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If $A$ is a $2\times2$ matrix, what is $\det(4A)$ in terms of $\det(A)$?

This seems trivial, but I'm not sure exactly what they are asking. I'm guessing I have some matrix $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ where I know $\det(A) = ad - bc$. So if they want to know what is $\det(4A)$ wouldn't it just be $4A = 4\begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}4a&4b\\4c&4d\end{bmatrix} = \det(A) = 16ad - 16bc$?

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    $\begingroup$ This is not really an answer, but your confusion seems more based in the wording of the question than in properties of determinants. To generalize: write $\mathsf E$ in terms of $\mathsf V$ means the expected answer is an equation $\mathsf E = \text{RHS}$, where the right-hand side is an expression in only the variable(s) $\mathsf V$. $\endgroup$
    – lynn
    Commented Jul 14, 2016 at 10:25

7 Answers 7

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Let's consider your equation $$4A = 4\begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}4a&4b\\4c&4d\end{bmatrix} = \det(A) = 16ad - 16bc$$

The first three objects in this are $2\times 2$ matrices while the last 2 are numbers. So clearly these can't all be equal. What you really want to say here is

$$\color{red}{\det(}4A\color{red}{)} = \color{red}{\det\left(\color{black}{4\begin{bmatrix}a&b\\c&d\end{bmatrix}}\right)} = \color{red}{\begin{vmatrix}\color{black}{4a}&\color{black}{4b}\\\color{black}{4c}&\color{black}{4d}\end{vmatrix}} = 16ad - 16bc$$

Then just complete the logic with a final $$=16\det(A)$$ and you're done.

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    $\begingroup$ And, in general, for $n\times n$ matrix $A$, $$\det(kA) = k^n\det A$$ This is because a determinant is a sum of $n$–term products of the matrix items, so multipying the matrix by a numer $k$ causes each of items be multiplied by $k$, hence each term of the sum gets a $k^n$ factor. $\endgroup$
    – CiaPan
    Commented Jul 13, 2016 at 8:12
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More generally, if $A$ is $n \times n$, $\det(cA) = c^n\det(A)$ for any scalar $c$. This is because the determinant is a multilinear function of its columns.

In fact, one can define the determinant as the unique function from $n \times n$ matrices to scalars that is $n$-linear alternating in the columns, and takes the value $1$ for the identity matrix.

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    $\begingroup$ @Yusha see Leibniz formula of determinant if you want the detail. $\endgroup$ Commented Jul 13, 2016 at 6:43
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    $\begingroup$ @Yusha Do you know the "elementary row operation" that scales one single row of the matrix by a factor $c$? And know how the determinant behaves under this elementary row operation? Going from a matrix $A$ to the matrix $cA$ is exactly the same as performing such row scalings $n$ times. So you can see the formula $\det(cA) = c^n\det(A)$ in this perspective. There is an interesting special case, $$\det(-I) = (-1)^n,$$ where I have taken $c=-1$ and $A=I$, the identity matrix. This shows that reflection in the origin is a rotation (i.e. orientation-preserving) if and only if $n$ is even. $\endgroup$ Commented Jul 13, 2016 at 11:41
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In particular, $\det(4A) = 4^2 \det(A)$.

Another way to think about it is to remember that the determinant of an $n \times n$ matrix gives the (signed) volume of the $n$-dimensional box enclosed by its column vectors. Increasing each side length by a factor of $k$ will increase the volume by a factor of $k^n$.

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  • $\begingroup$ Could you please explain the line enclosed by its column vectors in your answer. $\endgroup$
    – sameerkn
    Commented Jul 13, 2016 at 10:08
  • $\begingroup$ @sameerkn, see the pictures in [ askamathematician.com/2013/05/… ], for example. $\endgroup$
    – Vectornaut
    Commented Jul 13, 2016 at 19:49
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determinant is a multilinear form so for all matrix $A\in M_n(\mathbb{R})$ and for all $\alpha\in \mathbb{R}$ $$ \det(\alpha A)=\alpha^n \det(A) $$

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It is indeed trivial. With $A$ as you have it, $\det(A)=ad-bc$.

Now, $\det(4A)=\det(\begin{bmatrix} 4a & 4b\\ 4c & 4d \end{bmatrix})$=$16ad-16bc=16(ad-bc)=16*\det(A)$.

Alternatively, just use the formula: if $A$ is a real valued $n \times n$ matrix, then $\det(cA) = c^n\det(A)$ $\forall$ $c\in \mathbb{R}$.

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Seeing the determinant as the volume of the unit cube $Q$ under the map $A$, one sees that the composition with the multiplication $x\mapsto kx$ will dialate the image $AQ$ by a factor $k$.

But as the volume scales like $k^n$ in $n-$dimensional space, you get that $$\det(kA)=vol(kA(Q))=k^n vol(A(Q))=k^n\det(A).$$

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Another view of the multilinear aspect of the determinant of a square matrix $A$ is the "recursive form" involving minors. If the matrix is $n\times n$, one can compute determinant with minors stemming from $(n-1)\times (n-1)$ matrices:

$$|A| = \sum (-1)^{i+j} a_{ij} M_{ij}\,.$$

The determinant of a $1\times 1$ matrix (a scalar) grows linearly with the factor $\lambda$ applied to the (unique) matrix element. Hence the determinant of a $2\times 2$ matrix grows twice faster, since you get one $\lambda$ for the $a_{ij}$ terms (for instance $a$ and $c$ for you), and one for the $M_{ij}$ terms, which derive from dimension $1\times 1$ matrices (for instance $b$ and $d$ for you), as you wrote with $\lambda=4$.

For a $3\times 3$ matrix, you get one $\lambda$ for the $a_{ij}$ terms, and $\lambda^2$ for the $M_{ij}$ terms, coming from dimension $2\times 2$ objects. More generally i, you get one $\lambda$ for the $a_{ij}$ terms, and $\lambda^{n-1}$ for the $M_{ij}$ terms, which are computed from dimension $(n-1)\times (n-1)$ matrices, hence $\lambda^n $ total.

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