Chapter 6.1, Problem 19PS

To determine

(a) To determine:

The rank of $B$

The rank of matrix $B=2$

Given:

   $B$ Is $3\times 3$ matrix with Eigen value $0,1,2.$

Concept used:

The rank of matrix is the number of independent rows.

Calculation:

Let ${\lambda }_1,{\lambda }_2,{\lambda }_3$ be the Eigen value.

   $\begin{array}{l}{\lambda }_1=0\\ {\lambda }_2=1\\ {\lambda }_3=2\end{array}$

   $\begin{array}{c}\text{The value of determinant} =0\times 1\times 2\\ =0\end{array}$

That means two rows are similar.

So,

   $\begin{array}{c}\text{The rank of matrix} =3-1\\ =2\end{array}$

Conclusion:

Copy solution part.

To determine

(b) To determine:

The determinant of $B^{T}B$

   $\det \left(B^{T}B\right)=0$

Given:

   $B$ Is $3\times 3$ matrix with Eigen value $0,1,2.$

Concept used:

The transpose of the matrix is the interchange of row to corresponding column.

Calculation:

We have to determine the determinant of matrix $B^{T}B$, as determinant of matrix $B$ is zero as explained in part $\left(a\right)$ so we get,

   $\det \left(B\right)=0\cdots \cdots \left(1\right)$

Also for its transpose we have,

   $\det \left(B^T\right)=0\cdots \cdots \left(1\right)$

So, for determinant of matrix $B^{T}B$ we have,

   $\begin{array}{c}\det \left(B^{T}B\right)=\left|B^T\right|\left|B\right|\\ =0\end{array}$

So, we have $\det \left(B^{T}B\right)=0$.

Conclusion:

   $\det \left(B^{T}B\right)=0$

To determine

(c) To determine:

The determinant of $B^{T}B$

It can not be determined.

Given:

   $B$ is $3\times 3$ matrix with Eigen value $0,1,2.$

Concept used:

The transpose of the matrix is the interchange of row to corresponding column.

The value of determinant is the product of its Eigen value.

Calculation:

Since, $\det (B^{T}B)=0$

That means the product of Eigen value is $0$.

It means, at least one of the Eigen value is $0.$ but it can't be determined.

Conclusion:

It can not be determined.

To determine

(d) To determine:

The eigenvalues of ${\left(B^2+I\right)}^{-1}$

The Eigen value of ${(B^2+I)}^{-1}$ are $1,0.5,\mathrm{0.2.}$

Given:

   $B$ is $3\times 3$ matrix with Eigen value $0,1,2.$

Concept used:

For matrix $B^2$ the eigenvalues are given as the square of the eigenvalue of matrix $B$.

Calculation:

For eigenvalues of matrix ${\left(B^2+I\right)}^{-1}$, we will first determine tile eigenvalues of matrix $B^2$.

For matrix $B^2$ the eigenvalues are given as the square of the eigenvalues of matrix $B$.

So, we get the eigenvalues of matrix $B^2$ as $0,1\text{ and 4}$.

For eigenvalues of $\left(B^2+I\right)$ we have to increase the eigenvalues of matrix $B^2\text{ by 1}$

So we get the eigenvalues of matrix $\left(B^2+I\right)$ as $1,2\text{ and }5$,

For matrix ${\left(B^2+I\right)}^{-1}$ the eigenvalues are given as the inverse of the eigenvalues of matrix

   $\left(B^2+I\right)$

So, we get the eigenvalues of matrix ${\left(B^2+I\right)}^{-1}$ as $1,0.5\text{ and }0.2$.

Conclusion:

The Eigen value of ${(B^2+I)}^{-1}$ are $1,0.5,\mathrm{0.2.}$