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By using the properties of determinants show that \[ \begin{array}{l} \begin{vmatrix} 1&x&x^2 \\ x^2&1&x \\ x&x^2&1 \end{vmatrix} = (1-x^3)^2 \end{array} \]

1 Answer

  • If each element of a row (or column) of a determinant is multiplied by a constant k ,then its value gets multiplied by k.
  • By this property we can take out any common factor from any one row or any one column of the given determinant.
  • Elementary transformations can be done by
  • 1. Interchanging any two rows or columns. rows.
  • 2. Mutiplication of the elements of any row or column by a non-zero number
  • 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
Let $\bigtriangleup=\begin{vmatrix}1& x & x^2\\x^2 & 1& x\\x &x^2 & 1\end{vmatrix}$
Let us add $R_1,R_2,R_3$,hence we apply $R_1 \rightarrow R_1+R_2+R_3$
Therefore $\bigtriangleup=\begin{vmatrix}1+x+x^2& 1+x+x^2 & 1+x+x^2\\x^2 & 1& x\\x &x^2 & 1\end{vmatrix}$
Let us take $1+x+x^2$ as the common factor from $R_1$
Therefore $\bigtriangleup=(1+x+x^2)\begin{vmatrix}1& 1 & 1\\x^2 & 1& x\\x &x^2 & 1\end{vmatrix}$
By applying $C_1\rightarrow C_1-C_2$ and $C_2\rightarrow C_2-C_3$
Therefore $\bigtriangleup=(1+x+x^2)\begin{vmatrix}0& 0& 1\\x^2-1 & 1-x& x\\x-x^2 &x^2-1 & 1\end{vmatrix}$
But $(x^2-1)=(x+1)(x-1)$
We can also write $x-x^2=x(1-x^2)$
Therefore $\bigtriangleup=(1+x+x^2)\begin{vmatrix}0& 0& 1\\x^2-1 & 1-x& x\\x(1-x)&(x-1)(x+1) & 1\end{vmatrix}$
Therefore $\bigtriangleup=(1+x+x^2)\begin{vmatrix}0& 0& 1\\(x+1)(x-1) & 1-x& x\\x(1-x)&(x-1)(x+1) & 1\end{vmatrix}$
let us take (1-x) as a common factor from $C_1$ and $C_2$.
$\bigtriangleup=(1+x+x^2)(1-x)(1-x)\begin{vmatrix}0& 0& 1\\-(x+1) & 1& x\\x &-(x+1) & 1\end{vmatrix}$
Now expanding along $R_1$
But we know $(1+x+x^2)(1-x)=1-x^3$
Hence $\bigtriangleup=(1-x^3)(1-x)[x^2+2x+1-x]$
Again $(1+x+x^2)(1-x)=1-x^3$
Therefore $\bigtriangleup=(1-x^3)(1-x^3)=(1-x^3)^2$
Hence proved.


answered Feb 22, 2013 by sreemathi.v
edited Feb 24, 2013 by vijayalakshmi_ramakrishnans

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