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If A= \begin{bmatrix} 2 &-3 &5 \\ 3& 2& -4\\ 1 & 1 & -2 \end{bmatrix},  find A-1  Hence using A-1  solve the system of equations 2x-3y+5z= 11, 3x+2y-4z= -5,x+y-2z= -3.

 

 

 

 
 
 
 
 

Answers (1)

we have, A= \begin{bmatrix} 2 &-3 & 5\\ 3 & 2 &-4 \\ 1& 1 & -2 \end{bmatrix}
\therefore \left | A \right |= \begin{vmatrix} 2 & -3 &5 \\ 3 & 2 & -4\\ 1& 1 & -2 \end{vmatrix}= 2\left ( -4+4 \right )+3\left ( -6+4 \right )+5\left ( 3-2 \right )
                                               = 2\left ( 0 \right )+3\left ( -2 \right )+5\left ( 1 \right )
                                                = -1\neq 0
Hence , A is invertible and A-1 exists
Let Aij be the cofactor of element aij in A= \left [ aij \right ]
Then A_{11}= \left ( -1 \right )^{1+1}\begin{vmatrix} 2 & -4\\ 1 & -2 \end{vmatrix}= 0   A_{12}= \left ( -1 \right )^{1+2}\begin{vmatrix} 3 & -4\\ 1 & -2 \end{vmatrix}= 2
A_{13}= \left ( -1 \right )^{1+3}\begin{vmatrix} 3 & 2\\ 1 & 1 \end{vmatrix}= 1, A_{21}= \left ( -1 \right )^{2+1}\begin{vmatrix} -3 & 5\\ 1 & -2 \end{vmatrix}= -1
A_{22}= \left ( -1 \right )^{2+2}\begin{vmatrix} 2 & 5\\ 1 & -2 \end{vmatrix}= -9     A_{23}= \left ( -1 \right )^{2+3}\begin{vmatrix} 2 & -3\\ 1 & 1 \end{vmatrix}= -5
A_{31}= \left ( -1 \right )^{3+1}\begin{vmatrix} -3 & 5\\ 2 & -4 \end{vmatrix}= 2        A_{32}= \left ( -1 \right )^{3+2}\begin{vmatrix} 2 & 5\\ 3 & -9 \end{vmatrix}= 23
A_{33}= \left ( -1 \right )^{3+3}\begin{vmatrix} 2 & -3\\ 3 & 2 \end{vmatrix}= 13
\therefore adj A= \begin{bmatrix} 0 & 2 & 1\\ -1& -9 & -5\\ 2 & 23 & 13 \end{bmatrix}^{T}= \begin{bmatrix} 0 & -1&2 \\ 2 & -9 &23 \\ 1 & -5 & 13 \end{bmatrix}
\therefore A^{-1}=\frac{1}{\left | A \right |} adj A= \frac{1}{-1} \begin{bmatrix} 0 & -1 & 2\\ 2& -9 & 23\\ 1 &-5 & 13 \end{bmatrix}                        
= \begin{bmatrix} 0 & 1 & -2\\ -2& 9 & -23\\ -1 &5 & -13 \end{bmatrix}
Now, the given system of equations is exprerrible as
\begin{bmatrix} 2 & -3 & 5\\ 3& 2 & -4\\ 1 &1 & -2 \end{bmatrix}\begin{bmatrix} x\\ y \\ z \end{bmatrix}= \begin{bmatrix} 11\\ -5 \\ 3 \end{bmatrix}
  or AX = B
where A= \begin{bmatrix} 2 & -3 & 5\\ 3& 2 & -4\\ 1 &1 & -2 \end{bmatrix},\begin{bmatrix} x\\ y \\ z \end{bmatrix}and B= \begin{bmatrix} 11\\ -5 \\ 3 \end{bmatrix}
Now  AX= B \Rightarrow A^{-1}AX= A^{-1}B
\therefore X = A^{-1}B
\Rightarrow X= \begin{bmatrix} 0 & 1 & -2\\ -2& 9&-23 \\ -1& 5 & -13 \end{bmatrix}\begin{bmatrix} 11\\ -5 \\ -3 \end{bmatrix}
X= \begin{bmatrix} x\\ y \\ z \end{bmatrix}= \begin{bmatrix} 1\\ 2 \\ 3 \end{bmatrix}\; \; \therefore x= 1,y= 2,z= 3

Posted by

Ravindra Pindel

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