# Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 934: 95

$\left( 1,-1 \right),\left( 1,1 \right),\left( -1,1 \right),\left( -1,-1 \right)$

#### Work Step by Step

Consider the given expression, ${{x}^{2}}-2{{y}^{2}}=-1$ …… (1) $2{{x}^{2}}-{{y}^{2}}=1$ …… (2) Now, we will multiply with $2$ in equation (1), \begin{align} & 2{{x}^{2}}-4{{y}^{2}}=-2 \\ & 2{{x}^{2}}-{{y}^{2}}=1 \end{align} …… (3) Using (2) and (3), we get, \begin{align} & 2{{x}^{2}}-4{{y}^{2}}=-2 \\ & 2{{x}^{2}}-{{y}^{2}}=1 \\ & -3{{y}^{2}}=-3 \end{align} Therefore, ${{y}^{2}}=\pm 1$ Now, substitute $y=1$ in equation (1). Then, \begin{align} & {{x}^{2}}-2{{y}^{2}}=-1 \\ & {{x}^{2}}-2{{\left( 1 \right)}^{2}}=-1 \\ & x=\pm 1 \end{align} So, the solution of the provided expression is, \begin{align} & x=\pm 1 \\ & y=\pm 1 \\ \end{align}

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