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Consider scalars $a,b,c,d,e,f$ such that

$\det\left( \begin{array}{ccc} a & 1 & d\\ b & 1 & e\\ c & 1 & f\\ \end{array} \right) = 7$ and $\det\left( \begin{array}{ccc} a & 1 & d\\ b & 2 & e\\ c & 3 & f\\ \end{array} \right) = 11$

What is $\det\left( \begin{array}{ccc} a & 3 & d\\ b & 3 & e\\ c & 3 & f\\ \end{array} \right)$ and $\det\left( \begin{array}{ccc} a & 3 & d\\ b & 4 & e\\ c & 5 & f\\ \end{array} \right)$?

For the first one, I was thinking that the determinant would be $3\cdot 7 = 21$, and I think the determinant of the second one would still just be $11$ (since it's just a multiple of the second determinant matrix). Does this seem right? How can I solve these kinds of problems?

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  • $\begingroup$ I think that the second one will be $11+2\cdot7=11+14=25$... $\endgroup$ – Larara Nov 26 '14 at 3:23
  • $\begingroup$ @Larara What makes you think that? $\endgroup$ – Caleb Nov 26 '14 at 3:26
  • $\begingroup$ Well, if you use Laplace's formula you'll see that, for example, the first term (that multiplying $a$), will be $a\cdot(4\cdot f - 5\cdot e)=a\cdot((2+2)\cdot f - (3+2)\cdot e)=a\cdot(2\cdot f + 3\cdot e) + 2\cdot a\cdot(f-e)$, and the first term in the previous sum is the term multiplying $a$ in the matrix whose determinant is 11; and the second term in the previous sum is 2 times the term multiplying $a$ in matrix whose determinant is 7. And the terms multiplying $b$ and $c$ will be analogous. $\endgroup$ – Larara Nov 26 '14 at 3:38
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Givan an $n \times n$ matrix $A = (a_1, a_2, \ldots, a_{n})$, where $a_{i}$ are column vectors. There are identities $$det(a_1, \ldots, ca_i, \ldots, a_n) = cdet(A)$$ and $$det(a_1, \ldots, a_{i-1}, a_i + v, a_{i+1} \ldots, a_n) = det(A) + det(a_1, \ldots, a_{i-1}, v, a_{i+1}, \ldots, a_n)$$, where $v$ is a column vector.

Therefore, returning to your questions, $$det\begin{pmatrix} a & 3 & d\\ b & 3 & e\\ c & 3 & f \end{pmatrix} = 3 \cdot 7 = 21$$ and $$det\begin{pmatrix} a & 3 & d\\ b & 4 & e\\ c & 5 & f \end{pmatrix} = 2 \cdot 7 + 11 = 25.$$

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hint: $\left(\begin{array}{l} 3 \cr 4 \cr 5\end{array} \right) = 2 \left( \begin{array}{l} 1 \cr 1 \cr 1 \end{array} \right) + \left( \begin{array}{l} 1 \cr 2 \cr 3\end{array}\right).$

determinant is a linear function of every column.

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