Check whether or not the matrix
A=[1 2 3]
[0 2 3]
[0 0 3]
Is diagonalisable. If it is find a matrix P so that P^-1AP is a diagonal matrix. If A is not diagonalisable, obtain its minimal polynomial.
Check whether or not the matrix
"A=\\begin{pmatrix}\n 1 & 2 &3\\\\\n 0 &2&3\\\\0 &0&3\n\\end{pmatrix}"
Is diagonalisable. If it is find a matrix P so that P^-1AP is a diagonal matrix. If A is not diagonalisable, obtain its minimal polynomial.
find the charateristics polynomial
"f(\\lambda) =det(A-\\lambda\\Iota)=0"
"\\begin{vmatrix}\n 1-\\lambda & 2 &3\\\\\n 0 &2-\\lambda&3\\\\0 &0&3-\\lambda\n\\end{vmatrix} = (1-\\lambda)[(2-\\lambda)(3-\\lambda)]=0"
"\\lambda=1 , \\lambda-2, \\lambda=3"
since we have three eigen values, then the matrix is diagonalisable
when "\\lambda=1"
"\\begin{pmatrix}\n 0 & 2 &3\\\\\n 0 &1&3\\\\0 &0&2\n\\end{pmatrix} \\begin{pmatrix}\n x_1 \\\\\n x_2\\\\x_3 \n\\end{pmatrix}=0"
"x_2 +3x_3=0......(i)"
"x_2+3x_3=0.......(ii)"
"x_3=0"
"\\implies x_1=x_2=x_3=0"
when "\\lambda=2"
"\\begin{pmatrix}\n -1 & 2 &3\\\\\n 0 &0&3\\\\0 &0&1\n\\end{pmatrix} \\begin{pmatrix}\n x_1 \\\\\n x_2\\\\x_3 \n\\end{pmatrix}=0"
"-x_1+2x_2+3x_3=0"
"x_3 =0"
since "x_3=0\\implies 2x_2=x_1"
since there is no restriction placed on x1
x1=r(arbitrary) where can be any number
let r=2
then x1 = 2
x2 = 1
when "\\lambda=3"
"\\begin{pmatrix}\n -2 & 2 &3\\\\\n 0 &-1&3\\\\0 &0&0\n\\end{pmatrix} \\begin{pmatrix}\n x_1 \\\\\n x_2\\\\x_3 \n\\end{pmatrix}=0"
bringing out the augmented matrix
"\\begin{pmatrix}\n -2 & 2 &3&0\\\\\n 0 &-1&3&0\\\\0 &0&0&0\n\\end{pmatrix}R_1\\mapsto R_1-R_2" "\\begin{pmatrix}\n -2 &3&0&0\\\\\n 0 &-1&3&0\\\\0 &0&0&0\n\\end{pmatrix}"
"\\therefore -2x_1+3x_2=0\\implies x_1={3 \\over 2}x_2"
"-1x_2 +4x_3=0\\implies x_2 =4x_3"
since there is no restriction placed on x1
x2=s (arbitrary) where can be any number
let s=2
then x1 = 3
x3 = "{1 \\over 2}"
Matrix P ="\\begin{pmatrix}\n 0 & 2 &3\\\\\n 0 & 1&2 \\\\0&0&{1 \\over 2}\n\\end{pmatrix}"
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