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Question: If \(X\) is a \(2 \times 3\) matrix such that \[\left| {{X^T}X} \right| \ne 0\] and \(A = {I_2} - X{...

If XX is a 2×32 \times 3 matrix such that ∣XTX∣≠0\left| {{X^T}X} \right| \ne 0 and A=I2−X(XTX)−1XTA = {I_2} - X{\left( {{X^T}X} \right)^{ - 1}}{X^T} and A2{A^2} is equal to :
(XT{X^T} denotes transpose of matrix XX)
A. AA
B. II
C. A−1{A^{ - 1}}
D. AXAX

Explanation

Solution

Use the property of inverse of matrices,(AB)−1=B−1A−1{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}} on the given condition A=I2−X(XTX)−1XTA = {I_2} - X{\left( {{X^T}X} \right)^{ - 1}}{X^T} to find the value of AA. Simplify the value of AA 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 Vn×m×Wm×l{V_{n \times m}} \times {W_{m \times l}}, then, the order of the product is (VW)n×l{\left( {VW} \right)_{n \times l}}. Then, find the value of A2{A^2}.

Complete step by step Answer:

We are given that XX is a 2×32 \times 3 matrix, where ∣XTX∣≠0\left| {{X^T}X} \right| \ne 0 and A=I2−X(XTX)−1XTA = {I_2} - X{\left( {{X^T}X} \right)^{ - 1}}{X^T}
We will first use the property of inverse simplify the expression, A=I2−X(XTX)−1XTA = {I_2} - X{\left( {{X^T}X} \right)^{ - 1}}{X^T}
It is known that (AB)−1=B−1A−1{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}
A=I2−XX−1(XT)−1XTA = {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 2×32 \times 3 then the order of its inverse is 3×23 \times 2, which is also the order of the transpose of the matrix.
A=I2−X2×3X3×2−1(XT)2×3−1XT3×2A = {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 Vn×m×Wm×l{V_{n \times m}} \times {W_{m \times l}}, then, the order of the product is (VW)n×l{\left( {VW} \right)_{n \times l}}
Therefore, we get the above expression as, A=I2−I2I2A = {I_2} - {I_2}{I_2}
Where I2{I_2} is an identity matrix of order 2×22 \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=0A = 0, then A2=0{A^2} = 0 which is equal to AA.
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=I2−I2I2 A=I2−I2 A=0  A = {I_2} - {I_2}{I_2} \\\ A = {I_2} - {I_2} \\\ A = 0 \\\