Question

If \[A\] is a square matrix such that \[{A^2} = I\], then \[{(A - I)^3} + {(A + I)^3} - 7A\] is equal to
\[A.\] \[A\]
\[B.\] \[I - A\]
\[C.\] \[I + A\]
\[D.\] \[3A\]

Answer 1

Hint: We have to simplify the given iteration by using cubic algebraic formula and the identity matrix property. After doing some simplification and using some formula. Then we will get the required answer.

Formula used: The algebraic formula of \[{(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)\] .
And, the algebraic formula of \[{(a - b)^3} = {a^3} - {b^3} - 3ab(a - b)\].
Multiplication of any identity matrix and any other matrix gives us the same matrix itself.
Let's say \[M\] is a square matrix.
Here\[M = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right)\]
If we multiply it by a \[2 \times 2\] identity matrix, then we will get the following derivation:
\[M \times I = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)\]
Using the multiplication and addition rule of a matrix:
\[M \times I = \left( {\begin{array}{*{20}{c}}
  {{a_{11}} \times 1 + 0 \times {a_{12}}}&{0 \times {a_{11}} + {a_{12}} \times 1} \\
  {{a_{21}} \times 1 + 0 \times {a_{22}}}&{0 \times {a_{21}} + {a_{22}} \times 1}
\end{array}} \right)\]
\[ \Rightarrow M \times I = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right)\]
\[ \Rightarrow M \times I = M.\]
And, also we need to apply that multiplication of two identity matrices of the same degree always gives the identity matrix of same degree.
\[I \times I = I.\]

Complete step-by-step solution:
We will apply the general formula of algebra to simplify the above equation:
\[{(A - I)^3} + {(A + I)^3} - 7A\]\[ = \{ {A^3} - {I^3} - 3AI(A - I)\} + \{ {A^3} + {I^3} + 3AI(A + I)\} - 7A\]
Now multiply the terms in brackets terms and simplify it further:
\[ \Rightarrow \{ {A^3} - {I^3} - 3{A^2}I + 3A{I^2}\} + \{ {A^3} + {I^3} + 3{A^2}I + 3A{I^2}\} - 7A\]
As we know that matrix multiplication of the same degree can be used the same as algebraic multiplication rule.
Now, cancel out the negative and positive terms of the quantities we get:
\[ \Rightarrow ({A^3} + {A^3}) + (3A{I^2} + 3A{I^2}) - 7A\]
Now, add the same terms,
\[ \Rightarrow 2{A^3} + 6A{I^2} - 7A\]
By simplifying further using the general algebraic properties of matrix, we get:
\[ \Rightarrow 2({A^2})A + 6AI - 7A\]
Now put the given value of \[{A^2} = I\],
\[ \Rightarrow 2(I)A + 6AI - 7A\]
Associative law of multiplication of two matrices of same degree states that \[AI = IA\].
So, we can rearrange the above equation like:
\[ \Rightarrow 2AI + 6AI - 7A\]
Let us add the term and we get
\[ \Rightarrow 8AI - 7A\]
We calculate that multiplication of an identity matrix with any matrix gives the result of the same matrix itself.
\[ \Rightarrow 8A - 7A\]
Let us subtract the term and we get
\[ \Rightarrow A\] .
Hence \[{(A - I)^3} + {(A + I)^3} - 7A = A\]
\[\therefore \] Option A is the correct choice.

Note: Cumulative law and Associative law of any matrix holds true as the general algebra property.
Multiplication of two same degree matrices gives the matrix of same degree.
Identity matrix also known as unit matrix
Multiplication with identity matrix by any matrix gives us the mirror of the same matrix only.