Question

The characteristic roots of the matrix \[A = \left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  0&{ - 4}&2 \\
  0&0&7
\end{array}} \right]\]
is given by :-
(A) \[{\mathbf{1}}, - {\mathbf{4}},{\text{ }}{\mathbf{7}}\]
(B) \[{\mathbf{1}},{\text{ }}{\mathbf{4}},{\text{ }} - {\text{ }}{\mathbf{7}}\]
(C) \[{\mathbf{1}},{\text{ }}{\mathbf{4}},{\text{ }}{\mathbf{7}}\]
(D) \[ - {\mathbf{1}},{\text{ }} - {\mathbf{4}},{\text{ }} - {\mathbf{7}}\]

Answer 1

Hint :- To solve this,first consider a square matrix and then proceed
If A be a Square Matrix
I be a unit matrix of same order
Then, $\left| {A - \lambda I} \right|$ is called the Characteristic Polynomial of Matrix.
And $\left| {A - \lambda I} \right| = 0$ is called Characteristic Roots of Matrix.


Complete step by step solution:
\[A = \left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  0&{ - 4}&2 \\
  0&0&7
\end{array}} \right];I = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]\]
Here, A is given matrix and I is used matrix of $3$ order because A is $3 \times 3$
Now, we multiply by $\lambda $with the unit matrix. $\lambda $

\[\lambda I = \left[ {\begin{array}{*{20}{c}}
  \lambda &0&0 \\
  0&\lambda &0 \\
  0&0&\lambda
\end{array}} \right]\]
Now from hint
$\left| {A - \lambda I} \right| = 0$

\[ \Rightarrow \left| {A - \lambda I} \right| = \left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  0&{ - 4}&2 \\
  0&0&7
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  \lambda &0&0 \\
  0&\lambda &0 \\
  0&0&\lambda
\end{array}} \right] = 0\]
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {1 - \lambda }&2&3 \\
  0&{ - 4 - \lambda }&2 \\
  0&0&{7 - \lambda }
\end{array}} \right] = 0\]
By solving
$(1 - \lambda )[( - 4 - \lambda )(7 - \lambda ) - 0] - 2[0 - 0] + 3[0 - 0] = 0$
$(1 - \lambda )( - 4 - \lambda )(7 - \lambda ) = 0$
Because we get cubic equation of $\lambda $
So, there will three values of $\lambda $
$1 - \lambda = 0; - 4 - \lambda = 0;7 - \lambda = 0$
$\lambda = 1;\lambda = - 4;\lambda = 7$
$\lambda = 1, - 4,7$
So right option is (A) $1, - 4,7$
Therefore characteristic roots of $A = 1, - 4,7$

Note – If A is on the Square Matrix of n then $\lambda $has maximum n values.
So, In this question we can find three values of $\lambda $. The number of values also depends on the matrix.So please give importance to the order of the matrix being considered