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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Examine the consistency of the system of equations:\[\] \[\quad 5x -y +4z = 5 \] \[\quad2x + 3y + 5z= 2\] \[\quad 5x - 2y + 6z = -1\]

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Toolbox:
  • (i)A matrix is said to be singular if |A|$\neq 0$.
  • (ii)If A is a non-singular matrix such that
  • AX=B.
  • then $X=A^{-1}X$
  • Using this we can solve the system of equation which has unique solution.
This can be written in the form AX=B.
 
$\begin{bmatrix}5 & -1& 4\\2 & 3& 5\\5 & -2 &6\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\2\\1\end{bmatrix}$
 
Where $A=\begin{bmatrix}5 & -1& 4\\2 & 3& 5\\5 & -2&6\end{bmatrix}\;X=\begin{bmatrix}x\\y\\z\end{bmatrix}\;B=\begin{bmatrix}5\\2\\1\end{bmatrix}$
 
Let us now calculate the determinant value of A by expanding along $R_1$
 
|A|=$5(3\times 6-5\times -2)-(-1)(2\times 6-5\times 5)+4(2\times -2-5\times 3)$
 
$\;\;\;=5(18+10)+1(12-25)+4(-4-15).$
 
$\;\;\;=5(28)+1(-13)+4(-19).$
 
$\;\;\;=140-13-76=51\neq 0.$
 
Therefore A is non-singular.
 
Hence inverse exists.
 
Hence the given system of equation is consistent.

 

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

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