If A= \begin{bmatrix} 0 & 1 & 2\\ 1& 2 &3 \\ 3& 1&1 \end{bmatrix}, find A-1. Using A-1 , solve the system of equations
y+2z= 5
x+2y+3z= 10
3x+y+z= 9

 

 

 

 
 
 
 
 

Answers (1)

A= \begin{bmatrix} 0 & 1 &2 \\ 1& 2 &3 \\ 3 & 1 &1 \end{bmatrix}       

  y+2z= 5

x+2y+3z= 10

 3x+y+z= 9
\therefore \left | A \right |= 0 \left ( 2-3 \right )-1\left ( 1-9 \right )+2\left ( 1-6 \right )
           = -1\left ( -8 \right )+2\left ( -5 \right )
            \Rightarrow 8-10
          = -2\: \: \: \therefore A^{-1}exist
Now,
A_{11}= -1\; \; A_{12}= 8\; \; A_{13}= -5
A_{21}= 1\; \; A_{22}= -6\; \; A_{23}= 3
A_{31}= -1\; \; A_{32}= 2\; \; A_{33}= -1
\therefore A^{-1}= \frac{1}{\left | A \right |}\left ( adj\: A \right )= \frac{-1}{2}\begin{bmatrix} -1 & 1 &-1 \\ 8& -6 &2 \\ -5& 3 & -1 \end{bmatrix}---(1)
Now the given system of the equation can be written in the form of AX = B where
A= \begin{bmatrix} 0 & 1 &2 \\ 1 & 2 &3 \\ 3 & 1 &1 \end{bmatrix}x= \begin{bmatrix} x\\ y \\z \end{bmatrix}\S B= \begin{bmatrix} 5\\ 10 \\9 \end{bmatrix}
The solution of the given of equation is given by
    X= A^{-1}B
\begin{bmatrix} x\\ y \\ z \end{bmatrix}= \frac{-1}{2}\begin{bmatrix} -1 & 1 &-1 \\ 8 & -6 &2 \\ -5& 3 &-1 \end{bmatrix}\begin{bmatrix} 5\\ 10 \\ 9 \end{bmatrix}\: Using (1)
        = \frac{1}{2}\begin{bmatrix} 1 & -1 & 1\\ -8& 6 & -2\\ 5 & -3 & 1 \end{bmatrix}\begin{bmatrix} 5\\ 10 \\ 9 \end{bmatrix}
       = \frac{1}{2}\begin{bmatrix} 1 -10 +9\\ -40 +60 -18\\ 25 -30 +9 \end{bmatrix}
Value of x,y,z are 0,1,2 respectively.
                                  

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