\( \left\{\begin{array}{l}\frac{16}{a^{2}}-\frac{5}{b^{2}}=-1 \\\frac{1}{a^{2}}-\frac{5}{\frac{4}{b^{2}}}=-1\end{array}\right. \)

To solve the system of equations \[ \left\{ \begin{array}{l} \frac{16}{a^{2}}-\frac{5}{b^{2}}=-1 \\ \frac{1}{a^{2}}-\frac{5}{\frac{4}{b^{2}}}=-1 \end{array} \right. \] let's start by introducing substitutions that simplify our equations. Let \( x = \frac{1}{a^{2}} \) and \( y = \frac{1}{b^{2}} \). Then the equations become: 1. \( 16x - 5y = -1 \) 2. \( x - 5(4y) = -1 \) which simplifies to \( x - 20y = -1 \) Now we can rewrite these two equations: 1. \( 16x - 5y = -1 \) (Equation 1) 2. \( x - 20y = -1 \) (Equation 2) Now, we'll solve for \( x \) and \( y \) using these two equations. From Equation 2, we can express \( x \): \[ x = 20y - 1 \] Now, we can substitute this expression for \( x \) into Equation 1: \[ 16(20y - 1) - 5y = -1 \] \[ 320y - 16 - 5y = -1 \] \[ 315y - 16 = -1 \] \[ 315y = 15 \] \[ y = \frac{15}{315} = \frac{1}{21} \] Next, substitute \( y \) back into the expression for \( x \): \[ x = 20\left(\frac{1}{21}\right) - 1 \] \[ x = \frac{20}{21} - 1 \] \[ x = \frac{20}{21} - \frac{21}{21} = -\frac{1}{21} \] Now we have \( x = -\frac{1}{21} \) and \( y = \frac{1}{21} \). Since \( x = \frac{1}{a^2} \) and \( y = \frac{1}{b^2} \), we can find \( a^2 \) and \( b^2 \): \[ \frac{1}{a^2} = -\frac{1}{21} \, \Rightarrow \, a^2 = -21 \quad (\text{Not possible for real } a) \] \[ \frac{1}{b^2} = \frac{1}{21} \, \Rightarrow \, b^2 = 21 \, \Rightarrow \, b = \sqrt{21} \text{ or } -\sqrt{21} \] Since \( a^2 \) cannot be negative, the system of equations has no real solutions for \( a \). Thus, we only have a valid solution for \( b \).