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Question: If a matrix \(A\) is such that \(3{A^3} + 2{A^2} + 5A + I = 0\) then \({A^{ - 1}}\) is equal to: A...

If a matrix AA is such that 3A3+2A2+5A+I=03{A^3} + 2{A^2} + 5A + I = 0 then A−1{A^{ - 1}} is equal to:
A) −(3A2+2A+5) - \left( {3{A^2} + 2A + 5} \right)
B) 3A2+2A+53{A^2} + 2A + 5
C) 3A2−2A−53{A^2} - 2A - 5
D) None of the above

Explanation

Solution

Given equation is a simple equation in a matrix. We will use properties of matrix multiplications to first simplify the given equation. Then we will post multiply by the inverse to neutralise the equation and obtain the answer in the required form.

Complete step by step solution:
We have given the equation 3A3+2A2+5A+I=03{A^3} + 2{A^2} + 5A + I = 0 .
Since the identity with respect to the given matrix AA exists, it is safe to assume that the inverse A−1{A^{ - 1}} also exists with respect to the given matrix AA .
Thus, we can find the inverse of the given matrix.
We know that the identity element is defined as follows for any matrix AA:
AA−1=I=A−1AA{A^{ - 1}} = I = {A^{ - 1}}A
Now we will use one of the two equalities to express the identity element in the given equation as follows:
3A3+2A2+5A+A−1A=03{A^3} + 2{A^2} + 5A + {A^{ - 1}}A = 0
We can transfer all the terms except A−1A{A^{ - 1}}A on the right-hand side as follows:
A−1A=−3A3−2A2−5A{A^{ - 1}}A = - 3{A^3} - 2{A^2} - 5A
Now we can express the above equation as follows:
A−1A=−3A2⋅A−2A⋅A−5A{A^{ - 1}}A = - 3{A^2} \cdot A - 2A \cdot A - 5A
Now post multiply throughout by the A−1{A^{ - 1}} in the given equation.
A−1(AA−1)=−3A2⋅(AA−1)−2A⋅(AA−1)−5(AA−1){A^{ - 1}}\left( {A{A^{ - 1}}} \right) = - 3{A^2} \cdot \left( {A{A^{ - 1}}} \right) - 2A \cdot \left( {A{A^{ - 1}}} \right) - 5\left( {A{A^{ - 1}}} \right)
We know that AA−1=IA{A^{ - 1}} = I .
Therefore,
A−1(I)=−3A2⋅(I)−2A⋅(I)−5(I){A^{ - 1}}\left( I \right) = - 3{A^2} \cdot \left( I \right) - 2A \cdot \left( I \right) - 5\left( I \right)
Now for any matrix BB we have BI=BBI = B .
Thus, the above equation is written as follows:
A−1=−3A2−2A−5{A^{ - 1}} = - 3{A^2} - 2A - 5
Taking the term −1 - 1 common from the right-hand side we can write,
A−1=−(3A2+2A+5){A^{ - 1}} = - \left( {3{A^2} + 2A + 5} \right)

Hence, the correct option is A.

Note:
Here we first checked whether the inverse exists or not as none of the above is also an option. Later we expressed the identity by using the definition of the identity element. We just finally rearranged the given equation in order to obtain the answer in the forms suitable to the given options. Note that we can either post multiply or pre multiply, it depends on how we are expressing the identity element. Also, it is important to note that the given equation is in matrix so we cannot just divide to obtain the answer.