====== 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. 

=====Question 6=====
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 $\beta = -1$. GOOD



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