Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that A

A matrix is a rectangular array or table of numbers, symbols, or expressions that are arranged in rows and columns to represent a mathematical object or a property of such an object in mathematics.
For instance, consider a matrix with two rows and three columns.

To find an invertible matrix P and a diagonal matrix D such that A = PDP⁻¹:

Let [tex]\lambda[/tex] be an eigenvalue of A. Then:

[tex]\begin{aligned}0 &=|A-\lambda I| \\&=\left|\begin{array}{ccc}-11-\lambda & 3 & -9 \\0 & -5-\lambda & 0 \\6 & -3 & 4-\lambda\end{array}\right| \\&=-(0)\left|\begin{array}{cc}3 & -9 \\-3 & 4-\lambda\end{array}\right|+(-5-\lambda)\left|\begin{array}{cc}-11-\lambda & -9 \\6 & 4-\lambda\end{array}\right|-(0)\left|\begin{array}{cc}-11-\lambda & 3 \\6 & -3\end{array}\right| \\&=(-5-\lambda)[(-11-\lambda)(4-\lambda)+54] \\&=-(5+\lambda)^{2}(2+\lambda)\end{aligned}[/tex]

Therefore, [tex]\lambda = -2[/tex] or [tex]\lambda = -5[/tex].

Let [tex]\mathbf{v}=\left[\begin{array}{l}a \\b \\c\end{array}\right][/tex] be an eigenvector associated with the eigenvalue [tex]\lambda[/tex].[tex]\lambda = -2[/tex]: From [tex](A+2 I) \mathbf{v}=\mathbf{0}[/tex] we obtain:

[tex]\left[\begin{array}{ccc}-9 & 3 & -9 \\0 & -3 & 0 \\6 & -3 & 6\end{array}\right]\left[\begin{array}{l}a \\b \\c\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right][/tex]

Applying elementary row operations to find the reduced echelon form of the coefficient matrix, we obtain:

[tex]\left[\begin{array}{ccc}-9 & 3 & -9 \\0 & -3 & 0 \\6 & -3 & 6\end{array}\right] \stackrel{\left(-\frac{1}{9}\right) R_{1} \rightarrow R_{1}}{\longrightarrow}\left[\begin{array}{rrr}1 & -\frac{1}{3} & 1 \\0 & -3 & 0 \\6 & -3 & 6\end{array}\right] \stackrel{(-6) R_{1}+R_{3} \rightarrow R_{3}}{\longrightarrow}\left[\begin{array}{rrr}1 & -\frac{1}{3} & 1 \\0 & -2 & 0 \\0 & -1 & 0\end{array}\right] \stackrel{\left(-\frac{1}{2}\right) R_{2} \rightarrow R_{2}}{\longrightarrow}[/tex]

[tex]\left[\begin{array}{ccc}1 & -\frac{1}{3} & 1 \\0 & 1 & 0 \\0 & -1 & 0\end{array}\right] \stackrel{\left(\frac{1}{3}\right) R_{2}+R_{1} \rightarrow R_{1}}{\stackrel{R_{2}+R_{3} \rightarrow R_{3}}{\longrightarrow}}\left[\begin{array}{ccc}1 & 0 & 1 \\0 & 1 & 0 \\0 & -1 & 0\end{array}\right] \stackrel{R_{2}+R_{3} \rightarrow R_{3}}{\longrightarrow}\left[\begin{array}{ccc}1 & 0 & 1 \\0 & 1 & 0 \\0 & 0 & 0\end{array}\right][/tex]

Hence, we have a + c = 0 and b = 0. Let c = -1 and a = 1.So, [tex]\left[\begin{array}{c}1 \\0 \\-1\end{array}\right][/tex] is an eigenvector associated with [tex]\lambda = -2[/tex].

[tex]\lambda = -5[/tex]: From [tex](A+5 I) \mathbf{v}=\mathbf{0}[/tex] we obtain:

[tex]\left[\begin{array}{ccc}-6 & 3 & -9 \\0 & 0 & 0 \\6 & -3 & 9\end{array}\right]\left[\begin{array}{l}a \\b \\c\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right][/tex]

Applying elementary row operations to find the reduced echelon form of the coefficient matrix, we obtain:

[tex]\left[\begin{array}{ccc}-6 & 3 & -9 \\0 & 0 & 0 \\6 & -3 & 9\end{array}\right] \stackrel{\left(-\frac{1}{6}\right) R_{1} \rightarrow R_{1}}{\longrightarrow}\left[\begin{array}{ccc}1 & -\frac{1}{2} & \frac{3}{2} \\0 & 0 & 0 \\6 & -3 & 9\end{array}\right] \stackrel{(-6) R_{1}+R_{3} \rightarrow R_{3}}{\longrightarrow}\left[\begin{array}{ccc}1 & -\frac{1}{2} & \frac{3}{2} \\0 & 0 & 0 \\0 & 0 & 0\end{array}\right][/tex]

Hence, we have a - 1/2b + 3/2c = 0. Let b = 2 and c = 0.Then a = 1. Let b = 0 and c = -2. Then a = 3. Therefore,

[tex]\left[\begin{array}{l}1 \\2 \\0\end{array}\right],\left[\begin{array}{c}3 \\0 \\-2\end{array}\right][/tex]

are two linearly independent vectors associated with [tex]\lambda = -5[/tex]

Matrices P and D: Let

[tex]P=\left[\begin{array}{ccc}1 & 1 & 3 \\0 & 2 & 0 \\-1 & 0 & -2\end{array}\right][/tex]

be the matrix whose columns are the eigenvectors obtained in the previous step. Set

[tex]Q=\left[\begin{array}{ccc}-2 & 0 & 0 \\0 & -5 & 0 \\0 & 0 & -5\end{array}\right][/tex]

to be the diagonal matrix whose diagonal entries are the eigenvalues.

Thus we have, A = PDP⁻¹.

Therefore an invertible matrix P and a diagonal matrix D such that A = PDP⁻¹ is shown.

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The correct question is given below:

Diagonalize matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that A = PDP⁻¹.

[tex]A=\left[\begin{array}{ccc}-11 & 3 & -9 \\0 & -5 & 0 \\6 & -3 & 4\end{array}\right][/tex]