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Let matrix $A$ be defined as \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{pmatrix}

And $det(A)=d$

Let $B=$ \begin{pmatrix} 2a_{11} & 2a_{12} & \cdots & 2a_{1n} \\ 2a_{21} & 2a_{22} & \cdots & 2a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ 2a_{n1} & 2a_{n2} & \cdots & 2a_{nn}\\ \end{pmatrix} $$+$$

\begin{pmatrix} 3a_{11} & 3a_{12} & \cdots & 3a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{pmatrix}

Find $det(B)$ using elementary row operations. I think the answer is $3^{(n-1)}*5*d$, but I'm not sure.

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    $\begingroup$ Seems correct to me. $\endgroup$ Jun 6, 2015 at 3:33
  • $\begingroup$ I think it should be $2^{n-1}$ $\endgroup$
    – user208649
    Jun 6, 2015 at 3:35
  • $\begingroup$ You are correct. Determinants are $n$-linear functions so each row's scalar contributes to the entire determinant's scalar with multiplicity. $\endgroup$
    – Iceman
    Jun 6, 2015 at 3:42

1 Answer 1

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$$\begin{pmatrix} 2a_{11} & 2a_{12} & \cdots & 2a_{1n} \\ 2a_{21} & 2a_{22} & \cdots & 2a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ 2a_{n1} & 2a_{n2} & \cdots & 2a_{nn}\\ \end{pmatrix} + \begin{pmatrix} 3a_{11} & 3a_{12} & \cdots & 3a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{pmatrix}=\begin{pmatrix} 5a_{11} & 5a_{12} & \cdots & 5a_{1n} \\ 3a_{21} & 3a_{22} & \cdots & 3a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ 3a_{n1} & 3a_{n2} & \cdots & 3a_{nn}\\ \end{pmatrix}$$

Since the determinant is a $n$-linear function (considering that each line is an element of $\mathbb R^n$), it follows that, since the first line of $B$ is the first line of $A$ multiplied by $5$ and since all the other $n-1$ lines of $B$ are the same $n-1$ lines of $A$ multiplied by $3$, it follows that

$\det(B)=5.3^{n-1}.det(A)=5.3^{n-1}.d$

Your answer is correct.

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