Question

If \[{(aA)^{ - 1}} = \dfrac{1}{a}{A^{ - 1}}\] where a is any real number and A is a square matrix.
A. True
B. False

Answer 1

Hint: In mathematics, a matrix is a rectangular array or table of numbers, symbols, or expressions, arranged in rows and columns. Here in this question we have to prove the given inequality. Let us consider the example and prove the inequality.

Complete step-by-step answer:
Now let us consider the matrix A be
\[A = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right]\]
let a = 2 be a scalar value
therefore, the aA can be determined by
\[ \Rightarrow aA = 2 \times \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right]\]
On multiplying we have
\[ \Rightarrow aA = \left[ {\begin{array}{*{20}{c}}
  2&4 \\
  4&6
\end{array}} \right]\]
Now we have to determine the \[{(aA)^{ - 1}}\]
As we know that the formula for the inverse. It is defined as \[{A^{ - 1}} = \dfrac{{adjA}}{{|A|}}\]
Therefore the \[{(aA)^{ - 1}} = \dfrac{{adj(aA)}}{{|aA|}}\]
Let \[A = [{a_{ij}}]\] be a square matrix of order n . The adjoint of a matrix A is the transpose of the cofactor matrix of A . It is denoted by adj A . An adjoint matrix is also called an adjugate matrix.

The adjoint of the matrix is \[\left[ {\begin{array}{*{20}{c}}
  6&{ - 4} \\
  { - 4}&2
\end{array}} \right]\]
The \[|aA| = 12 - 8 = 4\]
Therefore \[{(aA)^{ - 1}} = \dfrac{1}{4}\left[ {\begin{array}{*{20}{c}}
  6&{ - 4} \\
  { - 4}&2
\end{array}} \right]\]------ (1)
Now let we find the value of \[\dfrac{1}{a}{A^{ - 1}}\]
The Matrix and scalar value will be same
Now let us consider the matrix A be
\[A = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right]\]
let a = 2 be a scalar value
As we know that the formula for the inverse. It is defined as \[{A^{ - 1}} = \dfrac{{adjA}}{{|A|}}\]
The adjoint of the matrix is \[\left[ {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  { - 2}&1
\end{array}} \right]\]
The \[|A| = 3 - 4 = - 1\]
Therefore \[{(A)^{ - 1}} = \dfrac{{ - 1}}{1}\left[ {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  { - 2}&1
\end{array}} \right]\]
Therefore
\[\dfrac{1}{a}{A^{ - 1}} = \dfrac{{ - 1}}{2}\left[ {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  { - 2}&1
\end{array}} \right]\] ----- (2)
The equation (1) and the equation (2) are not the same.
Therefore \[{(aA)^{ - 1}} \ne \dfrac{1}{a}{A^{ - 1}}\]

So, the correct answer is “Option B”.

Note: Let \[A = [{a_{ij}}]\] be a square matrix of order n . The adjoint of a matrix A is the transpose of the cofactor matrix of A . It is denoted by adj A . An adjoint matrix is also called an adjugate matrix. The determinant of a matrix is a special number that can be calculated from a square matrix.