Answer to Question #282043 in Linear Algebra for Peyt

Question #282043
"16\\begin{bmatrix}\n a & b \\\\\n c & d\n\\end{bmatrix}\\begin{matrix}\n a & b \\\\\n c & d\n\\end{matrix}\\begin{pmatrix}\n a & b \\\\\n c & d\n\\end{pmatrix}\\begin{matrix}\n a & b \\\\\n c & d\n\\end{matrix}"


1
Expert's answer
2021-12-23T06:09:13-0500

Solution:

"16\\begin{pmatrix}a&b\\\\ \\:c&d\\end{pmatrix}\\begin{pmatrix}a&b\\\\ \\:c&d\\end{pmatrix}\\begin{pmatrix}a&b\\\\ \\:c&d\\end{pmatrix}\\begin{pmatrix}a&b\\\\ \\:c&d\\end{pmatrix}\n\\\\=16\\begin{pmatrix}a^2+bc&ab+bd\\\\ ca+dc&bc+d^2\\end{pmatrix}\\begin{pmatrix}a^2+bc&ab+bd\\\\ ca+dc&bc+d^2\\end{pmatrix}\n\\\\=16\\begin{pmatrix}b^2c^2+3a^2bc+bcd^2+2abcd+a^4&2ab^2c+2b^2cd+a^3b+bd^3+abd^2+a^2bd\\\\ 2abc^2+2bc^2d+a^3c+cd^3+acd^2+a^2cd&b^2c^2+a^2bc+3bcd^2+2abcd+d^4\\end{pmatrix}"


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