Question

If $A = \left[ {\begin{array}{*{20}{c}}
  0&1 \\
  0&0
\end{array}} \right]$ and $AB = O$ , then $B = $
A. $\left[ {\begin{array}{*{20}{c}}
  1&1 \\
  1&1
\end{array}} \right]$
B. $\left[ {\begin{array}{*{20}{c}}
  0&1 \\
  { - 1}&0
\end{array}} \right]$
C. $\left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  1&0
\end{array}} \right]$
D. $\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  0&0
\end{array}} \right]$

Answer 1

Hint: Here, we suppose some value for matrix $B$ . Then, use the matrix multiplication rule of the matrices to get the value of a few of the unknowns in matrix $B$. After that, we can say that the product of the matrices $A$ and $B$ equals the zero matrices if all the unknowns of the matrix $B$ are zero, which is not in the options. So, we have to consider values according to that.

Formula Used:
Let $A = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
  {{b_{11}}}&{{b_{12}}} \\
  {{b_{21}}}&{{b_{22}}}
\end{array}} \right]$ , then $AB = C = \left[ {\begin{array}{*{20}{c}}
  {{c_{11}}}&{{c_{12}}} \\
  {{c_{21}}}&{{c_{22}}}
\end{array}} \right]$ .
Here, ${c_{11}} = {a_{11}}{b_{11}} + {a_{12}}{b_{21}}$ ; ${c_{12}} = {a_{11}}{b_{12}} + {a_{12}}{b_{22}}$ ; ${c_{21}} = {a_{21}}{b_{11}} + {a_{22}}{b_{21}}$ ; ${c_{22}} = {a_{21}}{b_{12}} + {a_{22}}{b_{22}}$ .

Complete step by step Solution:
Given, matrix $A = \left[ {\begin{array}{*{20}{c}}
  0&1 \\
  0&0
\end{array}} \right]$ .
We have here that $AB = O$, where the matrix $B$ is unknown and $O$ is a zero matrix.
Suppose, the matrix $B = \left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]$ .
Therefore, we have $AB = O$ as
$AB = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  0&1 \\
  0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right]$ …………………(1.1)
Multiply both the matrices on the left side, and we get
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  c&d \\
  0&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right]$
Equalising both the sides, we get $c = d = 0$ .
If we substitute these values in equation (1.1), we would get
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  0&1 \\
  0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  a&b \\
  0&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right]$
Out of the four given options, the best fitted values of $a$ and $b$ will be: $a = - 1$ and $b = 0$ .
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  0&1 \\
  0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  0&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right]$
So, we have our matrix $B = \left[ {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  0&0
\end{array}} \right]$ .

Therefore, the correct option is (D).

Note: The product of matrices is only possible if the order of the given matrices is equal or not. If the given matrix is not a square matrix i.e., does not have an equal number of rows and columns, then the product would only be possible if the number of rows of one matrix is equal to the number of columns of the other.