Q.$A=\begin{bmatrix}\alpha&0\\1&1\end{bmatrix}$, $B=\begin{bmatrix}1&0\\2&1\end{bmatrix}$ find $\alpha$, if $A^2=B$.

Given $A=\begin{bmatrix}\alpha&0\\1&1\end{bmatrix}$, $B=\begin{bmatrix}1&0\\2&1\end{bmatrix}$. (The book's printed symbol '∝' here is a font-rendering artifact for the Greek letter $\alpha$, consistent with the question asking to 'find α'.)

Compute $A^2=A\cdot A$:

$(A^2)_{11}=\alpha(\alpha)+0(1)=\alpha^2$

$(A^2)_{12}=\alpha(0)+0(1)=0$

$(A^2)_{21}=1(\alpha)+1(1)=\alpha+1$

$(A^2)_{22}=1(0)+1(1)=1$

$A^2=\begin{bmatrix}\alpha^2&0\\\alpha+1&1\end{bmatrix}$

Set $A^2=B$:

$\begin{bmatrix}\alpha^2&0\\\alpha+1&1\end{bmatrix}=\begin{bmatrix}1&0\\2&1\end{bmatrix}$

By equality of matrices: …

Compute A² symbolically in terms of α, equate it entry-by-entry to B, and sol …

• Using only $\alpha^2=1$ and concluding $\alpha=\pm1$ without checking the second equation $\alpha+1=2$, which selects the unique valid value. …