Question

Find matrices A and B such that A + B = $\left[\begin{array}{ccc}5&4\\7&3\end{array}\right]$ AND A - B = $\left[\begin{array}{ccc}11&2\\-1&7\end{array}\right]$

Answer 1

Answer:

★Given:-

$A + B = \left[\begin{array}{ccc}5&4\\7&3\end{array}\right]$

$A - B = \left[\begin{array}{ccc}11&2\\-1&7\end{array}\right]$

★Find:-

• Find matrices A and B

★Solution:-

By doing A + B,

$: { \implies{A + B + A - B = \left[\begin{array}{ccc}5&4\\7&3\end{array}\right] + \left[\begin{array}{ccc}11&2\\-1&7\end{array}\right]}}$

$: { \implies{A + A = \left[\begin{array}{ccc}5 + 11&4 + 2\\7 + ( - 1)&3 + 7\end{array}\right] }}$

$: { \implies{2A = \left[\begin{array}{ccc}16&6\\6&10\end{array}\right] }}$

$: { \implies{ \frac{2A}{2} = \left[\begin{array}{ccc} \frac{16}{2} & \frac{6}{2} \\ \frac{6}{2} & \frac{10}{2} \end{array}\right] }}$

${ \bf{ : { \implies{ A = \left[\begin{array}{ccc} 8& 3\\ 3 & 5 \end{array}\right] }}}}$

By Substituting A matrix in A + B,

$: { \implies{\left[\begin{array}{ccc}8&3\\3&5\end{array}\right]+ B = \left[\begin{array}{ccc}5&4\\7&3\end{array}\right]}}$

$: { \implies{B = \left[\begin{array}{ccc}5&4\\7&3\end{array}\right] - \left[\begin{array}{ccc}8&3\\3&5\end{array}\right]}}$

$: { \implies{B = \left[\begin{array}{ccc}5 - 8&4 - 3\\7 - 3&3 - 5\end{array}\right] }}$

$: {\bf{ \implies{B = \left[\begin{array}{ccc} - 3&1\\4& - 2\end{array}\right] }}}$

Therefore,

• $\sf{Matrix \: { A = \left[\begin{array}{ccc} 8& 3\\ 3 & 5 \end{array}\right] }}$
• ${ \sf{Matrix \: B = \left[\begin{array}{ccc} - 3&1\\4& - 2\end{array}\right] }}$

Hope it's Helps you ☻

Answer 2

Step-by-step explanation:

Given :-

A + B = $\pmb{\bf{\left[\begin{array}{ccc}5&4\\7&3\\\end{array}\right] }}$

A - B = $\pmb{\bf{\left[\begin{array}{ccc}11&2\\-1&7\\\end{array}\right] }}$

To find :-

A and B ?

Solution :-

A + B = $\pmb{\bf{\left[\begin{array}{ccc}5&4\\7&3\\\end{array}\right] }}$ - (1)

A - B = $\pmb{\bf{\left[\begin{array}{ccc}11&2\\-1&7\\\end{array}\right] }}$ - (2)

Adding (1) & (2) :

=> (A + B) + (A - B) =  $\pmb{\bf{\left[\begin{array}{ccc}5&4\\7&3\\\end{array}\right] }}$ +  $\pmb{\bf{\left[\begin{array}{ccc}11&2\\-1&7\\\end{array}\right] }}$

=> 2A = $\pmb{\bf{\left[\begin{array}{ccc}16&6\\6&10\\\end{array}\right] }}$

=> A = $\pmb{\bf{\left[\begin{array}{ccc}(\frac{16}{2} )&(\frac{6}{2} )\\(\frac{6}{2} )&(\frac{10}{2} )\end{array}\right] }}$

=> A = $\pmb{\bf{\left[\begin{array}{ccc}8&3\\3&5\\\end{array}\right] }}$

∴ A = $\pmb{\bf{\left[\begin{array}{ccc}8&3\\3&5\\\end{array}\right] }}$

Substitute value of A in (1) :

=> A + B = $\pmb{\bf{\left[\begin{array}{ccc}5&4\\7&3\\\end{array}\right] }}$

=>  $\pmb{\bf{\left[\begin{array}{ccc}8&3\\3&5\\\end{array}\right] }}$ + B = $\pmb{\bf{\left[\begin{array}{ccc}5&4\\7&3\\\end{array}\right] }}$

=> B = $\pmb{\bf{\left[\begin{array}{ccc}5&4\\7&3\\\end{array}\right] }}$  -  $\pmb{\bf{\left[\begin{array}{ccc}8&3\\3&5\\\end{array}\right] }}$

=> B = $\pmb{\bf{\left[\begin{array}{ccc}-3&1\\4&-2\\\end{array}\right] }}$

∴ B = $\pmb{\bf{\left[\begin{array}{ccc}-3&1\\4&-2\\\end{array}\right] }}$

Verification :-

=> A + B = $\pmb{\bf{\left[\begin{array}{ccc}5&4\\7&3\\\end{array}\right] }}$

=>  $\pmb{\bf{\left[\begin{array}{ccc}8&3\\3&5\\\end{array}\right] }}$ +  $\pmb{\bf{\left[\begin{array}{ccc}-3&1\\4&-2\\\end{array}\right] }}$ = $\pmb{\bf{\left[\begin{array}{ccc}5&4\\7&3\\\end{array}\right] }}$

=> $\pmb{\bf{\left[\begin{array}{ccc}5&4\\7&3\\\end{array}\right] }}$ = $\pmb{\bf{\left[\begin{array}{ccc}5&4\\7&3\\\end{array}\right] }}$

∴ L.H.S = R.H.S