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Mathematics Question on Matrices

Let A=[1−1\2α] and B=[β1\10],α,β∈R\begin{array}{l}A=\begin{bmatrix}1 & -1 \\\2 & \alpha \\\\\end{bmatrix}\ \text{and}\ B=\begin{bmatrix}\beta & 1 \\\1 & 0 \\\\\end{bmatrix},\alpha, \beta \in R\end{array}.

Let α1α1 be the value of α which satisfies
(A+B)2=A2+[22\22]\begin{array}{l}(A + B)^2 =A^2 + \begin{bmatrix}2 & 2 \\\2 & 2 \\\\\end{bmatrix}\end{array}

and α2α2 be the value of α which satisfies

(A+B)2=B2.\begin{array}{l}\left(A + B\right)^2 = B^2.\end{array}

Then ∣α1–α2∣|α1 – α2| is equal to _________.

Answer

∵(A+B)2=A2+B2+AB+BA\begin{array}{l}\left(A + B\right)^2 = A^2 + B^2 + AB + BA\end{array}

=A2+[22\22]\begin{array}{l}=A^2 + \begin{bmatrix}2 & 2 \\\2 & 2 \\\\\end{bmatrix}\end{array}

∴B2+AB+BA=[22\22]   …(1)\begin{array}{l}\therefore B^2 + AB + BA =\begin{bmatrix}2 & 2 \\\2 & 2 \\\\\end{bmatrix}\ \ \ …(1)\end{array}

AB=[1−1\2α][β1\10]=[β−11α+2β2]\begin{array}{l}AB =\begin{bmatrix}1 & -1 \\\2 & \alpha \\\\\end{bmatrix}\begin{bmatrix}\beta & 1 \\\1 & 0 \\\\\end{bmatrix}= \begin{bmatrix}\beta-1 & 1 \\\\\alpha + 2\beta & 2 \\\\\end{bmatrix}\end{array}

BA=[β1\10][1−1\2α]=[β+2α–β\1−1]\begin{array}{l}BA = \begin{bmatrix}\beta & 1 \\\1 & 0 \\\\\end{bmatrix}\begin{bmatrix}1 & -1 \\\2 & \alpha \\\\\end{bmatrix}= \begin{bmatrix}\beta+2 & \alpha – \beta \\\1 & -1 \\\\\end{bmatrix}\end{array}

B2=[β1\10][β1\10]=[β2+1ββ1]\begin{array}{l}B^2 = \begin{bmatrix}\beta & 1 \\\1 & 0 \\\\\end{bmatrix}\begin{bmatrix}\beta & 1 \\\1 & 0 \\\\\end{bmatrix}= \begin{bmatrix}\beta^2 + 1 & \beta \\\\\beta & 1 \\\\\end{bmatrix}\end{array}

By (1) we get

[β2+2β+2α+1α+3β+12]=[22\22]\begin{array}{l}\begin{bmatrix} \beta^2 + 2\beta + 2& \alpha + 1 \\\\\alpha + 3\beta + 1 & 2 \\\\\end{bmatrix}=\begin{bmatrix}2 & 2 \\\2 & 2 \\\\\end{bmatrix}\end{array}

∴α=1,β=0,⇒α1=1\begin{array}{l}\therefore \alpha = 1, \beta = 0 , \Rightarrow \alpha_ 1 = 1\end{array}, Similarly If A 2 + AB + BA = 0 then

(A2=[1−1\2α][1−1\2α]=[−1−1−α\2+2αα2−2])\begin{array}{l}\left(A^2 = \begin{bmatrix} 1& -1 \\\2 & \alpha \\\\\end{bmatrix}\begin{bmatrix}1 & -1 \\\2 & \alpha \\\\\end{bmatrix}=\begin{bmatrix}-1 & -1-\alpha \\\2+2\alpha & \alpha^2-2 \\\\\end{bmatrix}\right)\end{array}

[2βα–β+1−1−αα+2β+1+2+2αα2−2+1]=[00\00]\begin{array}{l}\begin{bmatrix} 2\beta& \alpha – \beta + 1 -1 -\alpha \\\\\alpha + 2\beta + 1 +2 + 2\alpha & \alpha^2-2+1 \\\\\end{bmatrix}=\begin{bmatrix}0 & 0 \\\0 & 0 \\\\\end{bmatrix}\end{array}

⇒β=0 and α=–1⇒α2=–1\begin{array}{l}\Rightarrow \beta = 0 ~\text{and}~ \alpha = – 1 \Rightarrow \alpha_2 = – 1\end{array}

∴∣α1–α2∣=∣2∣=2\begin{array}{l}\therefore |\alpha_1 – \alpha_2|=|2|=2\end{array}