# 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

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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.

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.

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