Questions & Answers

Question

Answers

3&0&3 \\

0&3&0 \\

3&0&3

\end{array}} \right]\]. Then the roots of the equation \[\det \left( {A - \lambda {I_3}} \right) = 0\] (where \[{I_3}\] is the identity matrix of order 3) are ?

A. 3, 0, 3

B. 0, 3, 6

C. 1, 0, -6

D. 3, 3, 6

Answer
Verified

Hint:- We had to only put the value of \[{I_3}\] in the equation \[\det \left( {A - \lambda {I_3}} \right) = 0\] as \[{I_3}\] is the identity matrix of order 3. And after solving this equation we will get the required value of \[\lambda \].

__Complete step-by-step answer:__

Now as we know that the identity matrix has its all diagonal elements equal to 1 and all other elements are equal to 0.

So, the matrix \[{I_3}\] will be an identity matrix of order 3.

So, \[{I_3} = \left[ {\begin{array}{*{20}{c}}

1&0&0 \\

0&1&0 \\

0&0&1

\end{array}} \right]\]

Now as we know that when we multiply any constant value to any matrix then that value is multiplied with each element of that matrix.

So, \[\lambda {I_3} = \left[ {\begin{array}{*{20}{c}}

\lambda &0&0 \\

0&\lambda &0 \\

0&0&\lambda

\end{array}} \right]\] (1)

Now, as the order of the matrix A and \[\lambda {I_3}\] is the same (i.e. 3). So, when we subtract these two matrices then each corresponding element will get subtracted.

So, \[A - \lambda {I_3} = \left[ {\begin{array}{*{20}{c}}

3&0&3 \\

0&3&0 \\

3&0&3

\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}

\lambda &0&0 \\

0&\lambda &0 \\

0&0&\lambda

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{3 - \lambda }&0&3 \\

0&{3 - \lambda }&0 \\

3&0&{3 - \lambda }

\end{array}} \right]\] (2)

So, now we have to find the determinant of the matrix \[A - \lambda {I_3}\] and put that equal to zero. To find the value of \[\lambda \].

As we know that if \[X = \left[ {\begin{array}{*{20}{c}}

a&b&c \\

d&e&f \\

g&h&i

\end{array}} \right]\] is a matrix then the determinant of the matrix X is calculated as,

\[ \Rightarrow \det \left( X \right) = a\left( {e \times i - f \times h} \right) - b\left( {d \times i - f \times g} \right) + c\left( {d \times h - e \times g} \right)\]

So, now let us find the determinant of the matrix \[A - \lambda {I_3}\].

\[ \Rightarrow \det \left( {A - \lambda {I_3}} \right) = \left( {3 - \lambda } \right)\left( {\left( {3 - \lambda } \right) \times \left( {3 - \lambda } \right) - 0} \right) - 0\left( {0 \times \left( {3 - \lambda } \right) - 0 \times 3} \right) + 3\left( {0 \times 0 - \left( {3 - \lambda } \right) \times 3} \right)\]

\[ \Rightarrow \det \left( {A - \lambda {I_3}} \right) = {\left( {3 - \lambda } \right)^3} - 9\left( {3 - \lambda } \right) = \left( {3 - \lambda } \right)\left( {{3^2} + {\lambda ^2} - 6\lambda - 9} \right)\]

\[ \Rightarrow \det \left( {A - \lambda {I_3}} \right) = \left( {3 - \lambda } \right)\left( {{\lambda ^2} - 6\lambda } \right) = \lambda \left( {3 - \lambda } \right)\left( {\lambda - 6} \right)\] -----(3)

So, now as it is given in the question that the \[\det \left( {A - \lambda {I_3}} \right) = 0\]. So, equating equation 3 with zero.

\[ \Rightarrow \lambda \left( {3 - \lambda } \right)\left( {\lambda - 6} \right) = 0\]

So, \[\lambda = 0,3,6\]

Hence, the correct option will be B.

Note:- Whenever we come up with this type of problem then we should remember that \[{I_n}\] is an identity matrix of order \[n \times n\], with all its diagonal elements equal to 1 and all other elements equal to zero. And when we multiply any constant term \[\lambda \] with any matrix then each element of the matrix is multiplied by \[\lambda \]. So, first we had to find the value of \[\lambda {I_3}\] and then subtract that with the matrix A to find the value of matrix A – \[\lambda {I_3}\]. And then we had to find the determinant formula to find the value of det(A – \[\lambda {I_3}\]) and then equate that with zero to find the value of \[\lambda \]. This will be the easiest and efficient way to find the solution of the problem.

