Browse Questions

# Solve by matrix inversion method following system of linear equation:$\;2x\;-\;y\;=7\;,\;3x\;-2y\;=11$

Toolbox:
• Let A be a square matrix of order 2. $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Then $adj\: A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ ( shortcut for finding the adjoint )
• 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 equation is $AX = B$ with $A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}, X = \begin{bmatrix} x \\ y \end{bmatrix}, B = \begin{bmatrix} 7 \\ 11 \end{bmatrix}$
$|A| = -4+3 = -1 \neq 0$ The system can be solved by matrix inversion. $X = A^{-1} B$
Step 2
To find $A^{-1} = \large\frac{1}{|A|} adj\: A$
$adj\: A = \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix}$
$A^{-1} = \large\frac{1}{-1} \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$
Step 3
$X = A^{-1}B = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix} \begin{bmatrix} 7 \\ 11 \end{bmatrix} = \begin{bmatrix} 14-11 \\ 21-22 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$
$\therefore x = 3, y = -1$

edited May 27, 2013

+1 vote