Question

If \( A = \left[ \begin{array}{cc} \alpha & 2 \\ 1 & 2 \end{array} \right] \), \( B = \left[ \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right] \) and \( A^2 - 4A + 2I = 0, B^2 - 2B + I = 0 \), then \( \text{adj}(A^3 - B^3) \) is equal to

• A. 11
• B. 7
• C. -11
• D. 121

Hint:

When dealing with matrix powers and adjugates, always look for simplifications and apply standard identities like the difference of cubes to reduce the complexity.

Answer 1

The Correct Option is A

To solve the given problem, we must first analyze the given matrices \(A\) and \(B\) and their respective equations.

**Step 1: Analyze the matrix \(A\) and its equation.**

Matrix \(A\) is given as:

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

The matrix equation given is:

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

First, compute \(A^2\):

\(A^2 = \begin{bmatrix} \alpha & 2 \\ 1 & 2 \end{bmatrix} \cdot \begin{bmatrix} \alpha & 2 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} \alpha^2+2 & 2\alpha + 4 \\ \alpha + 2 & 4+1 \end{bmatrix} = \begin{bmatrix} \alpha^2 + 2 & 2\alpha + 4 \\ \alpha + 2 & 5 \end{bmatrix}\)

Substitute \(A^2\) into the equation:

\(\begin{bmatrix} \alpha^2 + 2 & 2\alpha + 4 \\ \alpha + 2 & 5 \end{bmatrix} - 4 \begin{bmatrix} \alpha & 2 \\ 1 & 2 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)

Expands to:

\(\begin{bmatrix} \alpha^2 + 2 - 4\alpha + 2 & 2\alpha + 4 - 8 \\ \alpha + 2 - 4 & 5 - 8 + 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)

Thus, solving:

• First equation: \( \alpha^2 - 4\alpha + 4 = 0 \Rightarrow (\alpha-2)^2 = 0 \Rightarrow \alpha = 2 \)

**Step 2: Verify matrix \(B\) and its properties.**

Matrix \(B\) is given as:

\(B = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\)

The matrix equation given is:

\(B^2 - 2B + I = 0\)

Compute \(B^2\):

\(B^2 = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}\)

Substitute \(B^2\) into the equation:

\(\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} - 2 \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)

It perfectly balances, showing the equation for \(B\) holds as well.

**Step 3: Calculate \(A^3 - B^3\) and its adjugate.**

We previously found \(\alpha = 2\) so:

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

Compute \(A^3\): First find \(A^2\) again with \(\alpha = 2\)

\(A^2 = \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix} \cdot \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 4 & 6 \end{bmatrix}\)

Then compute \(A^3\):

\(A^3 = A \cdot A^2 = \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix} \cdot \begin{bmatrix} 6 & 8 \\ 4 & 6 \end{bmatrix} = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix}\)

Now compute \(B^3\):

\(B^3 = B \cdot B^2 = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 4 & 4 \end{bmatrix}\)

Compute \(A^3 - B^3\):

\(A^3 - B^3 = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix} - \begin{bmatrix} 4 & 4 \\ 4 & 4 \end{bmatrix} = \begin{bmatrix} 16 & 24 \\ 10 & 16 \end{bmatrix}\)

The adjugate of a 2x2 matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) is \(\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\).

Here, we find the adjugate of the resultant matrix \(A^3 - B^3\):

\(\text{adj}(A^3 - B^3) = \begin{bmatrix} 16 & -24 \\ -10 & 16 \end{bmatrix}\)

Now compute the determinant of \(A^3 - B^3\):