anonymous
  • anonymous
**WILL GIVE MEDALS, SUPER EASY QUESTION** https://gyazo.com/93f26f1944e2f6f171354d917159af95
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jim_thompson5910
  • jim_thompson5910
based on these properties http://math.kennesaw.edu/~plaval/math3260/det2.pdf if we multiply any column by a constant k, then det(A) = k*det(A)
anonymous
  • anonymous
okay , how does this help me in this question?
anonymous
  • anonymous
is the first one 12? I think i got the first one, second is tricky

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jim_thompson5910
  • jim_thompson5910
\[ \det\left[ \begin{array}{ccc} a & 6 & d\\ b & 6 & e\\ c & 6 & f\\ \end{array} \right] = \det\left[ \begin{array}{ccc} a & 6*1 & d\\ b & 6*1 & e\\ c & 6*1 & f\\ \end{array} \right] \] \[ \det\left[ \begin{array}{ccc} a & 6 & d\\ b & 6 & e\\ c & 6 & f\\ \end{array} \right] = 6*\det\left[ \begin{array}{ccc} a & 1 & d\\ b & 1 & e\\ c & 1 & f\\ \end{array} \right] \] \[ \det\left[ \begin{array}{ccc} a & 6 & d\\ b & 6 & e\\ c & 6 & f\\ \end{array} \right] = 6*2 \] \[ \det\left[ \begin{array}{ccc} a & 6 & d\\ b & 6 & e\\ c & 6 & f\\ \end{array} \right] = 12 \]
anonymous
  • anonymous
yes i got it! can you help me with the second one?
anonymous
  • anonymous
i can't seem to get the second one for some reason
jim_thompson5910
  • jim_thompson5910
still thinking on that one
anonymous
  • anonymous
i still can't figure it out lol
jim_thompson5910
  • jim_thompson5910
I think I might have something. One second
jim_thompson5910
  • jim_thompson5910
\[ \det\left[ \begin{array}{ccc} a & -4 & d\\ b & -7 & e\\ c & -10 & f\\ \end{array} \right] = \det\left[ \begin{array}{ccc} a & -3-1 & d\\ b & -6-1 & e\\ c & -9-1 & f\\ \end{array} \right] \] \[ \det\left[ \begin{array}{ccc} a & -4 & d\\ b & -7 & e\\ c & -10 & f\\ \end{array} \right] = \det\left[ \begin{array}{ccc} a & -3 & d\\ b & -6 & e\\ c & -9 & f\\ \end{array} \right] - \det\left[ \begin{array}{ccc} a & 1 & d\\ b & 1 & e\\ c & 1 & f\\ \end{array} \right] \] \[ \det\left[ \begin{array}{ccc} a & -4 & d\\ b & -7 & e\\ c & -10 & f\\ \end{array} \right] = \det\left[ \begin{array}{ccc} a & -3*1 & d\\ b & -3*2 & e\\ c & -3*3 & f\\ \end{array} \right] - \det\left[ \begin{array}{ccc} a & 1 & d\\ b & 1 & e\\ c & 1 & f\\ \end{array} \right] \] \[ \det\left[ \begin{array}{ccc} a & -4 & d\\ b & -7 & e\\ c & -10 & f\\ \end{array} \right] = -3*\det\left[ \begin{array}{ccc} a & 1 & d\\ b & 2 & e\\ c & 3 & f\\ \end{array} \right] - \det\left[ \begin{array}{ccc} a & 1 & d\\ b & 1 & e\\ c & 1 & f\\ \end{array} \right] \] \[ \det\left[ \begin{array}{ccc} a & -4 & d\\ b & -7 & e\\ c & -10 & f\\ \end{array} \right] = -3*(-5)-(2) \] \[ \det\left[ \begin{array}{ccc} a & -4 & d\\ b & -7 & e\\ c & -10 & f\\ \end{array} \right] = 13 \]
anonymous
  • anonymous
perfect, thanks man
jim_thompson5910
  • jim_thompson5910
no problem

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