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Solve system of linear equations, using matrix method:\[\] \[4x-3y=3\] \[3x-5y=7\]

$\begin{array}{1 1} X=-6/11,Y=-19/11 \\X=12/11,Y=19/11 \\ X=-6/11,Y=19/11 \\ X=-12/11,Y=-19/11 \end{array} $

1 Answer

  • A matrix is said to be singular if |A|= 0.
  • A matrix is said to be invertible if |A|$\neq 0$.
  • If A is a non-singular matrix,AX=B,then $X=A^{-1}B.$
  • Using this we can solve the system of equation which has unique solution.
This can be written in the form AX=B.
$\begin{bmatrix}4 & -3\\3 & -5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\7\end{bmatrix}$
Where $A=\begin{bmatrix}4 & -3\\3 & -5\end{bmatrix}\;X=\begin{bmatrix}x\\y\end{bmatrix}\;B=\begin{bmatrix}3\\7\end{bmatrix}$
Let us see if matrix A is singular and non-singular.Let us find the value of the determinant |A|
$|A|=4\times -5-3\times -3=-20+9=-11\neq 0$
Hence it is non -singular.
So it is invertible.
We know $A^{-1}=\frac{1}{|A|}(adj \;A)$
Now let us find out adj A.
This can be obtained by interchanging the element of $a_{11}$ and $a_{22}$ and interchanging the symbols of elements of $a_{21}$ and $a_{12}$
Therefore $adj A=\begin{bmatrix}-5 & 3\\-3 & 4\end{bmatrix}$ and |A|=-11.
$A^{-1}=\frac{1}{-11}\begin{bmatrix}-5 & 3\\-3 & 4\end{bmatrix}$
Matrix multiplication can be obtained by multiplying the rows of A with the column of B.
Now AX=B$\Rightarrow X=A^{-1}B.$
$X=\frac{1}{11}\begin{bmatrix}-5 & 3\\-3 & 4\end{bmatrix}\begin{bmatrix}3\\7\end{bmatrix}$
$\Rightarrow \begin{bmatrix}x\\y\end{bmatrix}= \begin{bmatrix}-6/11\\-19/11\end{bmatrix}$
Hence x=-6/11 and y=-19/11.


answered Feb 27, 2013 by sreemathi.v
edited Feb 27, 2013 by vijayalakshmi_ramakrishnans
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