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

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 us multiply $R_3$ by b and by applying $R_1\rightarrow R_1+bR_3$ we get
let $\bigtriangleup=\begin{vmatrix}1+a^2+b^2& 0 & -b(1+a^2+b^2)\\2ab & 1-a^2+b^2 & 2a\\2b &-2a & 1-a^2-b^2\end{vmatrix}$
Now multiply $R_3$ by a and apply $R_2\rightarrow R_2-aR_3$
$\bigtriangleup=\begin{vmatrix}1+a^2+b^2& 0 & -b(1+a^2+b^2)\\0 & 1+a^2+b^2 & a(1+a^2+b^2)\\2b &-2a & 1-a^2-b^2\end{vmatrix}$
Now take $(1+a^2+b^2)$ as a common factor from $R_1$ and $R_2$.
$\bigtriangleup=(1+a^2+b^2)^2\begin{vmatrix}1& 0 & -b\\0 &1 & a\\2b &-2a & 1-a^2-b^2\end{vmatrix}$
Now expanding along $R_1$ we get,
$\bigtriangleup=(1+a^2+b^2)^2[1\times 1(1-a^2-b^2)-a\times -2a-(0)-b(0\times -2a-1\times 2b]$
Hence $\bigtriangleup=\begin{vmatrix}1+a^2-b^2& 2ab & -2b\\2ab & 1-a+b^2 & 2a\\2b &-2a & 1-a^2-b^2\end{vmatrix}=(1+a^2+b^2)^3$
Hence proved.


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

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