Question

If $ A = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right) $ and $ B = \left( {\begin{array}{*{20}{c}}
  a&0 \\
  0&b
\end{array}} \right) $ , where a,b are natural numbers, then which one of the following are correct?
(a) There exists more than one but finite number of $ B's $ such that $ AB = BA $
(b) There exist exactly one $ B $ such that $ AB = BA $
(c) There exist infinitely many $ B's $ such that $ AB = BA $
(d) There cannot exist any B such that $ AB = BA $

Answer 1

Hint: Matrix multiplication is possible only when the number of columns in one matrix is equal to the number of rows in the second matrix. We do matrix multiplication by multiplying the column value with the corresponding row value in the other matrix and summing them up. The resultant matrix will have an equal number of columns as that of the first matrix and equal number of rows as that of the second matrix.

Complete step-by-step answer:
We know that
 $ A = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right) $ and
 $ B = \left( {\begin{array}{*{20}{c}}
  a&0 \\
  0&b
\end{array}} \right) $
Now , we will calculate $ AB $
By matrix multiplication we will get,
 $
  AB = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  a&0 \\
  0&b
\end{array}} \right) \\
  AB = \left( {\begin{array}{*{20}{c}}
  {1 \times a + 2 \times 0}&{1 \times 0 + 2 \times b} \\
  {3 \times a + 4 \times 0}&{3 \times 0 + 4 \times b}
\end{array}} \right) \\
  AB = \left( {\begin{array}{*{20}{c}}
  a&{2b} \\
  {3a}&{4b}
\end{array}} \right) \\
  $
Now, we will calculate BA
By matrix multiplication we will get,
 $
  BA = \left( {\begin{array}{*{20}{c}}
  a&0 \\
  0&b
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right) \\
  BA = \left( {\begin{array}{*{20}{c}}
  {a \times 1 + 0 \times 3}&{a \times 2 + 0 \times 4} \\
  {0 \times 1 + b \times 3}&{0 \times 2 + b \times 4}
\end{array}} \right) \\
  A = \left( {\begin{array}{*{20}{c}}
  a&{2a} \\
  {3b}&{4b}
\end{array}} \right) \\
  $
We know that
 $ AB = BA $
 $ \left( {\begin{array}{*{20}{c}}
  a&{2b} \\
  {3a}&{4b}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  a&{2a} \\
  {3b}&{4b}
\end{array}} \right) $
Comparing the equations we get,
 $
  a = a - (i) \\
  2b = 2a - (ii) \\
  3a = 3b - (iii) \\
  4b = 4b - (iv) \;
  $
Equation $ (i) $ and $ (iii) $ are already equal
When we compare $ (ii) $ and $ (iv) $ we see that the equations are equal only when $ a = b $ .
Since there are infinitely many values of $ a $ that can be equal to $ b $ . Hence $ B $ can infinitely many values such that $ AB = BA $
Hence, option $ (c) $ is correct. There exists infinitely many B’s such that AB=BA.
So, the correct answer is “OPTION C”.

Note: In a matrix multiplication, $ AB = BA $ (commutative property ) is only possible when either $ AorB $ is an identity element or else both the matrices are equal. In this question, both the matrices are equal.