\1 & 17& -4\end{pmatrix}[/tex]

[tex]\implies \begin{pmatrix}6x+y & 2+0 & 10-4\\0+1 & 4x-3y & -6+2\end{pmatrix} = \begin{pmatrix}31 & 2 & 6\\1 & 17& -4 \end{pmatrix}\\\\\\\implies \begin{pmatrix}6x+y & 2 & 6\\1 & 4x-3y & -4\end{pmatrix} = \begin{pmatrix}31 & 2 & 6\\1 & 17& -4 \end{pmatrix}\\\\\text{System of equations},\\\\6x +y =31,~~ 4x -3y = 17\\\\\\(iii)\\\\6x +y-31 =0 \\\\4x-3y -17=0\\\\\text{Solving by cross multiplication method}\\\\\\\dfrac x{1(-17) - (-31)(-3)} = \dfrac y{4(-31) - 6(-17)} = \dfrac 1{6(-3) - 4(1)}[/tex]

[tex]\implies \dfrac{x}{-17 -93} = \dfrac{y}{-124 +102} = \dfrac 1{-18-4}\\\\\\\implies -\dfrac{x}{110} = -\dfrac{y}{22} = -\dfrac 1{22}\\\\ \implies \dfrac{x}{110} = \dfrac{y}{22} = \dfrac 1{22}\\\ \\\text{Hence,}\\\\\\x=\dfrac{110}{22}=5,~~~~ y= \dfrac{22}{22} =1[/tex]

sqdancefan

Answer:

  (i)

  [tex]2A=\left[\begin{array}{ccc}6x&2&10\\0&4x&-6\end{array}\right][/tex]

  (ii) 6x+y=31; 4x-3y=17

  (iii) (x, y) = (5, 1)

Step-by-step explanation:

(i)

The matrix 2A is written by multiplying each element of A by 2.

  [tex]2A=\left[\begin{array}{ccc}6x&2&10\\0&4x&-6\end{array}\right][/tex]

__

(ii)

Variables are seen in elements (1, 1) and (2, 2) of the matrices A and B, so the simultaneous equations can be chosen from those elements of the equation 2A+B=C.

  • 6x+y=31
  • 4x-3y=17

__

(iii)

A graphing calculator shows the solution to be (x, y) = (5, 1).

View image sqdancefan
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