If $\begin{bmatrix$}[r] x & y \end{bmatrix}\begin{bmatrix}[r] x \ y \end{bmatrix} = \begin{bmatrix}[r] 25 \end{bmatrix} \text{ and } \begin{bmatrix}[r] -x & y \end{bmatrix}\begin{bmatrix}[r] 2x \ y \end{bmatrix} = \begin{bmatrix}[r] -2 \end{bmatrix};

find x and y, if :

(i) x, y ∈ W (whole numbers)

(ii) x, y ∈ Z (integers)

Given,

$\Rightarrow \begin{bmatrix$}[r] x & y \end{bmatrix}\begin{bmatrix}[r] x \ y \end{bmatrix} = \begin{bmatrix}[r] 25 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] x \times x + y \times y \end{bmatrix} = \begin{bmatrix}[r] 25 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] x^2 + y^2 \end{bmatrix} = \begin{bmatrix}[r] 25 \end{bmatrix}

By definition of equality of matrices we get,

x² + y² = 25
⇒ x² = 25 - y² ……(i)

Given,

$\Rightarrow \begin{bmatrix$}[r] -x & y \end{bmatrix}\begin{bmatrix}[r] 2x \ y \end{bmatrix} = \begin{bmatrix}[r] -2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] -x \times 2x + y \times y \end{bmatrix} = \begin{bmatrix}[r] -2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] -2x^2 + y^2 \end{bmatrix} = \begin{bmatrix}[r] -2 \end{bmatrix}

By definition of equality of matrices we get,

-2x² + y² = -2 ……(ii)

Substituting value of x² from (i) in (ii) we get,

⇒ -2(25 - y²) + y² = -2
⇒ -50 + 2y² + y² = -2
⇒ 3y² = -2 + 50
⇒ 3y² = 48
⇒ y² = 16
⇒ y = ± 4.

⇒ x² = 25 - y²
⇒ x² = 25 - 16
⇒ x² = 9
⇒ x = ± 3.

(i) Since, x, y ∈ W

∴ x = 3, y = 4.

Hence, x = 3 and y = 4.

(ii) Since, x, y ∈ Z

∴ x = ±3, y = ±4.

Hence, x = ±3 and y = ±4.