
If it is a \[3 \times 3\] matrix whose characteristic polynomial equation \[{\lambda ^3} - 3{\lambda ^2} \pm 4 = 0\] then roots of polynomial is?
A.0
B.-2
C.4
D.-3
Answer
568.5k+ views
Hint: We have given a \[3 \times 3\]metric and its characteristic polynomial we have to find the roots of this polynomial. The polynomial is cubic so it will have three roots. One of its roots will be calculated by hit and trial method. Once the root is calculated. We take it as a factor and divide the cubic polynomials with it. In the result we get quadratic polynomials. The remaining two roots can be calculated by the quadratic polynomials by quadratic formula method.
Complete step-by-step answer:
We have given a matrix of order\[3 \times 3\]. Also we have given a characteristic polynomial equation\[{\lambda ^3} - 3{\lambda ^2} + 4 = 0\]
Firstly we will find a root by hint and trial method.
Putting \[\lambda = - 1\] in (i)
\[{( - 1)^2} - 3{( - 1)^2} + 4 = 0\]
$\Rightarrow$ \[ - 1 - 3 + 4 = 0\]
$\Rightarrow$ \[ - 3 + 3 = 0\]
$\Rightarrow$ \[\lambda = - 1\]
Satisfies the equation.
So \[\lambda = - 1\] is one of its roots and \[(\lambda + 1) = 0\] is one of its factors. Now dividing \[{\lambda ^3} - 3{\lambda ^2} + 4\]by \[(\lambda + 1)\]
So we get an equation \[{\lambda ^2} - 4\lambda + 4 = 0\](ii) this is a quadratic equation we can find the value of \[\lambda \] by quadratic formula method.
Comparing (ii) with \[a{\lambda ^2} + b\lambda + c = 0\]
\[a = 1,b = - 4,c = 4\]
\[
\Rightarrow \lambda = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 1 \times 4} }}{{2 \times 1}} \\
\Rightarrow \lambda = \dfrac{{4 \pm \sqrt {16 - 16} }}{2} = \dfrac{{4 \pm 0}}{2} \\
\]
\[
\Rightarrow \lambda = \dfrac{{{4}}}{{{2}}},\dfrac{{{4}}}{{{2}}} \\
\Rightarrow \lambda = 2,2 \\
\]
So the roots of the polynomial is \[ - 1,2,2\].
Note: Matrix is a set of numbers arranged in row and columns so as to form a rectangular array. The numbers are called elements or entries of the matrix. It has various applications in various branches of mathematics.
Characteristic equation of matrix -: A matrix polynomial or a matrix equation is a polynomial which gives information about the matrix. It is closely related to determinant of matrix and helps in finding the given value of matrix.
Complete step-by-step answer:
We have given a matrix of order\[3 \times 3\]. Also we have given a characteristic polynomial equation\[{\lambda ^3} - 3{\lambda ^2} + 4 = 0\]
Firstly we will find a root by hint and trial method.
Putting \[\lambda = - 1\] in (i)
\[{( - 1)^2} - 3{( - 1)^2} + 4 = 0\]
$\Rightarrow$ \[ - 1 - 3 + 4 = 0\]
$\Rightarrow$ \[ - 3 + 3 = 0\]
$\Rightarrow$ \[\lambda = - 1\]
Satisfies the equation.
So \[\lambda = - 1\] is one of its roots and \[(\lambda + 1) = 0\] is one of its factors. Now dividing \[{\lambda ^3} - 3{\lambda ^2} + 4\]by \[(\lambda + 1)\]
So we get an equation \[{\lambda ^2} - 4\lambda + 4 = 0\](ii) this is a quadratic equation we can find the value of \[\lambda \] by quadratic formula method.
Comparing (ii) with \[a{\lambda ^2} + b\lambda + c = 0\]
\[a = 1,b = - 4,c = 4\]
\[
\Rightarrow \lambda = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 1 \times 4} }}{{2 \times 1}} \\
\Rightarrow \lambda = \dfrac{{4 \pm \sqrt {16 - 16} }}{2} = \dfrac{{4 \pm 0}}{2} \\
\]
\[
\Rightarrow \lambda = \dfrac{{{4}}}{{{2}}},\dfrac{{{4}}}{{{2}}} \\
\Rightarrow \lambda = 2,2 \\
\]
So the roots of the polynomial is \[ - 1,2,2\].
Note: Matrix is a set of numbers arranged in row and columns so as to form a rectangular array. The numbers are called elements or entries of the matrix. It has various applications in various branches of mathematics.
Characteristic equation of matrix -: A matrix polynomial or a matrix equation is a polynomial which gives information about the matrix. It is closely related to determinant of matrix and helps in finding the given value of matrix.
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