Question

If $X$ is a $2 \times 3$ matrix such that \[\left| {{X^T}X} \right| \ne 0\] and $A = {I_2} - X{\left( {{X^T}X} \right)^{ - 1}}{X^T}$ and ${A^2}$ is equal to :
(${X^T}$ denotes transpose of matrix $X$)
A. \[A\]
B. $I$
C. ${A^{ - 1}}$
D. $AX$

Answer 1

Hint: Use the property of inverse of matrices,\[{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}\] on the given condition $A = {I_2} - X{\left( {{X^T}X} \right)^{ - 1}}{X^T}$ to find the value of $A$. Simplify the value of $A$ using the properties such as, if we multiply any matrix with its inverse, we get the identity and if two matrices are multiplied such that ${V_{n \times m}} \times {W_{m \times l}}$, then, the order of the product is ${\left( {VW} \right)_{n \times l}}$. Then, find the value of ${A^2}$.

Complete step by step Answer:

We are given that $X$ is a $2 \times 3$ matrix, where \[\left| {{X^T}X} \right| \ne 0\] and $A = {I_2} - X{\left( {{X^T}X} \right)^{ - 1}}{X^T}$
We will first use the property of inverse simplify the expression, $A = {I_2} - X{\left( {{X^T}X} \right)^{ - 1}}{X^T}$
It is known that \[{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}\]
$A = {I_2} - X{X^{ - 1}}{\left( {{X^T}} \right)^{ - 1}}{X^T}$
Also, if we multiply any matrix with its inverse, we get the identity.
And if the matrix is of order \[2 \times 3\] then the order of its inverse is \[3 \times 2\], which is also the order of the transpose of the matrix.
\[A = {I_2} - {X_{2 \times 3}}{X_{3 \times 2}}^{ - 1}{\left( {{X^T}} \right)_{2 \times 3}}^{ - 1}{X^T}_{3 \times 2}\]
If two matrices are multiplied such that ${V_{n \times m}} \times {W_{m \times l}}$, then, the order of the product is ${\left( {VW} \right)_{n \times l}}$
Therefore, we get the above expression as, \[A = {I_2} - {I_2}{I_2}\]
Where \[{I_2}\] is an identity matrix of order $2 \times 2$
$
  A = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] \\
  A = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] \\
  A = 0 \\
$
If $A = 0$, then ${A^2} = 0$ which is equal to $A$.
Hence, option A is correct.

Note: Inverse of a matrix only exists if determinant of the matrix is non-zero. Such a matrix is said to be an invertible matrix. Any matrix multiplied, added, or subtracted from the identity matrix results in the same matrix. Therefore, in this question, we can directly write
$
  A = {I_2} - {I_2}{I_2} \\
  A = {I_2} - {I_2} \\
  A = 0 \\
$