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Solve by matrix inversion method following system of linear equation $\;x+y+z=9\;,\;2x+5y+7z=72\;,2x+y-z=0$

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Toolbox:
  • 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 matrix form of the given equations is
$ AX = B, \: \: \: A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 9 \\ 52 \\ 0 \end{bmatrix}$
$ |A| = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \end{vmatrix}= 1(-5-7)\: -1(-2-14)+1(2-10)$
$ = -12+16-8 = -4 \neq 0$
The system can be solved by matrix inversion. $ X = A^{-1}B$
Step 2
To find $ A^{-1}$
$ [A_{ij}]$ = cofactor matrix = $\begin{bmatrix} (-5-7) & -(-2-14) & (2-10) \\ -(-1-1) & (-1-2) & -(1-2) \\ (7-5) & -(7-2) & (5-2) \end{bmatrix}$
$ = \begin{bmatrix} -12 & 16 & -8 \\ 2 & -3 & 1 \\ 2 & -5 & 3 \end{bmatrix}$
$ A^{-1} = \large\frac{1}{|A|} adj\: A = \large\frac{1}{|A|} [A_{ij}]^T$
$ = \large\frac{1}{-4} = \begin{bmatrix} -12 & 2 & 2 \\ 16 & -3 & 5 \\ -8 & 1 & 3 \end{bmatrix}$
$ = \large\frac{1}{4} \begin{bmatrix} 12 & -2 & -2 \\ -16 & 3 & -5 \\ 8 & -1 & -3 \end{bmatrix}$
Step 3
$ X = A^{-1}B = \large\frac{1}{4} \begin{bmatrix} 12 & -2 & -2 \\ -16 & 3 & -5 \\ 8 & -1 & -3 \end{bmatrix} \begin{bmatrix} 9 \\ 52 \\ 0 \end{bmatrix}$
$ = \large\frac{1}{4} \begin{bmatrix} 108 & -104 & +0 \\ -144 & +156 & +0 \\ -172 & -52 & +0 \end{bmatrix} = \large\frac{1}{4} \begin{bmatrix} 4 \\ 12 \\ -20 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix} $
$ \therefore x = 1, y = 3, z = 5$

 

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

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