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Using the property of determinants and without expanding, prove that determinant -a^2 ab ac ba -b^2 bc ca cb -c^2 = 4a^2b^2c^2 Ex 4.2 Q: 7

Q : 7          Using the property of determinants and without expanding, prove that 

                  \begin{vmatrix} -a^2 &ab &ac \\ ba &-b^2 &bc \\ ca & cb & -c^2 \end{vmatrix}=4a^2b^2c^2

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Given determinant : \dpi{100} \begin{vmatrix} -a^2 &ab &ac \\ ba &-b^2 &bc \\ ca & cb & -c^2 \end{vmatrix}

\triangle = \begin{vmatrix} -a^2 &ab &ac \\ ba &-b^2 &bc \\ ca & cb & -c^2 \end{vmatrix}

As we can easily take out the common factors a,b,c from rows R_{1},R_{2},R_{3} respectively.

So, get then:

=abc \begin{vmatrix} -a &b &c \\ a &-b &c \\ a & b & -c \end{vmatrix}

Now, taking common factors a,b,c from the columns C_{1},C_{2},C_{3} respectively.

=a^2b^2c^2 \begin{vmatrix} -1 &1 &1 \\ 1 &-1 &1 \\ 1 & 1 & -1 \end{vmatrix}

Now, applying rows transformations R_{1} \rightarrow R_{1} + R_{2}   and then R_{3} \rightarrow R_{2} + R_{3} we have;

\triangle = a^2b^2c^2\begin{vmatrix} 0 &0 &2 \\ 1&-1 &1 \\ 2& 0 &0 \end{vmatrix}

Expanding to get R.H.S.

\triangle = a^2b^2c^2 \left ( 2\begin{vmatrix} 1 &-1 \\ 2& 0 \end{vmatrix} \right ) = 2a^2b^2c^2(0+2) =4a^2b^2c^2

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