Question

5x-6y + 4z=15
7x+4y - 3z=19
2x + y + 6z=46 answer this in matrix inversion method​

Answer 1

Answer: Matrix inversion Method (solving for "x")

$\left[\begin{array}{ccc}x\\y\\y\end{array}\right]=\left[\begin{array}{ccc}4.67\\4\\6\end{array}\right]$

Step-by-step explanation:

According to Matrix inversion Method:

A*X=B

$I_{3} *X=B$           (⇒ $I_{3} =A^{-1}*A$ )

$X=A^{-1}*B$

Firstly, Equation should be written in Matrix to calculate adjoint of Matrix A :

A*X=B

$\left[\begin{array}{ccc}5&-6&4\\7&4&-3\\2&1&6\end{array}\right]\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}15\\19\\46\end{array}\right]$

2- Now finding the Determinant of Mat[A] (3*3):

$det (A) = 5\left[\begin{array}{ccc}4&-3\\1&6\end{array}\right] -(-6) \left[\begin{array}{ccc}7&3\\2&6\end{array}\right] -4\left[\begin{array}{ccc}7&4\\2&1\end{array}\right]$

$det(A)=5(24-(-3))+6(42-(-6))+4(7-8)  =5(24+3)+6(42+6)+4(-1)\\=5(27)+6(48)-4\\=135+288-4\\=135+284\\=419$

3- Taking Transpose of the matrix (changing rows into columns &columns into rows) :

$A^{t} = \left[\begin{array}{ccc}5&7&2\\-6&4&1\\4&-3&6\end{array}\right]$

4(a)- Taking Adjoint of transpose matrix ( if first digit is selected, leaving the row and column , determinant is taken of the remaining digits ... and then these determinants form the Adjoint Matrix) :

$\left[\begin{array}{ccc}4&1\\-3&6\end{array}\right] =27\\\left[\begin{array}{ccc}-6&1\\4&6\end{array}\right] =-40\\\left[\begin{array}{ccc}-6&4\\4&-3\end{array}\right] =2\\\left[\begin{array}{ccc}7&2\\-3&6\end{array}\right] =48\\\left[\begin{array}{ccc}5&2\\4&6\end{array}\right] =22\\\left[\begin{array}{ccc}5&7\\4&-3\end{array}\right] =-43\\\left[\begin{array}{ccc}7&2\\4&1\end{array}\right] =-1\\\left[\begin{array}{ccc}5&2\\-6&1\end{array}\right] =17\\\left[\begin{array}{ccc}5&7\\-6&4\end{array}\right]=62$

$=\left[\begin{array}{ccc}27&-40&2\\48&22&-43\\-1&17&62\end{array}\right]$

4(b)- Changing the Signs of the Adjoint :

$=\left[\begin{array}{ccc}27&-40&2\\48&22&-43\\-1&17&62\end{array}\right] *\left[\begin{array}{ccc}+&-&+\\-&+&-\\+&-&+\end{array}\right] \\\\Adj(A)=\left[\begin{array}{ccc}27&40&2\\-48&22&43\\-1&-17&62\end{array}\right]$

5- Taking the Inverse of Matrix (A):

Formula: $A^{-1}=\frac{Adj(A)}{det(A)}$

$=\left[\begin{array}{ccc}\frac{27}{419} &\frac{40}{419} &\frac{2}{419} \\\frac{-48}{419} &\frac{22}{419} &\frac{43}{419} \\\frac{-1}{419} &\frac{-17}{419} &\frac{62}{419} \end{array}\right]$

6- Finding the value of Matrix[X]:

$X=A^{-1}*B$

Multiplying every digit of row with the respective number.

$\left[\begin{array}{ccc}\frac{27}{419} &\frac{40}{419} &\frac{2}{419} \\\frac{-48}{419} &\frac{22}{419} &\frac{43}{419} \\\frac{-1}{419} &\frac{-17}{419} &\frac{62}{419} \end{array}\right] *\left[\begin{array}{ccc}15\\19\\46\end{array}\right]$

$=\left[\begin{array}{ccc}(\frac{27}{419}*15)+(\frac{40}{419}*19)+(\frac{2}{419}*46)  \\(\frac{-48}{419}*15)+( \frac{22}{419}*19)+(\frac{43}{419}*46)\\(\frac{-1}{419}*15)+(\frac{-17}{419}*19)+(\frac{62}{419}*46)\end{array}\right]$

Simplifying :

$=\left[\begin{array}{ccc}\frac{405}{419}+\frac{760}{419}+\frac{92}{419}\\\frac{-720}{419}+\frac{418}{419}+\frac{1978}{419}\\\frac{-15}{419}+\frac{-323}{419}+\frac{2852}{419}\end{array}\right]$

$\left[\begin{array}{ccc}\frac{405+760+92}{419} \\\frac{-720+418+1978}{419} \\\frac{-15-323+2852}{419}\end{array}\right]$

$A^{-1}*B=\left[\begin{array}{ccc}\frac{1957}{419} \\\frac{1676}{419} \\\frac{2514}{419} \end{array}\right]$

Hence,

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}4.67\\4\\6\end{array}\right]$

Hope you find it convenient to understand.     :)