Question

matrix inversion method
x+y = 75, x-4y = 0​

Answer 1

Answer:

this very simple but you need to have more careful to learn

Step-by-step explanation:

given,

   x+y=75 , x-4y=0

The matrix form of the system is    AX=B

$\left[\begin{array}{cc}1&1&\\1&-4\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{cc}75\\0\end{array}\right]$  where,

$\left[\begin{array}{cc}1&1&\\1&-4\end{array}\right] = A$   $\left[\begin{array}{c}x\\y\end{array}\right] = X$  $\left[\begin{array}{cc}75\\0\end{array}\right] = B$  

$AX=B$

post multiply my $A^{-1}$

$(A^{-1}A)X = A^{-1}B$

$X=A^{-1}B$             $[\because A^{-1}A = 1]$

we need to find $A^{-1}$ so we use formula

$A^{-1} = \frac{1}{|A|} (adj A)$

$|A| = \left|\begin{array}{cc}1&1\\1&-4\end{array}\right|\\ = -4-1\\=-5$

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

so we use $A^{-1}$ formula,

$A^{-1} = \frac{1}{-5} \left[\begin{array}{cc}-4&-1\\-1&1\end{array}\right]$

we founded $A^{-1}$

$X=A^{-1}B=\frac{1}{-5} \left[\begin{array}{cc}-4&-1\\-1&1\end{array}\right] \left[\begin{array}{cc}75\\0\end{array}\right]$

                  $=\frac{1}{-5} \left[\begin{array}{ccc}-300\\-75\end{array}\right]$

                  $=\left[\begin{array}{ccc}\frac{-300}{-5}\\\frac{-75}{-5}\end{array}\right]$

              $X=\left[\begin{array}{ccc}60\\15\end{array}\right]$

         $\left[\begin{array}{ccc}x\\y\end{array}\right]=\left[\begin{array}{ccc}60\\15\end{array}\right]$

∴ x=60 and y=15

I hope it very useful

All the Best  :-)

Answer 2

Answer:

Step-by-step explanation:

X+y=75

x-4y=0