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Prove that $\det(kA) = k^n \det(A)$ for and ($n \times n$) matrix.

I have tried looking at this a couple of ways but can't figure out where to start. It's confusing to me since the equation for a determinant is such a weird summation.

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    $\begingroup$ Can you explain how you have defined determinant? $\endgroup$ Commented Apr 1, 2015 at 8:46
  • $\begingroup$ Also, do you know any other features of the determinant, e.g., if you have produce a matrix $A'$ by replacing a row of a matrix $A$ with $k$ times that row, do you know how $\det A$ and $\det A'$ are related? $\endgroup$ Commented Apr 1, 2015 at 8:48
  • $\begingroup$ You can prove this directly, just by using the definition of the determinant. $\endgroup$ Commented Apr 1, 2015 at 8:52

4 Answers 4

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A mnemonic using more advanced stuff, just to recover it from the few determinant rules I manage to remember is:

$Det(kA)=Det(kI_nA)=Det(kI_n)Det(A)=Det\left( \begin{bmatrix} k & 0 & ... & 0 \\ 0 & k & ... & 0 \\ ... & ... & ... & ...\\ 0 & 0 & ... & k \end{bmatrix} \right) Det(A) = k^n Det(A)$

Where I have used $Det(AB)=Det(A)Det(B)$ and a formula for the determinant of diagonal matrices.

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To elaborate on the above answer, the formula for the determinant is $$\operatorname{det}(A)=\sum_{\sigma\in S_n}sign(\sigma)\Pi_{i=1}^na_{i,\sigma_i}$$ so $$\operatorname{det}(kA)=\sum_{\sigma\in S_n}sign(\sigma)\Pi_{i=1}^nka_{i,\sigma_i}$$ $$=k^n\operatorname{det}(A)$$

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  • $\begingroup$ what if the matrix $A$ is orthogonal? I'm confused because if $A$ is orthogonal won't $kA$ also be orthogonal and hence $det(kA)=+1$ or $-1$ ? $\endgroup$
    – momo
    Commented Sep 18, 2018 at 11:54
  • $\begingroup$ @momo No: if $A$ is orthogonal then $A^T A = I$, so then $(kA)^T(kA) = k^2 I$. $\endgroup$ Commented Jan 21, 2019 at 15:59
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Note that the matrix $kA$ has elements $[kA]_{ij}=kA_{ij}$, where $A_{ij}$ are the elements of $A$. If we were to calculate the determinant expression formula, each term has the factor $k$ appearing $n$ times, where $n$ is the dimension of the matrix. You can factor these out from the entire expression, and you're left with something proportional to $\det A$.

Example:

\begin{align} \det \left(k\begin{bmatrix} A_{11} & A_{12}\\ A_{21} & A_{22}\\ \end{bmatrix}\right) =& \det \begin{bmatrix} kA_{11} & kA_{12}\\ kA_{21} & kA_{22}\\ \end{bmatrix}\\ = & kA_{11}kA_{22}-kA_{21}kA_{12}\\ = & k^2(A_{11}A_{22}-A_{21}A_{12})\\ =& k^2\det A \end{align}

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In the most simplest of case, say a matrix $A$ which is a $2 \times 2$ matrix, multiplying it by a constant $k$ gives the following general setup. \begin{equation} k \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ka & kb \\ kc & kd \end{pmatrix} \end{equation} The determinant is therefore (writing $B=kA$); \begin{eqnarray} \text{det}B &=& (ka)(kd)-(kb)(kc)\\ &=& k^{2}ad-k^{2}bc \\ &=& k^{2}(ad-bc) \\ &=& k^{2} \text{det}A \end{eqnarray} Notice the exponent on $k$ is of order $2$ for the $2 \times 2$ case; a conjecture is that that is would be $3$ for the $3 \times 3$ case. It is now a case of trying to expand that notion to the $n \times n$ case.

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