Question

Let \(\begin{array}{l}A=\begin{bmatrix}1 & -1 \\2 & \alpha \\\end{bmatrix}\ \text{and}\ B=\begin{bmatrix}\beta & 1 \\1 & 0 \\\end{bmatrix},\alpha, \beta \in R\end{array}\). 

Let \(α1\) be the value of α which satisfies
\(\begin{array}{l}(A + B)^2 =A^2 + \begin{bmatrix}2 & 2 \\2 & 2 \\\end{bmatrix}\end{array}\)

and \(α2\) be the value of α which satisfies

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

Then \(|α1 – α2|\) is equal to _________.

Answer 1

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

\(\begin{array}{l}=A^2 + \begin{bmatrix}2 & 2 \\2 & 2 \\\end{bmatrix}\end{array}\)

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

\(\begin{array}{l}AB =\begin{bmatrix}1 & -1 \\2 & \alpha \\\end{bmatrix}\begin{bmatrix}\beta & 1 \\1 & 0 \\\end{bmatrix}= \begin{bmatrix}\beta-1 & 1 \\\alpha + 2\beta & 2 \\\end{bmatrix}\end{array}\)

\(\begin{array}{l}BA = \begin{bmatrix}\beta & 1 \\1 & 0 \\\end{bmatrix}\begin{bmatrix}1 & -1 \\2 & \alpha \\\end{bmatrix}= \begin{bmatrix}\beta+2 & \alpha – \beta \\1 & -1 \\\end{bmatrix}\end{array}\)

\(\begin{array}{l}B^2 = \begin{bmatrix}\beta & 1 \\1 & 0 \\\end{bmatrix}\begin{bmatrix}\beta & 1 \\1 & 0 \\\end{bmatrix}= \begin{bmatrix}\beta^2 + 1 & \beta \\\beta & 1 \\\end{bmatrix}\end{array}\)

By (1) we get

\(\begin{array}{l}\begin{bmatrix} \beta^2 + 2\beta + 2& \alpha + 1 \\\alpha + 3\beta + 1 & 2 \\\end{bmatrix}=\begin{bmatrix}2 & 2 \\2 & 2 \\\end{bmatrix}\end{array}\)

\(\begin{array}{l}\therefore \alpha = 1, \beta = 0 , \Rightarrow \alpha_ 1 = 1\end{array}\), Similarly If A² + AB + BA = 0 then

\(\begin{array}{l}\left(A^2 = \begin{bmatrix} 1& -1 \\2 & \alpha \\\end{bmatrix}\begin{bmatrix}1 & -1 \\2 & \alpha \\\end{bmatrix}=\begin{bmatrix}-1 & -1-\alpha \\2+2\alpha & \alpha^2-2 \\\end{bmatrix}\right)\end{array}\)

\(\begin{array}{l}\begin{bmatrix} 2\beta& \alpha – \beta + 1 -1 -\alpha \\\alpha + 2\beta + 1 +2 + 2\alpha & \alpha^2-2+1 \\\end{bmatrix}=\begin{bmatrix}0 & 0 \\0 & 0 \\\end{bmatrix}\end{array}\)

\(\begin{array}{l}\Rightarrow \beta = 0 ~\text{and}~ \alpha = – 1 \Rightarrow \alpha_2 = – 1\end{array}\)

\(\begin{array}{l}\therefore |\alpha_1 – \alpha_2|=|2|=2\end{array}\)