Question

Suppose A is any $3\times 3$ non – singular matrix and (A – 3I)( A – 5I) = O, where $I={{I}_{3}}$ and $O={{O}_{3}}$ , if $\alpha A+\beta {{A}^{-1}}=4I$ , then $\alpha +\beta $ is equals to:
( a ) 8
( b ) 12
( c ) 13
( d ) 7

Answer 1

Hint: To solve this question what we will do is first we will solve the equation (A – 3I)( A – 5I) = O, then we will divide obtained equation with matrix A, then using property of Identity matrix and by rearranging equation like $\alpha A+\beta {{A}^{-1}}=4I$, we will obtain the value of $\alpha ,\beta $, and then $\alpha +\beta $.

Complete step by step answer:
Before we start solving the given question let us see what non – singular matrix is and what does notation ${{I}_{n}}$, ${{A}^{-1}}$ and ${{O}_{n}}$ means.
If we have any matrix say, matrix A , then ${{A}^{-1}}$ represents the inverse of matrix A such that $A.{{A}^{-1}}=I$, where A and ${{A}^{-1}}$ are $n\times n$( square ) matrix also, $I$ is also $n\times n$( square ) matrix.
${{I}_{n}}$ is $n\times n$( square ) matrix, called an identity matrix whose elements of diagonal are 1 and rest elements are 0.
${{O}_{n}}$is $n\times n$( square ) matrix, called a null matrix whose all elements are 0.
An $n\times n$( square ) matrix is called non – singular matrix, if there exists an $n\times n$ matrix B such that AB = BA = ${{I}_{n}}$, where ${{I}_{n}}$ denotes the $n\times n$identity matrix.
$n\times m$ means n columns and m rows.
Also, if any matrix A is nonsingular, then the inverse of the matrix always exists.
Now, in question it is given that, A is any $3\times 3$ non – singular matrix and (A – 3I)( A – 5I) = O, where $I={{I}_{3}}$ and $O={{O}_{3}}$ , if $\alpha A+\beta {{A}^{-1}}=4I$ and we have to evaluate the value of $\alpha +\beta $.
So, we have (A – 3I)( A – 5I) = O
${{A}^{2}}-8A+15I=0$
Dividing above equation be A, we get
$\dfrac{1}{A}({{A}^{2}}-8A+15I)=0$
So, $A-8I+15I{{A}^{-1}}=0$, as $A.{{A}^{-1}}=I$
We can re – write $A-8I+15I{{A}^{-1}}=0$ as
$A+15{{A}^{-1}}=8I$
Dividing, equation $A+15{{A}^{-1}}=8I$ by 2, we get
$\dfrac{A}{2}+\dfrac{15{{A}^{-1}}}{2}=4I$
Now we can compare $\dfrac{A}{2}+\dfrac{15{{A}^{-1}}}{2}=4I$with $\alpha A+\beta {{A}^{-1}}=4I$,
We get, $\alpha =\dfrac{1}{2}$and $\beta =\dfrac{15}{2}$,
So, $\alpha +\beta $will be
$\alpha +\beta =\dfrac{1}{2}+\dfrac{15}{2}$
On solving, we get
$\dfrac{16}{2}=8$

So, the correct answer is “Option A”.

Note: To solve this question, we need to know the meaning and representation of identity matrix, null matrix and singular and non – singular matrix. Try to get hind from the question and then solve as it will make your understanding for the given question better.