Q64 / 150 · MHT CET 2026 (20 April Shift 1) · Mathematics
Let $A = \begin{bmatrix} a & 1 \\ 1 & b \end{bmatrix}$, 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$2\sqrt{2}$
- Correct answerC$4$
- D$1$
Solution
Given $A = \begin{bmatrix} a & 1 \\ 1 & b \end{bmatrix}$
Since $a$ and $b$ are roots of $x^2 - 4x + 2 = 0$, we have $a + b = 4$ and $ab = 2$.
The determinant of $A$ is $|A| = ab - 1 = 2 - 1 = 1$.
The inverse of $A$ is $A^{-1} = \dfrac{1}{|A|} \begin{bmatrix} b & -1 \\ -1 & a \end{bmatrix} = \begin{bmatrix} b & -1 \\ -1 & a \end{bmatrix}$.
Now, $A + A^{-1} = \begin{bmatrix} a & 1 \\ 1 & b \end{bmatrix} + \begin{bmatrix} b & -1 \\ -1 & a \end{bmatrix} = \begin{bmatrix} a+b & 0 \\ 0 & a+b \end{bmatrix}$.
Substituting $a + b = 4$, we get $A + A^{-1} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} = 4I_2$.
Comparing with $A + A^{-1} = kI_2$, we get $k = 4$.
Since $a$ and $b$ are roots of $x^2 - 4x + 2 = 0$, we have $a + b = 4$ and $ab = 2$.
The determinant of $A$ is $|A| = ab - 1 = 2 - 1 = 1$.
The inverse of $A$ is $A^{-1} = \dfrac{1}{|A|} \begin{bmatrix} b & -1 \\ -1 & a \end{bmatrix} = \begin{bmatrix} b & -1 \\ -1 & a \end{bmatrix}$.
Now, $A + A^{-1} = \begin{bmatrix} a & 1 \\ 1 & b \end{bmatrix} + \begin{bmatrix} b & -1 \\ -1 & a \end{bmatrix} = \begin{bmatrix} a+b & 0 \\ 0 & a+b \end{bmatrix}$.
Substituting $a + b = 4$, we get $A + A^{-1} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} = 4I_2$.
Comparing with $A + A^{-1} = kI_2$, we get $k = 4$.
Hence, the correct answer is option: 'C'.