Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Determinants

Question:

Choose the correct value for

\(\begin{vmatrix}(x + y)^2 & zx & zy\\zx & (z + y)^2 & xy\\zy & xy & (z + x)^2\end{vmatrix}\)

Options:

\(2(x + y + z)^3\)

\(2xyz(x + y + z)^3\)

\(xyz(x + y + z)^3\)

\(2xyz(x + y + z)^2\)

Correct Answer:

\(2xyz(x + y + z)^3\)

Explanation:

Given,

\(\begin{vmatrix}(x + y)^2 & zx & zy\\zx & (z + y)^2 & xy\\zy & xy & (z + x)^2\end{vmatrix}\)

Applying, \(R_1 \to zR_1 , R_2 \to xR_2, R_3 \to yR_3\), we get

\(= \frac{1}{xyz}\begin{vmatrix}z(x + y)^2 & z^2x & z^2y\\zx^2 & x(z + y)^2 & x^2y\\zy^2 & xy^2 & y(z + x)^2\end{vmatrix}\)

Applying, \(C_1 \to \frac{C_1}{z}, C_2 \to \frac{C_2}{x}, C_3 \to \frac{C_3}{y}\) we get

\(= \begin{vmatrix}(x + y)^2 & z^2 & z^2\\x^2 & (z + y)^2 & x^2\\y^2 & y^2 & (z + x)^2\end{vmatrix}\)

Applying, \(C_1 \to C_1 − C_3, C_2 \to C_2 − C_3\) we get

\(= (x + y + z)^2\begin{vmatrix}x + y − z & 0 & z^2\\ 0 & z + y − x & x^2\\y − z − x & y − z − x & (z + x)^2\end{vmatrix}\)

Applying, \(C_1 \to C_1 + \frac{C_3}{z}, C_2 \to C_2 + \frac{C_3}{x}\) we get

\(= (x + y + z)^2\begin{vmatrix}x + y  & \frac{z^2}{x} & z^2\\ \frac{x^2}{z} & z + y & x^2\\0 & 0 & 2xz\end{vmatrix}\)

\(= (x + y + z)^2[2xz(xz + xy + yz + y^2 −zx)]\)

\(= (x + y + z)^2 [2xyz(x + y + z)]\)

\(= 2xyz(x + y + z)^3\)