The inverse element of $$ \left|\begin{array}{lll} y & y & y \\ y & y & y \\ y & y & y \end{array}\right| $$ in group $$ M=|| \begin{array}{ccc} x & x & x \\ x & x & x \\ x & x & x \end{array}|/ x \in R, x \neq 0, \quad I=(1 / 3)| \begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}|| \text { w.r.t } $$ matrix multiplication is (a) \(\begin{array}{lll}(1 / \mathrm{y}) & (1 / \mathrm{y}) & (1 / \mathrm{y}) \\\ (1 / \mathrm{y}) & (1 / \mathrm{y}) & (1 / \mathrm{y}) \\ (1 / \mathrm{y}) & (1 / \mathrm{y}) & (1 / \mathrm{y})\end{array} \mid\) (b) \(\left|\begin{array}{lll}(1 / 3 y) & (1 / 3 y) & (1 / 3 y) \\ (1 / 3 y) & (1 / 3 y) & (1 / 3 y) \\ (1 / 3 y) & (1 / 3 y) & (1 / 3 y)\end{array}\right|\) (c) \(\left|\begin{array}{lll}(1 / 6 \mathrm{y}) & (1 / 6 \mathrm{y}) & (1 / 6 \mathrm{y}) \\ (1 / 6 \mathrm{y}) & (1 / 6 \mathrm{y}) & (1 / 6 \mathrm{y}) \\\ (1 / 6 \mathrm{y}) & (1 / 6 \mathrm{y}) & (1 / 6 \mathrm{y})\end{array}\right|\) (d) \(\left|\begin{array}{lll}(1 / 9 \mathrm{y}) & (1 / 9 \mathrm{y}) & (1 / 9 \mathrm{y}) \\ (1 / 9 \mathrm{y}) & (1 / 9 \mathrm{y}) & (1 / 9 \mathrm{y}) \\\ (1 / 9 \mathrm{y}) & (1 / 9 \mathrm{y}) & (1 / 9 \mathrm{y})\end{array}\right|\)

Step by step solution

01

Test Option (a)

Multiply given matrix by option (a) and see if it results in the identity matrix. Given Matrix: \( \begin{bmatrix} y & y & y \\ y & y & y \\ y & y & y \\ \end{bmatrix} \) Option (a): \( \begin{bmatrix} \frac{1}{y} & \frac{1}{y} & \frac{1}{y} \\ \frac{1}{y} & \frac{1}{y} & \frac{1}{y} \\ \frac{1}{y} & \frac{1}{y} & \frac{1}{y} \\ \end{bmatrix} \) Multiplication result: \( \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} \) This is not the identity matrix, so option (a) is not the inverse element.

02

Test Option (b)

Multiply given matrix by option (b) and see if it results in the identity matrix. Option (b): \( \begin{bmatrix} \frac{1}{3y} & \frac{1}{3y} & \frac{1}{3y} \\ \frac{1}{3y} & \frac{1}{3y} & \frac{1}{3y} \\ \frac{1}{3y} & \frac{1}{3y} & \frac{1}{3y} \\ \end{bmatrix} \) Multiplication result: \( \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} \) This is not the identity matrix, so option (b) is not the inverse element.

03

Test Option (c)

Multiply given matrix by option (c) and see if it results in the identity matrix. Option (c): \( \begin{bmatrix} \frac{1}{6y} & \frac{1}{6y} & \frac{1}{6y} \\ \frac{1}{6y} & \frac{1}{6y} & \frac{1}{6y} \\ \frac{1}{6y} & \frac{1}{6y} & \frac{1}{6y} \\ \end{bmatrix} \) Multiplication result: \( \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} \) This is not the identity matrix, so option (c) is not the inverse element.

04

Test Option (d)

Multiply given matrix by option (d) and see if it results in the identity matrix. Option (d): \( \begin{bmatrix} \frac{1}{9y} & \frac{1}{9y} & \frac{1}{9y} \\ \frac{1}{9y} & \frac{1}{9y} & \frac{1}{9y} \\ \frac{1}{9y} & \frac{1}{9y} & \frac{1}{9y} \\ \end{bmatrix} \) Multiplication result: \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \) This is the identity matrix, so option (d) is the inverse element. Final answer: (d) \(\begin{bmatrix} \frac{1}{9y} & \frac{1}{9y} & \frac{1}{9y} \\ \frac{1}{9y} & \frac{1}{9y} & \frac{1}{9y} \\ \frac{1}{9y} & \frac{1}{9y} & \frac{1}{9y} \end{bmatrix}\)

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