Key Concept: Orthogonal Matrices

A square matrix $A$ is said to be an orthogonal matrix if its transpose is equal to its inverse, i.e., $A^T = A^{-1}$. This property implies that $AA^T = I$ and $A^T A = I$, where $I$ is the identity matrix of the same order as $A$. Orthogonal matrices represent transformations that preserve length and angle (like rotations and reflections).

Problem Statement

We are given a matrix $A = \left[ {\begin{matrix} 1 & { - \alpha } \\ \alpha & \beta \\ \end{matrix} } \right]$ and the condition $AA^T = I_2$. Our goal is to find the value of ${\alpha ^4} + {\beta ^4}$.

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Step-by-Step Solution

1. Determine the Transpose of Matrix A ($A^T$)

The transpose of a matrix is obtained by interchanging its rows and columns. Given $A = \left[ {\begin{matrix} 1 & { - \alpha } \\ \alpha & \beta \\ \end{matrix} } \right]$, its transpose $A^T$ is: $A^T = \left[ {\begin{matrix} 1 & \alpha \\ { - \alpha } & \beta \\ \end{matrix} } \right]$ Explanation: The first row of $A$ (1, $-\alpha$) becomes the first column of $A^T$, and the second row of $A$ ($\alpha$, $\beta$) becomes the second column of $A^T$.

2. Perform the Matrix Multiplication $AA^T$

Now we multiply matrix $A$ by its transpose $A^T$: $A A^T = \left[ {\begin{matrix} 1 & { - \alpha } \\ \alpha & \beta \\ \end{matrix} } \right] \left[ {\begin{matrix} 1 & \alpha \\ { - \alpha } & \beta \\ \end{matrix} } \right]$ To multiply two matrices, we take the dot product of the rows of the first matrix with the columns of the second matrix. Let the product matrix be $C = AA^T = \left[ {\begin{matrix} c_{11} & c_{12} \\ c_{21} & c_{22} \\ \end{matrix} } \right]$.

• For $c_{11}$ (first row, first column element): Multiply the first row of $A$ by the first column of $A^T$: $c_{11} = (1)(1) + (-\alpha)(-\alpha) = 1 + \alpha^2$

• For $c_{12}$ (first row, second column element): Multiply the first row of $A$ by the second column of $A^T$: $c_{12} = (1)(\alpha) + (-\alpha)(\beta) = \alpha - \alpha\beta$

• For $c_{21}$ (second row, first column element): Multiply the second row of $A$ by the first column of $A^T$: $c_{21} = (\alpha)(1) + (\beta)(-\alpha) = \alpha - \alpha\beta$

• For $c_{22}$ (second row, second column element): Multiply the second row of $A$ by the second column of $A^T$: $c_{22} = (\alpha)(\alpha) + (\beta)(\beta) = \alpha^2 + \beta^2$

So, the product matrix $AA^T$ is: $A A^T = \left[ {\begin{matrix} {1 + {\alpha ^2}} & {\alpha - \alpha \beta } \\ {\alpha - \alpha \beta } & {{\alpha ^2} + {\beta ^2}} \\ \end{matrix} } \right]$

3. Equate $AA^T$ to the Identity Matrix $I_2$

We are given the condition $AA^T = I_2$. The identity matrix of order 2 is $I_2 = \left[ {\begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} } \right]$. Therefore, we set the calculated product equal to $I_2$: $\left[ {\begin{matrix} {1 + {\alpha ^2}} & {\alpha - \alpha \beta } \\ {\alpha - \alpha \beta } & {{\alpha ^2} + {\beta ^2}} \\ \end{matrix} } \right] = \left[ {\begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} } \right]$

4. Formulate and Solve a System of Equations

For two matrices to be equal, their corresponding elements must be equal. This gives us a system of equations:

• Equation 1 (from element $c_{11}$): $1 + \alpha^2 = 1$ Explanation: Equating the top-left elements of both matrices.

• Equation 2 (from element $c_{12}$ or $c_{21}$): $\alpha - \alpha\beta = 0$ Explanation: Equating the top-right (or bottom-left) elements.

• Equation 3 (from element $c_{22}$): $\alpha^2 + \beta^2 = 1$ Explanation: Equating the bottom-right elements.

Now, let's solve these equations:

• From Equation 1: $1 + \alpha^2 = 1$ Subtracting 1 from both sides gives: $\alpha^2 = 0$

• From Equation 3: $\alpha^2 + \beta^2 = 1$ Substitute the value of $\alpha^2 = 0$ into this equation: $0 + \beta^2 = 1$ $\beta^2 = 1$

• (Optional check with Equation 2): $\alpha - \alpha\beta = 0$ Factor out $\alpha$: $\alpha(1 - \beta) = 0$ Since $\alpha^2 = 0$, it implies $\alpha = 0$. Substituting $\alpha = 0$ into $\alpha(1 - \beta) = 0$ gives $0(1-\beta) = 0$, which is $0=0$. This equation is consistent with our values and doesn't provide new information for $\beta$ itself, but confirms $\alpha=0$. If $\alpha \neq 0$, then $1-\beta=0 \implies \beta=1$. But we found $\alpha=0$, so this is not the case.

5. Calculate ${\alpha ^4} + {\beta ^4}$

We have found: $\alpha^2 = 0$ $\beta^2 = 1$

To find $\alpha^4$ and $\beta^4$, we simply square these values: $\alpha^4 = (\alpha^2)^2 = (0)^2 = 0$ $\beta^4 = (\beta^2)^2 = (1)^2 = 1$

Finally, we calculate the required sum: ${\alpha ^4} + {\beta ^4} = 0 + 1 = 1$

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Tips and Common Mistakes

• Matrix Multiplication Order: Always be careful with the order of matrix multiplication. $AB$ is generally not equal to $BA$. Here, $AA^T$ is used, which is specific.
• Orthogonal Matrix Properties: Remember that if $A$ is orthogonal, then $\det(A)^2 = 1$, which means $\det(A) = \pm 1$. For this matrix $A$, $\det(A) = (1)(\beta) - (-\alpha)(\alpha) = \beta + \alpha^2$. Since $\alpha^2=0$ and $\beta=1$ (or $\beta=-1$), $\det(A) = \pm 1$. This property provides a good way to cross-check results.
• Equating Elements: When equating two matrices, ensure you equate corresponding elements correctly. A common mistake is to mix up positions.
• Solving Equations: Pay attention to algebraic manipulations. Simple errors can lead to incorrect values. For example, $\alpha^2=0$ implies $\alpha=0$, but $\beta^2=1$ implies $\beta = \pm 1$. However, for $\beta^4$, both $\beta=1$ and $\beta=-1$ yield $\beta^4=1$.

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Summary and Key Takeaway

By systematically calculating the product $AA^T$ and equating it to the identity matrix $I_2$, we derived a system of equations for $\alpha$ and $\beta$. Solving these equations revealed $\alpha^2 = 0$ and $\beta^2 = 1$. From these values, we easily found $\alpha^4 = 0$ and $\beta^4 = 1$, leading to the final answer ${\alpha ^4} + {\beta ^4} = 1$. This problem effectively tests your understanding of matrix transpose, matrix multiplication, and the properties of orthogonal matrices.

The final answer is $\boxed{\text{1}}$.