Question & Answer
QUESTION

Which of the following is NOT left inverse of matrix $\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  1&1 \\
  2&3
\end{array}} \right]$ ?
A. $\left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{2}}&{\dfrac{1}{2}}&0 \\
  { - \dfrac{1}{2}}&{\dfrac{1}{2}}&0
\end{array}} \right]$
B. $\left[ {\begin{array}{*{20}{c}}
  2&{ - 7}&3 \\
  { - \dfrac{1}{2}}&{\dfrac{1}{2}}&0
\end{array}} \right]$
C. $\left[ {\begin{array}{*{20}{c}}
  { - \dfrac{1}{2}}&{\dfrac{1}{2}}&0 \\
  { - \dfrac{1}{2}}&{\dfrac{1}{2}}&0
\end{array}} \right]$
D.\[\]$\left[ {\begin{array}{*{20}{c}}
  0&3&{ - 1} \\
  { - \dfrac{1}{2}}&{\dfrac{1}{2}}&0
\end{array}} \right]$ .

ANSWER Verified Verified
Hint :In this question, we will have the properties of the inverse of a matrix .To solve this question ,we have to check all the matrices that are given in the options that either are left inverse of the given matrix or not left inverse of that matrix.

Complete step-by-step answer:
To check whether the given matrix is left inverse or not, it must satisfy the following condition :
$ \Rightarrow LA = \left[ {{A^{ - 1}}} \right]\left[ A \right] = I$ .
Where $A$ is the given matrix and ${A^{ - 1}}$is the left inverse of the matrix $A$ and $I$ is the identity matrix.
$A = \left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  1&1 \\
  2&3
\end{array}} \right]{\text{ and }}I = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$ .
1. For option A.
Left inverse, $LA = \left[ {{A^{ - 1}}} \right]\left[ A \right] = I$
$
   \Rightarrow LA = {\text{ }}\left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{2}}&{\dfrac{1}{2}}&0 \\
  { - \dfrac{1}{2}}&{\dfrac{1}{2}}&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  1&1 \\
  2&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{2} \times 1 + \dfrac{1}{2} \times 1 + 0}&{\dfrac{1}{2} \times - 1 + \dfrac{1}{2} \times 1 + 0} \\
  { - \dfrac{1}{2} \times 1 + \dfrac{1}{2} \times 1 + 0}&{ - \dfrac{1}{2} \times - 1 + \dfrac{1}{2} \times 1 + 0}
\end{array}} \right] \\
    \\
   \Rightarrow LA = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] = I \\
$
Hence ,option A is left inverse of matrix A .
2. For option B .
$
   \Rightarrow LA = {\text{ }}\left[ {\begin{array}{*{20}{c}}
  2&{ - 7}&3 \\
  { - \dfrac{1}{2}}&{\dfrac{1}{2}}&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  1&1 \\
  2&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {2 \times 1 + - 7 \times 1 + 3 \times 2}&{2 \times - 1 + - 7 \times 1 + 3 \times 3} \\
  { - \dfrac{1}{2} \times 1 + \dfrac{1}{2} \times 1 + 0}&{ - \dfrac{1}{2} \times - 1 + \dfrac{1}{2} \times 1 + 0}
\end{array}} \right] \\
    \\
   \Rightarrow LA = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] = I \\
 $
Hence, option B is left inverse of matrix A.
3. For option C .
$
   \Rightarrow LA = {\text{ }}\left[ {\begin{array}{*{20}{c}}
  { - \dfrac{1}{2}}&{\dfrac{1}{2}}&0 \\
  { - \dfrac{1}{2}}&{\dfrac{1}{2}}&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  1&1 \\
  2&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - \dfrac{1}{2} \times 1 + \dfrac{1}{2} \times 1 + 0}&{ - \dfrac{1}{2} \times - 1 + \dfrac{1}{2} \times 1 + 0} \\
  { - \dfrac{1}{2} \times 1 + \dfrac{1}{2} \times 1 + 0}&{ - \dfrac{1}{2} \times - 1 + \dfrac{1}{2} \times 1 + 0}
\end{array}} \right] \\
    \\
   \Rightarrow LA = \left[ {\begin{array}{*{20}{c}}
  0&1 \\
  0&1
\end{array}} \right] \ne I \\
 $
Hence , option C is not the left inverse of matrix A .
Here , we got the correct answer for the question by verifying only 3 options so we will not check the other options .
Therefore ,the correct answer is option C.
Note : In this type of question first we have to remember the properties of the inverse of a matrix. Then we will select the options one by one and apply those properties with the matrix given in the question . After that we will select the right option which is satisfying the statement and if some option gets left after we get the correct option then we will not check those options . through this we will get our result.