Answer with Step-by-step explanation:
We are given that a matrix B .
The eigenvalues of matrix are 0, 1 and 2.
a.We know that
Rank of matrix B=Number of different eigenvalues
We have three different eigenvalues
Therefore, rank of matrix B=3
b.
We know that
Determinant of matrix= Product of eigenvalues
Product of eigenvalues=[tex]0\times 1\times 2=0[/tex]
After transpose , the value of determinant remain same.
[tex]\mid B^TB\mid=\mid B^T\mid \mid B\mid =0\times 0=0[/tex]
c.Let Â
B=[tex]\left[\begin{array}{ccc}0&-&-\\-&1&-\\-&-&2\end{array}\right][/tex]
Transpose of matrix:Rows change into columns or columns change into rows.
After transpose of matrix B
[tex]B^T=\left[\begin{array}{ccc}0&-&-\\-&1&-\\-&-&2\end{array}\right][/tex]
[tex]B^TB=\left[\begin{array}{ccc}0^2&-&-\\-&1^2&-\\-&-&2^2\end{array}\right][/tex]
[tex]B^TB=\left[\begin{array}{ccc}0&-&-\\-&1&-\\-&-&4\end{array}\right][/tex]
Hence, the eigenvalues of [tex]B^TB[/tex] are 0, 1 and 4.
d.Eigenvalue of Identity matrix are 1, 1 and 1.
Eigenvalues of [tex]B^2+I=(0+1),(1+1),(2^2+1)=1,2,5[/tex]
We know that if eigenvalue of A is [tex]\lambda[/tex]
Then , the eigenvalue of [tex]A^{-1}[/tex] is [tex]\frac{1}{\lambda}[/tex]
Therefore, the eigenvalues of [tex](B^2+I)^{-1}[/tex] are Â
[tex]\frac{1}{1},\frac{1}{2},\frac{1}{5}[/tex]
The eigenvalues of [tex](B^2+I)^{-1}[/tex] are 1,[tex]\frac{1}{2}[/tex] and [tex]\frac{1}{5}[/tex]
