Question #271606

Use Cayley hamilton theorem to find the values of the matrix

A^8-5A^7+7A^6-3A^5+8A^4-5A^3+8A^2-2A+I

Expert's answer

Given matrix isÂ "A = \\begin{bmatrix}\n 2 & 1 & 1 \\\\\n 0 & 1 & 0 \\\\\n 1 & 1 & 2\n\\end{bmatrix}"

Then according to Cayley Hamilton theorem,

"|A - \\lambda I| = 0"

So we will have,

"A - \\lambda I = \\begin{bmatrix}\n 2 & 1 & 1 \\\\\n 0 & 1 & 0 \\\\\n 1 & 1 & 2\n\\end{bmatrix} - \\lambda\\begin{bmatrix}\n 1 & 0 & 0 \\\\\n 0 & 1 & 0 \\\\\n 0 & 0 & 1\n\\end{bmatrix}"

"|A - \\lambda I| = \\begin{vmatrix}\n 2-\\lambda & 1 & 1 \\\\\n 0 & 1-\\lambda & 0 \\\\\n 1 & 1 & 2-\\lambda\n\\end{vmatrix} = 0"

Then equation will be

"\\lambda^3 - 5\\lambda ^2+7\\lambda -3=0"

According to Cayley Hamilton theorem,

Every matrix is the root of it's eigen matrix.

then,Â "A^3 - 5A^2+7A -3=0"Â (1)

Given equation isÂ "A^8 \u2212 5A^7 + 7A^6 \u2212 3A^5 + A^4 \u2212 5A^3 + 8A^2 \u2212 2A + I"

This equation can be written as,

"A^8 \u2212 5A^7 + 7A^6 \u2212 3A^5 + A^4 \u2212 5A^3 + 8A^2 \u2212 2A + I = (A^3 - 5A^2+7A -3)(A^5+A) + (A^2+A+I)"

From equation (1), above equation will be modified as,

"A^8 \u2212 5A^7 + 7A^6 \u2212 3A^5 + A^4 \u2212 5A^3 + 8A^2 \u2212 2A + I = A^2+A+I"

So putting value ofÂ "A^2,A,I"

we will get,

"A^8 \u2212 5A^7 + 7A^6 \u2212 3A^5 + A^4 \u2212 5A^3 + 8A^2 \u2212 2A + I "

"= \\begin{bmatrix}\n 5 & 4 & 4 \\\\\n 0 & 1 & 0 \\\\\n 4 & 4 & 5\n\\end{bmatrix} + \\begin{bmatrix}\n 2 & 1 & 1 \\\\\n 0 & 1 & 0 \\\\\n 1 & 1 & 2\n\\end{bmatrix} + \\begin{bmatrix}\n 1 & 0 & 0 \\\\\n 0 & 1 & 0 \\\\\n 0 & 0 & 1\n\\end{bmatrix} = \\begin{bmatrix}8&5&5\\\\ 0&3&0\\\\ 5&5&8\\end{bmatrix}"

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