If A=\left[\begin{array}{rr}1 & -1 \\ 2 & -1\end{array}\right], B=\left[\begin{array}{rr}a & -1 \\ b & -1\end{array}\right] and (A+B)^{2}=\left(A^{2}+B^{2}\right) then find the values of a and b.
If A=\left[\begin{array}{rr}1 & -1 \\ 2 & -1\end{array}\right], B=\left[\begin{array}{rr}a & -1 \\ b & -1\end{array}\right] and (A+B)^{2}=\left(A^{2}+B^{2}\right) then find the values of a and b.

Solution:

We have A=\left(\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right), B=\left(\begin{array}{ll}a & -1 \\ b & -1\end{array}\right) and (A+B)^{2}=\left(A^{2}+\right. \left.B^{2}\right)
We need to find matrix \boldsymbol{a}, \boldsymbol{b}.
We’ll use the inverse application here by modification of the equation we have as \boldsymbol{A}=\boldsymbol{C B}^{-\mathbf{1}}

(i) The of \boldsymbol{A}+\boldsymbol{B} is computed as follow:
We have A+B=\left(\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right)+\left(\begin{array}{ll}a & -1 \\ b & -1\end{array}\right)=\left(\begin{array}{ll}1+a & -1-1 \\ 2+b & -1-1\end{array}\right)= \left(\begin{array}{ll}a+1 & -2 \\ b+2 & -2\end{array}\right)

(ii)We can have A^{2} as
A^{2}=A A=\left(\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right)\left(\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right)=\left(\begin{array}{cc} -1 & 0 \\ 0 & 0 \end{array}\right)

(iii)We can have \boldsymbol{B}^{2} as
B^{2}=B B=\left(\begin{array}{ll} a & -1 \\ b & -1 \end{array}\right)\left(\begin{array}{ll} a & -1 \\ b & -1 \end{array}\right)=\left(\begin{array}{ll} a^{2}-b & -a+1 \\ a b-b & -b+1 \end{array}\right)

(iv) We can have A^{2}+B^{2} as
\begin{aligned} A^{2}+B^{2} &=\left(\begin{array}{cc} -1 & 0 \\ 0 & 0 \end{array}\right)+\left(\begin{array}{cc} a^{2}-b & -a+1 \\ a b-b & -b+1 \end{array}\right) \\ &=\left(\begin{array}{cc} -1+a^{2}-b & -a+1 \\ a b-b & -b+1 \end{array}\right) \end{aligned}

(v)We can have \boldsymbol{A}+\boldsymbol{B})^{2} as
\begin{array}{l} (A+B)^{2}=(A+B)(A+B) \\ =\left(\begin{array}{ll} a+1 & -2 \\ b+2 & -2 \end{array}\right)\left(\begin{array}{ll} a+1 & -2 \\ b+2 & -2 \end{array}\right) \\ =\left(\begin{array}{cc} a^{2}+2 a-2 b-3 & 2-2 a \\ 2 a-b+a b-2 & -2 b \end{array}\right) \end{array}

(vi) We are given equation as (\boldsymbol{A}+\boldsymbol{B})^{2}=\left(\boldsymbol{A}^{2}+\boldsymbol{B}^{2}\right)
\left(\begin{array}{cc} a^{2}+2 a-2 b-3 & 2-2 a \\ 2 a-b+a b-2 & -2 b \end{array}\right)=\left(\begin{array}{cc} -1+a^{2}-b & -a+1 \\ a b-b & -b+1 \end{array}\right)
Thus we have the relations as follow:
-2 b=-b+1 ; 2-2 a=-a+1
Therefore on solving the equations we obtain
a=1 ; b=-1