Question

If\[A = \left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  2&{ - 1}
\end{array}} \right]\],\[B = \left[ {\begin{array}{*{20}{c}}
  a&1 \\
  b&{ - 1}
\end{array}} \right]\]and \[{(A + B)^2} = {A^2} + {B^2}\] then the value of a and b are
A. \[a = 4,b = 1\]
B. \[a = 1,b = 4\]
C. \[a = 0,b = 4\]
D. \[a = 2,b = 4\]

Answer 1

Hint: To solve the given problem we first find the value of\[{(A + B)^2}\] and \[{A^2} + {B^2}\]from matrices A and B. Since \[{(A + B)^2} = {A^2} + {B^2}\], we equate the matrices thus obtained and hence equate the corresponding elements of the two matrices to find the value of a and b.

Formula used:
If \[A = {[{a_{ij}}]_{m \times n}}\] and \[B = {[{b_{ij}}]_{n \times p}}\] then we can say that \[A \times B = C\] where the value of C is
\[C = {[{c_{ij}}]_{m \times p}}\]
Here \[{c_{ij}} = \mathop \sum \limits_{j = 1}^m {a_{ij}}{b_{jk}} = {a_{i1}}{b_{1k}} + {a_{i2}}{b_{2k}} + ........ + {a_{im}}{b_{mk}}\]

Complete step by step solution:
We are given that,
\[A = \left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  2&{ - 1}
\end{array}} \right]\]and \[B = \left[ {\begin{array}{*{20}{c}}
  a&1 \\
  b&{ - 1}
\end{array}} \right]\]
On adding the two matrices we get,
\[A + B = \left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  2&{ - 1}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  a&1 \\
  b&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {1 + a}&0 \\
  {2 + b}&{ - 2}
\end{array}} \right]\]
On squaring we get,
\[{(A + B)^2} = \left[ {\begin{array}{*{20}{c}}
  {1 + a}&0 \\
  {2 + b}&{ - 2}
\end{array}} \right].\left[ {\begin{array}{*{20}{c}}
  {1 + a}&0 \\
  {2 + b}&{ - 2}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {1 + {a^2} + 2a}&0 \\
  {2 + 2a + b + ab - 4 - 2b}&4
\end{array}} \right]\]\[ = \left[ {\begin{array}{*{20}{c}}
  {{a^2} + 2a + 1}&0 \\
  {2a - b + ab - 2}&4
\end{array}} \right]\]
On evaluating \[{A^2} + {B^2}\]we get,
\[{A^2} + {B^2} = \left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  2&{ - 1}
\end{array}} \right].\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  2&{ - 1}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  a&1 \\
  b&{ - 1}
\end{array}} \right].\left[ {\begin{array}{*{20}{c}}
  a&1 \\
  b&{ - 1}
\end{array}} \right]\]\[ = \left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  0&{ - 1}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  {{a^2} + b}&{a - 1} \\
  {ab - b}&{b + 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {{a^2} + b - 1}&{a - 1} \\
  {ab - b}&b
\end{array}} \right]\]
We are given that \[{(A + B)^2} = {A^2} + {B^2}\]
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {{a^2} + 2a + 1}&0 \\
  {2a - b + ab - 2}&4
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {{a^2} + b - 1}&{a - 1} \\
  {ab - b}&b
\end{array}} \right]\]
Equating the corresponding elements, we get
\[{a^2} + 2a + 1 = {a^2} + b - 1\]
\[ \Rightarrow 2a - b = - 2\]…(1)
\[a - 1 = 0\]
\[ \Rightarrow a = 1\]…(2)
\[2a - b + ab - 2 = ab - b\]
\[ \Rightarrow 2a - 2 = 0\]…(3)
\[b = 4\]…(4)
\[a = 1,b = 4\] satisfies all the four equations (1),(2),(3) and (4).
Hence \[a = 1,b = 4\]

Option B. is the correct answer.

Note: To solve the given problem, one must know to add and multiply two matrices. One must make sure that the terms are added before giving the resultant value in each position of the resultant matrix after matrix multiplication.