Matrix Polynomial Equation: Detailed Solution

The problem asks us to determine the value of a scalar $k$. We are given a specific matrix $A$ and a matrix polynomial equation that it satisfies.

The matrix $A$ is defined as:

\( A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \)

The matrix polynomial equation is:

\( A^2 - 4 + kI_2 = 0 \)

Here, $\(I_2\)$ denotes the 2x2 identity matrix, \( I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), and $0$ represents the 2x2 zero matrix, \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).

\(A^2\) Calculation: Step-by-Step Matrix Multiplication

The first step is to calculate \(A^2\), which means multiplying matrix A by itself.

\( A^2 = A \times A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \)

To find the elements of \(A^2\), we perform the dot product of rows of the first matrix with the columns of the second matrix:

• Element (1,1): \( (2)(2) + (-1)(-1) = 4 + 1 = 5 \)
• Element (1,2): \( (2)(-1) + (-1)(2) = -2 - 2 = -4 \)
• Element (2,1): \( (-1)(2) + (2)(-1) = -2 - 2 = -4 \)
• Element (2,2): \( (-1)(-1) + (2)(2) = 1 + 4 = 5 \)

Therefore, the matrix \(A^2\) is:

\( A^2 = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix} \)

Equation Interpretation: Clarifying the Matrix Polynomial Form

The provided equation is \( A^2 - 4 + kI_2 = 0 \). The term '-4' stands alone as a scalar, which is unusual in standard matrix polynomial notation. Typically, such terms are implicitly multiplied by the identity matrix, or the equation might involve the matrix A itself.

A common theorem in linear algebra is the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation. Let's find the characteristic equation for matrix A:

• The trace of A is \( \text{tr}(A) = 2 + 2 = 4 \).
• The determinant of A is \( \det(A) = (2)(2) - (-1)(-1) = 4 - 1 = 3 \).
• The characteristic equation is \( \lambda^2 - \text{tr}(A)\lambda + \det(A) = 0 \), substituting the values gives \( \lambda^2 - 4\lambda + 3 = 0 \).
• According to the Cayley-Hamilton theorem, the matrix A satisfies this equation: \( A^2 - 4A + 3I_2 = 0 \).

Comparing the standard form \( A^2 - 4A + 3I_2 = 0 \) with the given equation \( A^2 - 4 + kI_2 = 0 \), it appears most likely that the intended equation was \( A^2 - 4A + kI_2 = 0 \). This interpretation allows us to solve for k consistently.

Solving for \(k\): Substitution in the Matrix Equation

Using the likely intended equation \( A^2 - 4A + kI_2 = 0 \), we substitute the known matrices:

• \( A^2 = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix} \)
• \( 4A = 4 \times \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix} \)
• \( kI_2 = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} \)

Substituting these into the equation \( A^2 - 4A + kI_2 = 0 \):

\( \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix} - \begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix} + \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

Combine the elements of the matrices on the left side:

\( \begin{bmatrix} (5 - 8 + k) & (-4 - (-4) + 0) \\ (-4 - (-4) + 0) & (5 - 8 + k) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

This simplifies to:

\( \begin{bmatrix} -3 + k & 0 \\ 0 & -3 + k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

For the matrices to be equal, their corresponding elements must be equal. This gives us the equation:

\( -3 + k = 0 \)

Solving for \(k\):

\( k = 3 \)

Final Value of \(k\): Conclusion

By calculating \(A^2\) and interpreting the matrix polynomial equation in the context of the Cayley-Hamilton theorem, we find that the value of $k$ is 3.