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Solve by matrix inversion method following system of linear equation $2x-y+z=7\;,3x+y-5z=13\;,x+y+z=5$

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  • Let $ A = [ a_{ij} ] $ be a square matrix x. Let $ A_{ij}$ be the cofactor of $ a_{ij}$. Then $ [ A_{ij}]$ is the matrix of cofactors and $ adj\: A $ ( or adjoint of the matrix A) is given by $ adj\: A=[A_{ij}]^T$
  • The inverse of a non-singular square matrix A is given by $ A^{-1} = \frac{1}{|A|} adj\: A.$ A non-singular matrix is one whose determinant value is nonzero.
  • A system of x linear non-homogeneous equations in x unknowns given by
  • $ a_{11}x_1 +a_{12}x_2 \dotsc + a_{1n}x_n = b_1$
  • $ a_{22}x_1 +a_{22}x_2 \dotsc + a_{2n}x_n = b_2$
  • $\: \: \: \: \vdots \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \vdots $
  • $ a_{n1}x_1 +a_{n2}x_2 \dotsc + a_{nn}x_n = b_n$ can be written as $ AX = B$ in matrix form where $ A = [a_{ij}], x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$ and $ B = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix}$ If A is non-singular $ ( |A| \neq 0 )$ the inverse $A^{-1}$ exists and the system can be solved by matrix inversion $ X = A^{-1}B$
Step 1
The system is written as $ AX = B$ in matrix form
$ A = \begin{bmatrix} 2 & -1 & 1 \\ 3 & 1 & -5 \\ 1 & 1 & 1 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} , B = \begin{bmatrix} 7 \\ 13 \\ 5 \end{bmatrix}$
$ |A| = \begin{vmatrix} 2 & -1 & 1 \\ 3 & 1 & -5 \\ 1 & 1 & 1 \end{vmatrix} = 2(1+5)+1(3+5)-1(3-1)$
$ = 12+8+2=22 \neq 0$
$A^{-1}$ exists and the system can be solved by matrix inversion. $ X =A^{-1}B$
Step 2
To find $A^{-1} = \large\frac{1}{|A|} adj\: A$
$ [A_{ij}]$ = cofactor matrix = $ \begin{bmatrix} (1+5) & -(3+5) & (3-1) \\ -(-1-1) & (2-1) & -(2+1) \\ (5-1) & -(10-3) & (2+3) \end{bmatrix}$
$ = \begin{bmatrix}6 & -8 & 2 \\ 2 & 1 & -3 \\ 4 & 13 & 5 \end{bmatrix}$
$A^{-1} = \large\frac{1}{|A|} adj\: A = \large\frac{1}{|A|} [A_{ij}]^T = \large\frac{1}{22} \begin{bmatrix}6 & 2 & 4 \\ -8 & 1 & 13 \\ 2 & -3 & 5 \end{bmatrix}$
Step 3
$ X = A^{-1}B = \large\frac{1}{22} \begin{bmatrix}6 & 2 & 4 \\ -8 & 1 & 13 \\ 2 & -3 & 5 \end{bmatrix} \begin{bmatrix} 7 \\ 13 \\ 5 \end{bmatrix}$
$ = \large\frac{1}{22} \begin{bmatrix}42 & +26 & +20 \\ -56 & +13 & +65 \\ 14 & -39 & +25 \end{bmatrix}$
$ = \large\frac{1}{22} \begin{bmatrix} 88 \\ 22 \\ 0 \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix}$
$ \therefore x = 4, y = 1, z = 0 $

 

answered May 25, 2013 by thanvigandhi_1
edited May 27, 2013 by thanvigandhi_1
 

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