Question

# Let $A = \left[ {\begin{array}{*{20}{c}} 3&0&3 \\ 0&3&0 \\ 3&0&3 \end{array}} \right]$. 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, 3B. 0, 3, 6C. 1, 0, -6D. 3, 3, 6

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 $\lambda$.

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.