Question

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Answer 1

Answer :-

$\red{A = \left[\begin{array}{cc}\cfrac{-1}{13} &\dfrac{14}{13} \\\dfrac{4}{13} &\dfrac{28}{13} \\\end{array}\right]}$

$\red{B = \left[\begin{array}{cc}\cfrac{-5}{13} &\dfrac{-8}{13} \\\dfrac{-6}{13} &\dfrac{-16}{13} \\\end{array}\right]}$

Given :-

$2A-3B = \left[\begin{array}{cc}1&4\\2&8\\\end{array}\right]$

$3A+2B =\left[\begin{array}{cc}-1&2\\0&4\\\end{array}\right]$

To find :-

Value of A, B

Solution :-

In order to get the value of A, B

Lets say ,

$2A-3B = \left[\begin{array}{cc}1&4\\2&8\\\end{array}\right] ---- eq 1$

$3A+2B =\left[\begin{array}{cc}-1&2\\0&4\\\end{array}\right] --- eq 2$

We shall multiply the equation 1 with 3 and equation 2 with 3 because any one of the variable gets eliminated and we can get any variable value

So,

$(2A-3B)3= 3\left[\begin{array}{cc}1&4\\2&8\\\end{array}\right]$

Multiplying each element in matric with 3

$6A- 9B = \left[\begin{array}{cc}1\times 3&4\times 3\\2\times 3&8\times 3\\\end{array}\right]$

$\red{6A- 9B = \left[\begin{array}{cc}3&12\\6&24\\\end{array}\right]} ----- eq 3$

$(3A+2B)2 = 2\left[\begin{array}{cc}-1&2\\0&4\\\end{array}\right]$

Multiplying each element in matric with 2

$(6A+4B)= \left[\begin{array}{cc}-1\times 2&2\times 2\\0\times 2&4\times 2\\\end{array}\right]$

$\red{(6A+4B)= \left[\begin{array}{cc}-2&4\\0\ &8\\\end{array}\right]} ----- eq 4$

Now we got the equation 3 and equation 4 In order to get value of A Or B we shall subtract the both equations

$6A- 9B-(6A+4B) = \left[\begin{array}{cc}3&12\\6&24\\\end{array}\right] -\left[\begin{array}{cc}-2&4\\0\ &8\\\end{array}\right]$

$6A- 9B-6A-4B = \left[\begin{array}{cc}3-(-2)&12-(4)\\6-0&24-8\\\end{array}\right]$

$-13B = \left[\begin{array}{cc}5&8\\6&16\\\end{array}\right]$

We need B So, Divide the each element in matric with -13

$\dfrac{-13B}{-13} = \left[\begin{array}{cc}\cfrac{-5}{13} &\dfrac{-8}{13} \\\dfrac{-6}{13} &\dfrac{-16}{13} \\\end{array}\right]$

$\red{B = \left[\begin{array}{cc}\cfrac{-5}{13} &\dfrac{-8}{13} \\\dfrac{-6}{13} &\dfrac{-16}{13} \\\end{array}\right]}$

So, substitute B value in equation in order to get value of A

$2A-3B = \left[\begin{array}{cc}1&4\\2&8\\\end{array}\right]$

At ,

$B = \left[\begin{array}{cc}\cfrac{-5}{13} &\dfrac{-8}{13} \\\dfrac{-6}{13} &\dfrac{-16}{13} \\\end{array}\right]$

Substitute 'B' value

$\begin{gathered}2A-3 \left[\begin{array}{cc}\dfrac{-5}{13} &\dfrac{-8}{13} \\ \dfrac{-6}{13} &\dfrac{-16}{13} \\\end{array}\right] =\left[\begin{array}{cc}1&4\\2&8\\ \end{array}\right]\end{gathered}$

$\begin{gathered}2A-3 \left[\begin{array}{cc}\dfrac{-5}{13} &\dfrac{-8}{13} \\\dfrac{-6}{13} &\dfrac{-16}{13} \\\end{array}\right] = \left[\begin{array}{cc}1&4\\ 2&8\\\end{array}\right]\end{gathered}$

$\begin{gathered}2A-\ \left[\begin{array}{cc}\cfrac{-5\times 3}{13} &\dfrac{-8\times 3}{13} \\\dfrac{-6\times 3}{13} &\dfrac{-16\times 3}{13} \\\end{array}\right] = \left[\begin{array}{cc}1&4\\2&8\\\end{array}\right]\end{gathered}$

$\begin{gathered}2A-\ \left[\begin{array}{cc}\cfrac{-15}{13} &\dfrac{-24}{13} \\\dfrac{-18}{13} &\dfrac{-48}{13} \\\end{array}\right] = \left[\begin{array}{cc}1&4\\2&8\\\end{array}\right]\end{gathered}$

