Science:Math Exam Resources/Courses/MATH221/December 2009/Question 10 (b)

MATH221 December 2009

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Question 10 (b)
Find the numbers b,c which make the matrix below diagonalizable. (No need to diagonalize them.) ${\begin{bmatrix}2&1&b&3\\0&3&-1&c\\0&0&2&2\\0&0&0&3\end{bmatrix}}$

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Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Now let’s find the eigenvalues of b ${\text{det}}({\begin{bmatrix}2-\lambda &1&b&3\\0&3-\lambda &-1&c\\0&0&2-\lambda &2\\0&0&0&3-\lambda \end{bmatrix}})=-(\lambda -3)^{2}(\lambda -2)^{2}=0$ Solve the equation above to get $\lambda _{1}=\lambda _{2}=3,\lambda _{3}=\lambda _{4}=2$ As similar to a, we need to make sure both eigenvalues have exactly 2 corresponding eigenvectors. When $\lambda =3,A-\lambda I={\begin{bmatrix}-1&1&b&3\\0&0&-1&c\\0&0&-1&2\\0&0&0&0\end{bmatrix}}$ In order to get 2 free variables, we need to set ${\frac {-1}{-1}}={\frac {c}{2}}$ Thus $c=2$ When $\lambda =2,A-\lambda I={\begin{bmatrix}0&1&b&3\\0&1&-1&c\\0&0&0&2\\0&0&0&1\end{bmatrix}}$ In order to get 2 free variables, we need to set ${\frac {1}{1}}={\frac {b}{-1}}$ Thus $b=-1$

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