Q : 10        By using properties of determinants, show that:

                (i) \begin{vmatrix} x+4 &2x &2x \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}=(5x+4)(4-x)

Answers (1)

Given determinant: 

                                     \begin{vmatrix} x+4 &2x &2x \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}

Applying row transformation: R_{1} \rightarrow R_{1} + R_{2} + R_{3}  then we have;

\triangle = \begin{vmatrix} 5x+4 &5x+4 &5x+4 \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}

Taking a common factor: 5x+4

= (5x+4)\begin{vmatrix} 1 &1 &1 \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}

Now, applying column transformations C_{1} \rightarrow C_{1}- C_{2}  and C_{2} \rightarrow C_{2}- C_{3}

= (5x+4)\begin{vmatrix} 0 &0 &1 \\ x-4 & 4-x & 2x\\ 0 & x-4 & x+4 \end{vmatrix}

= (5x+4)(4-x)(4-x)\begin{vmatrix} 0 &0 &1 \\ 1 & 1 & 2x\\ 0 & 1 & x+4 \end{vmatrix}

= (5x+4)(4-x)^2

 

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