If A= \begin{bmatrix} 1 & 1& 1\\ 0& 1&3 \\ 1& -2 &1 \end{bmatrix}, find\: A^{-1}
Hence,solve the following system of equation:
x+y+z= 6,y+3z= 11\: and\: x-2y+z= 0

 

 

 

 
 
 
 
 

Answers (1)

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

cofailous
              A_{11}= 7\: \: \: A_{12}= 3\: \: \: A_{13}= -1
              A_{21}= -3\: \: \: A_{22}= 0\: \: \: A_{23}= 3
              A_{31}= 2\: \: \: A_{32}= -3\: \: \: A_{33}= 1
   A^{-1}= \frac{adj\left ( A \right )}{\left | A \right |}
Adj\left ( A \right )= \begin{bmatrix} 7 & 3&-1 \\ -3& 0 &3 \\ 2& -3 & 1 \end{bmatrix}\Rightarrow \begin{bmatrix} 7 & -3&2 \\ 3& 0 &-3 \\ -1& 3 & 1 \end{bmatrix}
\left | A \right |= 9, \: \: A^{-1}= \frac{1}{9}\begin{bmatrix} 7 &-3 &2 \\ 3& 0&-3 \\ -1& 3 &1 \end{bmatrix}
For syntax of equations
 AX= B
x= A^{-1}B
\begin{bmatrix} x\\ y \\ z \end{bmatrix}= \frac{1}{9}\begin{bmatrix} 7& -3 & 2\\ 3& 0 & -3\\ -1&3 &1 \end{bmatrix}\begin{bmatrix} 6\\ 11 \\ 0 \end{bmatrix}
\begin{bmatrix} x\\ y \\ z \end{bmatrix}= \frac{1}{9}\begin{bmatrix} 9\\ 18 \\ 27 \end{bmatrix}
x= 1,\: \: y= 2,\: \: z= 3
 

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