The original matrix is:

\begin{bmatrix} 1 &2 &0 &1 \\ 3 &3 &3 &9 \\ 1 & 4 & 1 &4 \\ 1 & 1 & 2 &0 \end{bmatrix}

Every time I reduced this to row echelon form, I got $\dfrac{1}{48}$ as the determinant when the actual determinant is $48$. Here are the row operations.

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The rows that I have highlighted are the ones that change the determinant since we are changing a row by a factor. All the other operations don't change the determinant and we never switch rows here.

So we get $\det(A) = \dfrac{-1}{3}\times\dfrac{1}{3}\times\dfrac{-3}{16}\times 1 \text{ (entries along diagonal)} = \dfrac{1}{48}$. But the actual determinant is $48$. Where have I went wrong? I did the row operations a couple of times but I am making the same mistake.

  • $\begingroup$ It is the opposite : entries along diagonal times scalar that you multiplied during the process = determinant. $\endgroup$ Mar 26, 2017 at 20:20
  • $\begingroup$ Multiplying a single row by a factor of $x$ changes the determinant by multiplication by $x$. So to get the original determinant, you'll have to divide by the factor, not multiply it. Hence you'll get 48. $\endgroup$ Mar 26, 2017 at 20:20
  • $\begingroup$ Note also that the determinant of a matrix where all entries are integers must be an integer, because it is obtained by taking only products and sums. $\endgroup$
    – mlc
    Mar 26, 2017 at 20:21
  • $\begingroup$ @SebastianSchulz why divide? If I have done those scalar multiplication to the row, shouldn't the determinant be the original determinant times these scalars? $\endgroup$
    – K Split X
    Mar 26, 2017 at 20:23
  • $\begingroup$ Rather than $\det(A) = \frac{-1}{3}\times\frac{1}{3}\times\frac{-3}{16}\times 1$, it should be $\det(A) \times\frac{-1}{3}\times\frac{1}{3}\times\frac{-3}{16}= 1$. To see why it goes that way and not the other, try with a simpler matrix, like $\left[\begin{smallmatrix}2&0\\0&1\end{smallmatrix}\right]$. $\endgroup$
    – Arthur
    Mar 26, 2017 at 20:31

1 Answer 1


Let's say we start with a matrix $A$ and perform a row operation of multiplying some row $R_i$ by a non-zero scalar $c$ and get a new matrix $A'$. What is the relation between $\det(A)$ and $\det(A')$? Since the determinant is a multilinear functions of the rows of $A$, we have

$$ \det(A') = c \det(A) \iff \det(A) = \frac{1}{c} \det(A'). $$

If we perform various row operations on $A$, the only operations which change the determinant are the multiplication operations. If we have performed multiplication operations by $c_1,\dots,c_k$ (and maybe other row operations) and arrived to the matrix $A'$, we have

$$ \det(A) = \frac{1}{c_1 \dots c_k} \det(A'). $$

In particular, if you reduced $A$ to an upper diagonal matrix $A'$ whose diagonal entries are $1$ then $\det(A') = 1$ and so

$$ \det(A) = \frac{1}{c_1 \dots c_k}. $$

  • 1
    $\begingroup$ Thank you very much for the explanation $\endgroup$
    – K Split X
    Mar 26, 2017 at 20:32

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