Question

If \( A = \begin{bmatrix} x & y & y y & x & y y & y & x \end{bmatrix} \) is a matrix such that \( 5A^{-1} = \begin{bmatrix} -3 & 2 & 2 2 & -3 & 2 2 & 2 & -3 \end{bmatrix} \), then \( A^2 - 4A = \)

• A. \( 5A^{-1} \)
• B. \( 5I \)
• C. \( 0 \)
• D. \( I \)

Hint:

For symmetric cyclic matrices of the form \(\begin{bmatrix} x & y & y y & x & y y & y & x \end{bmatrix}\), the off-diagonal elements of the product with another similar matrix quickly yield linear systems. Solving for \(x\) and \(y\) directly is often much faster and less error-prone than attempting algebraic matrix manipulations!

Answer 1

The Correct Option is B

Concept: The inverse of an invertible square matrix \(A\) satisfies the fundamental property: \[ A \cdot A^{-1} = A^{-1} \cdot A = I \] where \(I\) is the identity matrix of the same order. When given an explicit matrix expression for a scalar multiple of \(A^{-1}\), we can eliminate the inverse term from our target expression by multiplying through by \(A\). This allows us to work directly with matrix addition and scalar multiplication identities.

Step 1: Expressing the matrix \(5A^{-1}\) as a combination of a standard pattern matrix.
Let the given matrix for \(5A^{-1}\) be denoted as: \[ 5A^{-1} = \begin{bmatrix} -3 & 2 & 2 2 & -3 & 2 2 & 2 & -3 \end{bmatrix} \] Notice that we can rewrite this matrix by separating the diagonal elements from the off-diagonal elements: \[ 5A^{-1} = \begin{bmatrix} 2-5 & 2 & 2 2 & 2-5 & 2 2 & 2 & 2-5 \end{bmatrix} = \begin{bmatrix} 2 & 2 & 2 2 & 2 & 2 2 & 2 & 2 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0 0 & 5 & 0 0 & 0 & 5 \end{bmatrix} \] Thus, we can represent it neatly as: \[ 5A^{-1} = \begin{bmatrix} 2 & 2 & 2 2 & 2 & 2 2 & 2 & 2 \end{bmatrix} - 5I \quad \cdots (1) \]

Step 2: Multiplying both sides of equation (1) by matrix \(A\).
Post-multiplying both sides of the equation by \(A\), we get: \[ 5A^{-1} \cdot A = \begin{bmatrix} 2 & 2 & 2 2 & 2 & 2 2 & 2 & 2 \end{bmatrix} A - 5I \cdot A \] Using the property \(A^{-1}A = I\) and \(IA = A\): \[ 5I = \begin{bmatrix} 2 & 2 & 2 2 & 2 & 2 2 & 2 & 2 \end{bmatrix} \begin{bmatrix} x & y & y y & x & y y & y & x \end{bmatrix} - 5A \quad \cdots (2) \]

Step 3: Evaluating the matrix product and establishing the characteristic relation.
Alternatively, a more direct approach is to multiply the original matrices directly using \(A \cdot (5A^{-1}) = 5I\): \[ \begin{bmatrix} x & y & y y & x & y y & y & x \end{bmatrix} \begin{bmatrix} -3 & 2 & 2 2 & -3 & 2 2 & 2 & -3 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 0 0 & 5 & 0 0 & 0 & 5 \end{bmatrix} \] Let's compute the elements of the resulting product matrix:
• Element (1,1): \(-3x + 2y + 2y = -3x + 4y = 5 \quad \cdots (3)\)
• Element (1,2): \(2x - 3y + 2y = 2x - y = 0 \implies y = 2x \quad \cdots (4)\) Substituting equation (4) into equation (3): \[ -3x + 4(2x) = 5 \implies -3x + 8x = 5 \implies 5x = 5 \implies x = 1 \] Since \(y = 2x\), we have \(y = 2(1) = 2\). Thus, the matrix \(A\) is: \[ A = \begin{bmatrix} 1 & 2 & 2 2 & 1 & 2 2 & 2 & 1 \end{bmatrix} \]

Step 4: Computing \(A^2 - 4A\) using the determined matrix \(A\).
First, let's calculate \(A^2\): \[ A^2 = \begin{bmatrix} 1 & 2 & 2 2 & 1 & 2 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 2 & 1 & 2 2 & 2 & 1 \end{bmatrix} \] \[ A^2 = \begin{bmatrix} 1(1)+2(2)+2(2) & 1(2)+2(1)+2(2) & 1(2)+2(2)+2(1) 2(1)+1(2)+2(2) & 2(2)+1(1)+2(2) & 2(2)+1(2)+2(1) 2(1)+2(2)+1(2) & 2(2)+2(1)+1(2) & 2(2)+2(2)+1(1) \end{bmatrix} = \begin{bmatrix} 9 & 8 & 8 8 & 9 & 8 8 & 8 & 9 \end{bmatrix} \] Now, compute \(A^2 - 4A\): \[ A^2 - 4A = \begin{bmatrix} 9 & 8 & 8 8 & 9 & 8 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 8 & 4 & 8 8 & 8 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 0 0 & 5 & 0 0 & 0 & 5 \end{bmatrix} = 5I \]