# Question 6, Exercise 2.2

Solutions of Question 6 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

If $A=\left[\begin{array}{cc}2 & 1 \\ 3 & -3\end{array}\right]$ then find $\alpha$ and $\beta$ such that, $A^{2}+\alpha I=\beta A$.

Solution.

Given the matrix \begin{align} & A^{2}+\alpha I=\beta A\\ \implies &\begin{bmatrix} 2 & 1 \\ 3 & -3 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & -3 \end{bmatrix}+\alpha \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \beta \begin{bmatrix} 2 & 1 \\ 3 & -3 \end{bmatrix}\\ \implies & \begin{bmatrix} 4 + 3 & 2 - 3 \\ 6 - 9 & 3 + 9 \end{bmatrix}+\begin{bmatrix} \alpha & 0 \\ 0 & \alpha \end{bmatrix} = \begin{bmatrix} 2\beta & \beta \\ 3\beta & -3\beta \end{bmatrix}\\ \implies &\begin{bmatrix} 7 + \alpha & -1 \\ -3 & 12 + \alpha \end{bmatrix} = \begin{bmatrix} 2\beta & \beta \\ 3\beta & -3\beta \end{bmatrix}\end{align} By comparing corresponding elements in the matrices, we get: \begin{align}&7 + \alpha = 2\beta \cdots (1)\\ &-1 = \beta \cdots (2) \end{align}

Using $\beta = -1$ in (1), we get: \begin{align} & 7 + \alpha = 2(-1)\\ \implies & \alpha = -2 - 7\\ \implies & \alpha = -9\end{align}

Hence $\alpha = -9$ and $\beta = -1$.