Question

please prove
Topic-Determinants​

Answer 1

Question :-

Prove that

$\rm \: \begin{gathered}\sf \left | \begin{array}{ccc}x + y + 2z&x&y\\z&y + z + 2x&y\\z&x& z + x + 2y\end{array}\right | \end{gathered} = 2 {(x + y + z)}^{3}$

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: \begin{gathered}\sf \left | \begin{array}{ccc}x + y + 2z&x&y\\z&y + z + 2x&y\\z&x& z + x + 2y\end{array}\right | \end{gathered}$

$\qquad \: \: \: \: \: \: \boxed{\sf \pmb{  \:\rm \: OP \:C_1\: \to \: C_1 + C_2 + C_3 \: }}$

So, on applying this operation, we get

$\rm \: = \begin{gathered}\sf \left | \begin{array}{ccc}x + y + 2z + x + y&x&y\\z + y + z + 2x + y&y + z + 2x&y\\z + x + z + x + 2y&x& z + x + 2y\end{array}\right | \end{gathered}$

$\rm \: = \begin{gathered}\sf \left | \begin{array}{ccc}2(x + y + z)&x&y\\2(x + y + z)&y + z + 2x&y\\2(x + y + z)&x& z + x + 2y\end{array}\right | \end{gathered}$

Take out 2(x + y + z) common from first column , we get

$\rm \: = 2(x + y + z)\begin{gathered}\sf \left | \begin{array}{ccc}1&x&y\\1&y + z + 2x&y\\1&x& z + x + 2y\end{array}\right | \end{gathered}$

$\qquad \: \: \: \: \: \: \boxed{\sf \pmb{  \:\rm \: OP \:R_2 \: \to \: R_2 - R_1 \: }}$

and

$\qquad \: \: \: \: \: \: \boxed{\sf \pmb{  \:\rm \: OP \:R_3 \: \to \: R_3 - R_1 \: }}$

So, on applying these operations, we get

$\rm \: = 2(x + y + z)\begin{gathered}\sf \left | \begin{array}{ccc}1&x&y\\0&y + z + 2x - x&0\\0&0& z + x + 2y - y\end{array}\right | \end{gathered}$

$\rm \: = 2(x + y + z)\begin{gathered}\sf \left | \begin{array}{ccc}1&x&y\\0&y + z + x&0\\0&0& z + x + y\end{array}\right | \end{gathered}$

can be further rewritten as

$\rm \: = 2(x + y + z)\begin{gathered}\sf \left | \begin{array}{ccc}1&x&y\\0&x + y + z&0\\0&0& x + y + z\end{array}\right | \end{gathered}$

Now, Expanding along first column, we get

$\rm \: =  \: 2(x + y + z)\bigg( {(x + y + z)}^{2} - 0 \bigg)$

$\rm \: =  \: 2 {(x + y + z)}^{3}$

Hence,

$\rm \: \begin{gathered}\sf \left | \begin{array}{ccc}x + y + 2z&x&y\\z&y + z + 2x&y\\z&x& z + x + 2y\end{array}\right | \end{gathered} = 2 {(x + y + z)}^{3}$

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Additional Information :-

1. The determinant value remains unaltered if rows and columns are interchanged.

2. The determinant value is 0, if two rows or columns are identical.

3. The determinant value is multiplied by - 1, if successive rows or columns are interchanged.

4. The determinant value remains unaltered if rows or columns are added or subtracted.

Answer 2

Step-by-step explanation:

$\pmb{ \begin{aligned} \text { Solution: L.H.S. } &=\left|\begin{array}{ccc}x+y+2 z & x & y \\ \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right| \\ \\ \\ & \qquad\left[\begin{array}{c}\text { On applying } \\ C_{1} \rightarrow C_{1}+C_{2}+C_{3}\end{array}\right] \\ \\ \\ &=\left|\begin{array}{ccc}2 x+2 y+2 z & x & y \\ 2 x+2 y+2 z & y+z+2 x & y \\ 2 x+2 y+2 z & x & z+x+2 y\end{array}\right| \\ \\ \\ &=(2 x+2 y+2 z)\left|\begin{array}{ccc}1 & x & y \\ 1 & y+z+2 x & y \\ 1 & x & z+x+2 y\end{array}\right| \quad\left[\begin{array}{c}\text { On taking common } \\ (2 x+2 y+2 z) \text { from } C_{1}\end{array}\right] \\ \\ \\ &=2(x+y+z)\left|\begin{array}{ccc}1 & x & y \\ 0 & y+z+x & 0 \\ 0 & 0 & z+x+y\end{array}\right| \quad\left[\begin{array}{c}\text { On applying } \\ R_{2} \rightarrow R_{2}-R_{1} \\ R_{3} \rightarrow R_{3}-R_{1}\end{array}\right] \\ \\ &=2(x+y+z)\left[1\left\{(x+y+z)^{2}-0\right\}-0+0\right] \quad\left[\text { On expanding along } C_{1}\right] \\ \\ &=2(x+y+z)^{3} \end{aligned} }$