Question

Aratati ca det(B+2C)=detB  - detA pentru orice numar real b .
Daca este mult de scris , imi puteti da macar cateva indicatii cum sa procedez

Answer 1

    
$\displaystyle\\ A=\left(\begin{array}{ccc}1&2\\1&0\\\end{array}\right) \\\\ B=\left(\begin{array}{ccc}b&b\\0&1\\\end{array}\right) \\\\ C=\left(\begin{array}{ccc}1&0\\0&0\\\end{array}\right) \\\\ B+2C = \left(\begin{array}{ccc}b&b\\0&1\\\end{array}\right) + 2\times \left(\begin{array}{ccc}1&0\\0&0\\\end{array}\right) = \\\\ = \left(\begin{array}{ccc}b&b\\0&1\\\end{array}\right) +\left(\begin{array}{ccc}2\times1&2\times0\\2\times0&2\times0\\\end{array}\right)=$


[tex]\displaystyle\\ = \left(\begin{array}{ccc}b&b\\0&1\\\end{array}\right) + \left(\begin{array}{ccc}2&0\\0&0\\\end{array}\right) =\\\\ = \left(\begin{array}{ccc}b&b\\0&1\\\end{array}\right)+\left(\begin{array}{ccc}2&0\\0&0\\\end{array}\right) =\\\\ = \left(\begin{array}{ccc}b+2&b+0\\0+0&1+0\\\end{array}\right) =\\\\ = \left(\begin{array}{ccc}b+2&b\\0&1\\\end{array}\right)\\\\ \det(B+2C)= \left|\begin{array}{ccc}b+2&b\\0&1\\\end{array}\right| = (b+2)\times 1- b\times 0 = \boxed{\bf b+2}[/tex]


[tex]\displaystyle\\ \det B - \det A=\left|\begin{array}{ccc}b&b\\0&1\\\end{array}\right| -\left|\begin{array}{ccc}1&2\\1&0\\\end{array}\right| = \\\\ (b\times1 - b\times 0)-(1\times 0-2\times 1)=\\\\ =(b-0)-(0 -2) =b -(-2) = \boxed{\bf b+2}\\\\ \Longrightarrow~~\det(B+2C)=\det B - \det A = \boxed{\boxed{\bf b+2}}[/tex]