Question

show the matrix [ r1= -9 4 4 ,r2= -8 3 4,r3= -16 8 7,] of order 3*3 is diagonalizable.Obtain the diagonalizable matrix P.

Answer 1

Answer on Question #50729 - Math - Linear Algebra

Show the matrix $[r1 = -944, r2 = -834, r3 = -1687]$ of order $3^{*}3$ is diagonalizable. Obtain the diagonalizable matrix $P$ .

Solution

We have

$$A = \left( \begin{array}{c c c} - 9 & 4 & 4 \\ - 8 & 3 & 4 \\ - 1 6 & 8 & 7 \end{array} \right).$$

Compose the characteristic equation

$$\begin{array}{l} \left| \begin{array}{c c c} - 9 - \lambda & 4 & 4 \\ - 8 & 3 - \lambda & 4 \\ - 1 6 & 8 & 7 - \lambda \end{array} \right| = 0 \to | s u b t r a c t t h e f i r s t r o w f r o m t h e s e c o n d o n e | \\ \rightarrow \left|\begin{array}{c c c}- 9 - \lambda&4&4\\ \lambda + 1&- \lambda - 1&0\\- 1 6&8&7 - \lambda\end{array}\right| = 0 \rightarrow | f a c t o r o u t (\lambda + 1) f r o m t h e s e c o n d r o w | \\ \rightarrow (\lambda + 1) \left|\begin{array}{c c c}- 9 - \lambda&4&4\\ 1&- 1&0\\- 1 6&8&7 - \lambda\end{array}\right| = 0 \rightarrow | a d d t h e f i r s t c o l u m n t o t h e s e c o n d o n e | \\ \rightarrow (\lambda + 1) \left|\begin{array}{c c c}- 9 - \lambda&- \lambda - 5&4\\ 1&0&0\\- 1 6&- 8&7 - \lambda\end{array}\right| = 0 \rightarrow | e x p a n d a l o n g t h e s e c o n d r o w | \\ \rightarrow - (\lambda + 1) \left|\begin{array}{c c c}- \lambda - 5&4\\- 8&7 - \lambda\end{array}\right| = 0 \rightarrow - (\lambda + 1) \big ((\lambda + 5) (\lambda - 7) + 3 2 \big) = 0 \\ \rightarrow (\lambda + 1) (\lambda^ {2} - 2 \lambda - 3 5 + 3 2) = 0 \rightarrow (\lambda + 1) (\lambda^ {2} - 2 \lambda - 3) = 0 \\ \rightarrow (\lambda + 1) (\lambda^ {2} - 2 \lambda + 1 - 4) = 0 \rightarrow (\lambda + 1) (\lambda - 1 - 2) (\lambda - 1 + 2) = 0 \rightarrow \\ \end{array}$$ $$(\lambda - 3) (\lambda + 1) ^ {2} = 0, \text {h e n c e} \lambda_ {1} = 3, \lambda_ {2} = - 1, \lambda_ {3} = - 1.$$

If $\lambda_1 = 3$ , then

$$\begin{array}{l} \left( \begin{array}{c c c} - 9 - 3 & 4 & 4 \\ - 8 & 3 - 3 & 4 \\ - 1 6 & 8 & 7 - 3 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \to - 8 x + 4 z = 0 \to z = 2 x; \\ - 1 2 x + 4 y + 4 z = 0 \rightarrow - 1 2 x + 4 y + 8 x = 0 \rightarrow x = y. \\ \end{array}$$ $$\left( \begin{array}{c} x \\ y \\ z \end{array} \right) = C \left( \begin{array}{c} 1 \\ 1 \\ 2 \end{array} \right), \text {w h e r e} C \neq 0.$$

If $\lambda_{2} = -1$ or $\lambda_3 = -1$ , then

$$\left( \begin{array}{c c c} - 9 - \lambda & 4 & 4 \\ - 8 & 3 - \lambda & 4 \\ - 1 6 & 8 & 7 - \lambda \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \to \left( \begin{array}{c c c} - 8 & 4 & 4 \\ - 8 & 4 & 4 \\ - 1 6 & 8 & 8 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \to y + z = 2 x.$$

There are 3 variables and 1 equation.

If $y = 2, z = 0$ , then $x = 1$ . If $y = 0, z = 2$ , then $x = 1$ .

Therefore

$$\binom {x} {y} = a \binom {1} {2} + b \binom {1} {0}, \text{where vectors} \binom {1} {2} \text{ and } \binom {1} {0} \text{ are linearly independent in } \mathbb {R} ^ {3}.$$

Thus, matrix $A$ is diagonalizable,

$$P ^ {- 1} A P = \left( \begin{array}{ccc} 3 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{array} \right),$$

where

$$P = \left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 0 \\ 2 & 0 & 2 \end{array} \right).$$

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