Now as we know that the identity matrix has its all diagonal elements equal to 1 and all other elements are equal to 0.

So, the matrix \[{I_3}\] will be an identity matrix of order 3.

So, \[{I_3} = \left[ {\begin{array}{*{20}{c}}

1&0&0 \\

0&1&0 \\

0&0&1

\end{array}} \right]\]

Now as we know that when we multiply any constant value to any matrix then that value is multiplied with each element of that matrix.

So, \[\lambda {I_3} = \left[ {\begin{array}{*{20}{c}}

\lambda &0&0 \\

0&\lambda &0 \\

0&0&\lambda

\end{array}} \right]\] (1)

Now, as the order of the matrix A and \[\lambda {I_3}\] is the same (i.e. 3). So, when we subtract these two matrices then each corresponding element will get subtracted.

So, \[A - \lambda {I_3} = \left[ {\begin{array}{*{20}{c}}

3&0&3 \\

0&3&0 \\

3&0&3

\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}

\lambda &0&0 \\

0&\lambda &0 \\

0&0&\lambda

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{3 - \lambda }&0&3 \\

0&{3 - \lambda }&0 \\

3&0&{3 - \lambda }

\end{array}} \right]\] (2)

So, now we have to find the determinant of the matrix \[A - \lambda {I_3}\] and put that equal to zero. To find the value of \[\lambda \].

As we know that if \[X = \left[ {\begin{array}{*{20}{c}}

a&b&c \\

d&e&f \\

g&h&i

\end{array}} \right]\] is a matrix then the determinant of the matrix X is calculated as,

\[ \Rightarrow \det \left( X \right) = a\left( {e \times i - f \times h} \right) - b\left( {d \times i - f \times g} \right) + c\left( {d \times h - e \times g} \right)\]

So, now let us find the determinant of the matrix \[A - \lambda {I_3}\].

\[ \Rightarrow \det \left( {A - \lambda {I_3}} \right) = \left( {3 - \lambda } \right)\left( {\left( {3 - \lambda } \right) \times \left( {3 - \lambda } \right) - 0} \right) - 0\left( {0 \times \left( {3 - \lambda } \right) - 0 \times 3} \right) + 3\left( {0 \times 0 - \left( {3 - \lambda } \right) \times 3} \right)\]

\[ \Rightarrow \det \left( {A - \lambda {I_3}} \right) = {\left( {3 - \lambda } \right)^3} - 9\left( {3 - \lambda } \right) = \left( {3 - \lambda } \right)\left( {{3^2} + {\lambda ^2} - 6\lambda - 9} \right)\]

\[ \Rightarrow \det \left( {A - \lambda {I_3}} \right) = \left( {3 - \lambda } \right)\left( {{\lambda ^2} - 6\lambda } \right) = \lambda \left( {3 - \lambda } \right)\left( {\lambda - 6} \right)\] -----(3)

So, now as it is given in the question that the \[\det \left( {A - \lambda {I_3}} \right) = 0\]. So, equating equation 3 with zero.

\[ \Rightarrow \lambda \left( {3 - \lambda } \right)\left( {\lambda - 6} \right) = 0\]

So, \[\lambda = 0,3,6\]

Hence, the correct option will be B.

Note:- Whenever we come up with this type of problem then we should remember that \[{I_n}\] is an identity matrix of order \[n \times n\], with all its diagonal elements equal to 1 and all other elements equal to zero. And when we multiply any constant term \[\lambda \] with any matrix then each element of the matrix is multiplied by \[\lambda \]. So, first we had to find the value of \[\lambda {I_3}\] and then subtract that with the matrix A to find the value of matrix A – \[\lambda {I_3}\]. And then we had to find the determinant formula to find the value of det(A – \[\lambda {I_3}\]) and then equate that with zero to find the value of \[\lambda \]. This will be the easiest and efficient way to find the solution of the problem.

×

Sorry!, This page is not available for now to bookmark.