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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Solve system of linear equations, using matrix method:\[\] \[5x+2y=3\] \[3x+2y=5\]

$\begin{array}{1 1} x = 1, y = 4\\ x = -1, y = 4\\ x = 4, y = 1 \\ x = -4, y = -1 \end{array} $

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • 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.
 
$\begin{bmatrix}5 & 2\\3 & 2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\5\end{bmatrix}$
 
Where $A=\begin{bmatrix}5 & 2\\3 & 2\end{bmatrix}\;X=\begin{bmatrix}x\\y\end{bmatrix}\;B=\begin{bmatrix}3\\5\end{bmatrix}$
 
Let us now check whether A is a singular or non-singular matrix,by finding the value of the determinant
 
$|A|=(5\times 2-2\times 3)$
 
$\;\;\;=10-6=4\neq 0$
 
Hence it is non -singular and therefore inverse exists.
 
To find the inverse $A^{-1}$,let us first find the adjoint 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 $adj A=\begin{bmatrix}2 & -2\\-3 & 5\end{bmatrix}$ and |A|=4.
 
We know $A^{-1}=\frac{1}{|A|}(adj \;A)$
 
$\qquad\qquad=\frac{1}{4}\begin{bmatrix}2 & -2\\-3 & 5\end{bmatrix}$
 
Since AX=B
 
X=$A^{-1}B$
 
Let us find $A^{-1}B=\frac{1}{4}\begin{bmatrix}2 &-2\\-3 & 5\end{bmatrix}\begin{bmatrix}3\\5\end{bmatrix}$
 
We can do matrix multiplication by multiplying the rows of matrix A by the column of matrix B
 
$A^{-1}B=\frac{1}{4}\begin{bmatrix}6-10\\-9+25\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}$
 
$\Rightarrow \begin{bmatrix}x\\y\end{bmatrix}= \frac{1}{4}\begin{bmatrix}-4\\16\end{bmatrix}$
 
$\Rightarrow \begin{bmatrix}x\\y\end{bmatrix}= \begin{bmatrix}-1\\4\end{bmatrix}$
 
Hence x=-1 and y=4.

 

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

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