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  • If A = matrix 2 -3 5 3 2 -4 1 1 -2, find A^-1. Using A^-1, solve the system of equations 2x - 3y + 5z = 11, 3x + 2y - 4z = -5 , x + y - 2z = -3.

Q : 15        If   A=\begin{bmatrix} 2 &-3 &5 \\ 3 & 2 &-4 \\ 1 &1 &-2 \end{bmatrix} , find A^-^1. Using A^-^1 solve the system of equations 

                  2x-3y+5z=11

                  3x+2y-4z=-5

                     x+y-2z=-3

Answers (1)

best_answer

The given system of equations

 2x-3y+5z=11

  3x+2y-4z=-5

 x+y-2z=-3

can be written in the matrix form of AX =B, where

A=\begin{bmatrix} 2 &-3 &5 \\ 3 & 2 &-4 \\ 1 &1 &-2 \end{bmatrix}, X = \begin{bmatrix} x\\y \\z \end{bmatrix}  and\ B =\begin{bmatrix} 11\\ -5 \\ -3 \end{bmatrix}. 

we have, 

|A| =2(-4+4) +3(-6+4)+5(3-2) = 0-6+5 = -1 \neq0.

So, A is non-singular, Therefore, its inverse A^{-1} exists.

as we know A^{-1} = \frac{1}{|A|} (adjA)

Now, we will find the cofactors;

A_{11} =(-1)^{-4+4} = 0      A_{12} =(-1)^{1+2}(-6+4) = 2

A_{13} =(-1)^{1+3}(3-2) = 1      A_{21} =(-1)^{2+1}(6-5) = -1

A_{22} =(-1)^{2+2}(-4-5) = -9      A_{23} =(-1)^{2+3}(2+3) = -5

A_{31} =(-1)^{3+1}(12-10) = 2     A_{32} =(-1)^{3+2}(-8-15) = 23

A_{33} =(-1)^{3+3}(4+9) = 13

(adjA) =\begin{bmatrix} 0 &-1 &2 \\ 2& -9 & 23\\ 1&-5 & 13 \end{bmatrix}

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

So, the solutions can be found by X = A^{-1}B = \begin{bmatrix} 0 &1 &-2 \\ -2& 9 & -23\\ -1&5 & -13 \end{bmatrix}\begin{bmatrix} 11\\-5 \\ -3 \end{bmatrix}

\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 0-5+6\\-22-45+69 \\ -11-25+39 \end{bmatrix} = \begin{bmatrix} 1\\2 \\ 3 \end{bmatrix}

Hence the solutions of the given system of equations;

x =1,\ y =2,\ and\ \ z=3.

 

Posted by

Divya Prakash Singh

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