Question

If A is an idempotent matrix satisfying, ${{\left( I-0.4A \right)}^{-1}}=I-\alpha A$ where ‘ I’ is the unit matrix of same order as that of ‘A’, then the value of |9$\alpha $| is equal to?

Answer 1

Hint: Use the properties of matrix for solving this problem, also use the property of Idempotent matrix i.e. ‘If ‘X’ is an idempotent matrix then, ${{X}^{2}}=X$’ which will work as a key point is the solution.

We will write the given equation first,
$\therefore {{\left( I-0.4A \right)}^{-1}}=I-\alpha A$
We will multiply both sides by $\left( I-0.4A \right)$ therefore we will get,
$\therefore \left( I-0.4A \right)\times {{\left( I-0.4A \right)}^{-1}}=\left( I-0.4A \right)\times \left( I-\alpha A \right)$……………………………… (1)
To proceed further in the solution we should know the property of matrix which is given below,
Property:
$A\times {{A}^{-1}}=I$
Where, A be any matrix of the order $m\times n$
      ‘I’ be the identity matrix of the order $m\times n$
If we observe the equation (1) we say that in the term $\left( I-0.4A \right)$ there are some operations performed. As matrix ‘A’ is multiplied by 0.4 therefore it is going to give a matrix at the end and as given in the problem both ‘A’ and ‘I’ have same order therefore the whole term $\left( I-0.4A \right)$is going to give us a matrix at the end.
Therefore if we use the property given above in equation (1) we will get,
$\therefore I=\left( I-0.4A \right)\times \left( I-\alpha A \right)$
Now we are going to multiply the brackets on the right hand side of equation, therefore we will get,
$\therefore I={{I}^{2}}-\alpha \times A\times I-0.4\times A\times I+0.4\times \alpha \times {{A}^{2}}$………………………………………. (2)
To proceed further in the solution we should know the properties of matrix given below,
Properties:
${{I}^{2}}=I$
$A\times I=A$ Provided that the given matrix and identity matrix should have the same order.
By using above properties we can write equation (2) as,
$\therefore I=I-\alpha \times A-0.4\times A+0.4\times \alpha \times {{A}^{2}}$
By cancelling ‘I’ from both sides we will get,
$\therefore 0=0-\alpha \times A-0.4\times A+0.4\times \alpha \times {{A}^{2}}$
$\therefore 0=-\alpha \times A-0.4\times A+0.4\times \alpha \times {{A}^{2}}$
As ‘A’ is an Idempotent matrix therefore we can use the property of Idempotent matrix given below to solve the equation,
Property:
If ‘X’ is an idempotent matrix then, ${{X}^{2}}=X$
By using the above property we can write the above equation as,
$\therefore 0=-\alpha \times A-0.4\times A+0.4\times \alpha \times A$
$\therefore 0=-\alpha \times A+0.4\times \alpha \times A-0.4\times A$
By taking $\alpha \times A$ common we will get,
$\therefore 0=\alpha \times A\left( -1+0.4 \right)-0.4\times A$
Now, A can be cancelled out therefore we can write,
$\therefore 0=\alpha \left( -0.6 \right)-0.4$
By shifting 0.4 on the left hand side of the equation we will get,
$\therefore 0.4=\alpha \left( -0.6 \right)$
\[\therefore \dfrac{0.4}{-0.6}=\alpha \]
\[\therefore \alpha =\dfrac{0.4}{-0.6}\]
\[\therefore \alpha =-\dfrac{4}{6}\]
\[\therefore \alpha =-\dfrac{2}{3}\]
Multiplying by 9 on both sides of the equation we will get,
\[\therefore 9\alpha =9\left( -\dfrac{2}{3} \right)\]
Taking the modulus on both sides of the equation we will get,
\[\therefore |9\alpha |=\left| 9\left( -\dfrac{2}{3} \right) \right|\]
\[\therefore |9\alpha |=\left| -3\times 2 \right|\]
\[\therefore |9\alpha |=\left| -6 \right|\]
If we remove the modulus sign from the right hand side then we will get,
\[\therefore |9\alpha |=6\]
Therefore the value of \[|9\alpha |\] is equal to 6.

Note: Do remember that the value of \[A\times I\] is ‘A’ only when both the matrices have the same orders and if they are of different order then\[A\times I\ne A\]. This might confuse students during exams so always read the conditions carefully.