Question

If $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$ and $A^{-1} = \alpha I + \beta A$, $\beta \in R$ where I is the identity matrix of order 2, then $4(\alpha + \beta) =$}

• A. $\frac{10}{3}$
• B. $\frac{2}{3}$
• C. $\frac{8}{3}$
• D. $\frac{1}{3}$

Hint:

Cayley-Hamilton is the fastest way to express $A^{-1}$ in terms of $A$ and $I$.

Answer 1

The Correct Option is C

To solve the problem, we need to find the inverse of matrix \( A \) and express it in the form \( \alpha I + \beta A \) where \( A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} \).

Let's begin by calculating the inverse of matrix \( A \). The formula for the inverse of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by:

\[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

For matrix \( A \), \( a = 1 \), \( b = 2 \), \( c = -1 \), and \( d = 4 \). Therefore, we compute:

\[ ad - bc = (1)(4) - (2)(-1) = 4 + 2 = 6 \]

The inverse \( A^{-1} \) is:

\[ A^{-1} = \frac{1}{6} \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{4}{6} & -\frac{2}{6} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix} \]

Next, we express \( A^{-1} \) in the form \( \alpha I + \beta A \). The identity matrix \( I \) of order 2 is:

\[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

Now, suppose:

\[ \alpha I + \beta A = \alpha \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 2\beta \\ -\beta & \alpha + 4\beta \end{bmatrix} \]

Equating this with the calculated \( A^{-1} \):

\[ \begin{bmatrix} \alpha + \beta & 2\beta \\ -\beta & \alpha + 4\beta \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix} \]

We have the following equations:

• \(\alpha + \beta = \frac{2}{3}\)
• \(2\beta = -\frac{1}{3} \implies \beta = -\frac{1}{6}\)

Substitute \(\beta = -\frac{1}{6}\) into \(\alpha + \beta = \frac{2}{3}\):

\[ \alpha - \frac{1}{6} = \frac{2}{3} \implies \alpha = \frac{2}{3} + \frac{1}{6} = \frac{2}{3} + \frac{1}{6} = \frac{4}{6} + \frac{1}{6} = \frac{5}{6} \]

Finally, calculate \( 4(\alpha + \beta) \):

\[ 4(\alpha + \beta) = 4\left(\frac{2}{3}\right) = \frac{8}{3} \]

However, let's correct the process due to oversight in calculation:

When using \(\alpha = \frac{5}{6}\) and \(\beta = -\frac{1}{6}\), check the determination of \( 4(\alpha + \beta) \) again:

\[ \alpha + \beta = \frac{5}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \]

\[ 4(\alpha + \beta) = 4 \times \frac{2}{3} = \frac{8}{3} \]