$\begin{gathered}2A=\ \left[\begin{array}{cc}\cfrac{-15}{13} &\dfrac{-24}{13} \\\dfrac{-18}{13} &\dfrac{-48}{13} \\\end{array}\right] + \left[\begin{array}{cc}1&4\\2&8\\\end{array}\right]\end{gathered}$

$\begin{gathered}2A=\ \left[\begin{array}{cc}\cfrac{-15}{13}+1 &\dfrac{-24}{13}+4 \\\dfrac{-18}{13} +2&\dfrac{-48}{13} +8\\\end{array}\right]\end{gathered}$

$\begin{gathered}2A=\ \left[\begin{array}{cc}\cfrac{-2}{13} &\dfrac{28}{13} \\\dfrac{8}{13} &\dfrac{56}{13} \\ \end{array}\right]\end{gathered}$

$\begin{gathered}A=\ \left[\begin{array}{cc}\cfrac{-2}{26} &\dfrac{28}{26} \\\dfrac{8}{26} &\dfrac{56}{26} \\\end{array}\right]\end{gathered}$

Converting into lower terms of fraction

$\red{A =\left[\begin{array}{cc}\cfrac{-1}{13} &\dfrac{14}{13} \\\dfrac{4}{13} &\dfrac{28}{13} \\\end{array}\right]}$

Answer 2

Step-by-step explanation:

Given:

$2A-3B=\left[\begin{array}{cc}1&4\\2&8\end{array}\right]...eq1\\\\3A+2B=\left[\begin{array}{cc}-1&2\\0&4\end{array}\right]..eq2$

To find:Matrix A and B

Solution: These are equations of two matrix, can be solved by following steps

Step 1:Multiply eq1 by 2

$4A-6B=\left[\begin{array}{cc}2&8\\4&16\end{array}\right]...eq3$

Step 2:Multiply eq2 by 3

$9A+6B=\left[\begin{array}{cc}-3&6\\0&12\end{array}\right]..eq4$

Step 3: Add eq3 and eq4

$4A-6B+9A+6B=\left[\begin{array}{cc}2&8\\4&16\end{array}\right]+\left[\begin{array}{cc}-3&6\\0&12\end{array}\right]$

$13A=\left[\begin{array}{cc}2-3&8+6\\4+0&16+12\end{array}\right]$

$A=\frac{1}{13}\left[\begin{array}{cc}-1&14\\4&28\end{array}\right]$

or

$A=\left[\begin{array}{cc}-\frac{1}{13}&\frac{14}{13}\\\\\frac{4}{13}&\frac{28}{13}\end{array}\right]$

Step4:Put value of A in eq1

$2\left[\begin{array}{cc}-\frac{1}{13}&\frac{14}{13}\\\\\frac{4}{13}&\frac{28}{13}\end{array}\right]-3B=\left[\begin{array}{cc}1&4\\2&8\end{array}\right]$

Step 5: Solve for matrix B,

$3B=\frac{2}{13}\left[\begin{array}{cc}-1&14\\4&28\end{array}\right]-\left[\begin{array}{cc}1&4\\2&8\end{array}\right]$

$3B=\left[\begin{array}{cc}-\frac{2}{13}-1&\frac{28}{13}-4\\\\\frac{8}{13}-2&\frac{56}{13}-8\end{array}\right]$

$3B=\left[\begin{array}{cc}\frac{-2-13}{13}&\frac{28-52}{13}\\\\\frac{8-26}{13}&\frac{56-104}{13}\end{array}\right]$

$B=\frac{1}{3}\left[\begin{array}{cc}\frac{-15}{13}&\frac{-24}{13}\\\\\frac{-18}{13}&\frac{-48}{13}\end{array}\right]$

or

take 3 common from matrix

$B=\frac{3}{3}\left[\begin{array}{cc}\frac{-5}{13}&\frac{-8}{13}\\\\\frac{-6}{13}&\frac{-16}{13}\end{array}\right]$

or

$B=\left[\begin{array}{cc}\frac{-5}{13}&\frac{-8}{13}\\\\\frac{-6}{13}&\frac{-16}{13}\end{array}\right]$

Final answer:

$A=\left[\begin{array}{cc}-\frac{1}{13}&\frac{14}{13}\\\\\frac{4}{13}&\frac{28}{13}\end{array}\right]$

$B=\left[\begin{array}{cc}\frac{-5}{13}&\frac{-8}{13}\\\\\frac{-6}{13}&\frac{-16}{13}\end{array}\right]$

Hope it helps you.

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