Question

Let \(A = [\begin{array}{cc}a & 1 \\ 1 & b\end{array}]\), where \(a\) and \(b\) are the roots of the equation \(x^2-4x+2 = 0\). If \(A+A^{-1} = kI_2\), then the value of \(k\) is ____

• A. \(2\)
• B. \(2\sqrt{2}\)
• C. \(4\)
• D. \(1\)

Hint:

Use sum and product of roots a + b = 4 and ab = 2 to find the inverse.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a \(2 \times 2\) matrix \(\begin{pmatrix} a & 1 \\ 1 & b \end{pmatrix}\), the inverse is \(\frac{1}{ab - 1}\begin{pmatrix} b & -1 \\ -1 & a \end{pmatrix}\).

Step 2: Use the roots:
For \(x^2 - 4x + 2 = 0\), the sum of roots is \(a + b = 4\) and the product is \(ab = 2\). So \(ab - 1 = 1\) and
\[ A^{-1} = \begin{pmatrix} b & -1 \\ -1 & a \end{pmatrix} \]

Step 3: Add:
\[ A + A^{-1} = \begin{pmatrix} a + b & 0 \\ 0 & a + b \end{pmatrix} = 4I \]
So \(k = 4\).

Step 4: Why the other options are wrong.
Options 2 and 1 and \(2\sqrt2\) come from using the product (2) or its root instead of the sum of the roots, which is the number that appears on the diagonal.

Final Answer:
\(k = 4\), option (C). \[ \boxed{4} \]