Let "A = \\begin{pmatrix}\n 1 & 0 & 3\\\\\n 0 & 4 & 5 \\\\\n 1 & 2 & 6\n\\end{pmatrix}"
What is the contactor of the entry A23 = 5
The co-factor of 5 is obtained by eliminating the row and column of 5. By doing this, we obtain the matrix below
"\\begin{pmatrix} \n1 & 0 \\\\ 1&2\n\\end{pmatrix}"
the determinant of the matrix above is "1(2) -0=2"
Next, we check the sign associated with matrix. The matrix sign used to write the co-factor of each elements is given below;
"\\begin{pmatrix} \n+ & - & +\\\\ -&+&-\\\\+&-&+\n\\end{pmatrix}"
Therefore the sign associated with 5 is -1. Hence the co-factor of 5 is "-1(2) = -2"
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