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We know to get determinant in a matrix we use sarrus, laplace, x methods and so on

But question : WHAT is it, according to some pages is about permutations the products of element of the matrix , but WHAT IS IT ?

NIN
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    It is a single number that can tell you much about your matrix. – Randall Mar 04 '21 at 20:34
  • 3blue1brown explains his take on the determinant in [video 6](https://www.youtube.com/watch?v=Ip3X9LOh2dk) of his linear algebra series. I don't remember off the top of my head whether you need to see the preceeding 5, although if you have an hour to spare I do recommend you do that anyways. It conveys some pretty healthy ways to think about the basics of the subject. – Arthur Mar 04 '21 at 20:43
  • You'll likely be interested in the interpretation of a matrix determinant as the *signed hyper-volume* of a parallelopiped as described in the Answers to the proposed duplicate. That is closely related to the Jacobian appearing in multivariate change-of-variable integration formulas. But the determinant has many applications. – hardmath Mar 04 '21 at 20:49

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\begin{align} & \left[ \begin{array}{c} x \\ y \end{array} \right] \mapsto \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} ax+by \\ cx+dy \end{array} \right] \\[8pt] & \left[ \begin{array}{c} 1 \\ 0 \end{array} \right] \mapsto \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\left[ \begin{array}{c} 1 \\ 0 \end{array} \right] = \left[ \begin{array}{c} a \\ c \end{array} \right] \\[8pt] & \left[ \begin{array}{c} 0 \\ 1 \end{array} \right] \mapsto \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\left[ \begin{array}{c} 0 \\ 1 \end{array} \right] = \left[ \begin{array}{c} b \\ d \end{array} \right] \end{align}

Draw the arrows representing $\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]$ and $\left[ \begin{array}{c} 0 \\ 1 \end{array} \right].$ From the head of the arrow $\left[ \begin{array}{c} 1 \\ 0 \end{array} \right],$ draw another arrow parallel to $\left[ \begin{array}{c} 0 \\ 1 \end{array} \right]$ and from the head of the arrow $\left[ \begin{array}{c} 0 \\ 1 \end{array} \right]$ draw another parallel to $\left[ \begin{array}{c} 1 \\ 0 \end{array} \right].$ There you have a square whose area is $1.$

Then do the same with $\left[ \begin{array}{c} a \\ c \end{array} \right]$ and $\left[ \begin{array}{c} b \\ d \end{array} \right]$ and get a parallelogram. The area of that parallelogram is $\left|\det\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\right|.$ The determinant is positive if turning from $\left[ \begin{array}{c} a \\ c \end{array} \right]$ to $\left[ \begin{array}{c} b \\ d \end{array} \right]$ amounts to turning counterclockwise, and negative if clockwise. In other words, positive if the parallelogram has the same orientation as the square, and negative if the orientation is reversed.

Michael Hardy
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