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how can I show that the determinant divides by 13 without computing it? im given that each row is a multiple of 13.

1 2 7 4
5 9 4 1
1 6 2 5 
3 1 3 3

matrix 4x4

user113977
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3 Answers3

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1274=13*98 = 1000*1 + 100*2 + 10*7 + 4 5941=13*457= 1000*5 + 100*9 + 10*4 + 1 etc. Add to fourth column first one multiplied by 1000, second one multiplied by 100, third - by 10 and factor out 13 from the 4-th column.

Hrackadont
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You can use the property that, for constant $c$, and $M$ an $n \times n$ matrix , Det(cM)=$c^n Det(M)$.

user99680
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Call the $4\times 4$ matrix you gave $A$. Then $\det{(13A)} = 13^4\det(A)$, which is divisible by $13$.

user112167
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  • what did you use here? I assume its something new ive never seen before. if u could explain I would appreciate it – user113977 Dec 06 '13 at 22:44
  • I have used the fact that for any $n \times n$-matrix the determant of $c$ times a matrix $A$ is given by $c^n \det(A)$, which follows from the properties of the determinant. Most of the time this identity is given somewhere, if not have you got the identity that if you scale a row by a factor a factor $c$ that you will get $\det A=c\det(B)$, where $B$ is the matrix with the scaled row? – user112167 Dec 06 '13 at 22:49
  • no, i just started learning determinants recently. how can you say that the matrix is 13 times A? each number in column/row is a single digit number, they just told me that each row is a number that can be divided by 13 – user113977 Dec 06 '13 at 23:02
  • The probably mean that every number in every row can be divided by $13$. So for every row we can factor out a $13$. Because we have $4$ rows we get $13^4 \det{A}$. If you don't understand this, which rules you do know for determinants? – user112167 Dec 06 '13 at 23:12
  • now I get it, thanks man, I appreciate your patience – user113977 Dec 06 '13 at 23:19