Animated Solution for Mathematics - Matrices and Determinants: If for the matrix, $A=\left[\begin{array}{cc}1& -\alpha \\ \alpha & \beta \end{array}\right]$, $A{A}^T=I_2$, then the value of ${\alpha }^4+{\beta }^4$ is:

Unlocking the Matrix

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a matrix problem; we are peeling back the layers of a fundamental structure in linear algebra.

We are given a matrix $A=\left[\begin{array}{cc}1& -\alpha \\ \alpha & \beta \end{array}\right]$ and a condition that feels like a key turning in a lock: $A{A}^T=I_2$. Let us embark on this journey together.

The Anatomy of the Transpose

Before we can multiply, we must understand our components. The transpose, $A^T$, is the mirror image of our matrix $A$ across its main diagonal.

Imagine taking the rows of $A$ and standing them upright to become columns. The first row, $[1,-\alpha ]$, becomes the first column. The second row, $[\alpha ,\beta ]$, becomes the second column.

Thus, we define:

This simple act of reflection is the first step in revealing the hidden symmetry of the matrix.

The Collision of Matrices

Now, we perform the multiplication $A{A}^T$. This is where the magic happens.

We take the first row of $A$ and dot it with the first column of $A^T$. This gives us $(1)(1)+(-\alpha )(-\alpha )=1+{\alpha }^2$. This is our top-left entry.

Next, we take the first row of $A$ and dot it with the second column of $A^T$, yielding $(1)(\alpha )+(-\alpha )(\beta )=\alpha -\alpha \beta$. We repeat this for the bottom row, and we arrive at our resulting matrix:

The Identity Constraint

We are told that $A{A}^T=I_2$, where $I_2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$. This is the identity matrix, the 'neutral' element of the matrix world.

By equating our result to this identity, we are essentially saying that the transformation $A$ preserves the length of vectors—a hallmark of orthogonal matrices. We now have a system of equations:

The Final Revelation

Look at the first equation: $1+{\alpha }^2=1$. Subtracting $1$ from both sides, we find ${\alpha }^2=0$, which forces $\alpha =0$.

This is a moment of pure clarity! With $\alpha =0$, our second equation, ${\alpha }^2+{\beta }^2=1$, simplifies beautifully to $0+{\beta }^2=1$, meaning ${\beta }^2=1$.

The problem asks for ${\alpha }^4+{\beta }^4$. Since $\alpha =0$, ${\alpha }^4=0$. Since ${\beta }^2=1$, ${\beta }^4=({\beta }^2{)}^2=1^2=1$.

Adding them together, we get $0+1=1$.

We have arrived. The complexity of the matrix has dissolved into a simple, elegant result. The final answer is 1.