Menu Close

If-the-product-of-the-matrices-1-1-0-1-1-2-0-1-1-3-0-1-1-k-0-1-1-378-0-1-then-k-




Question Number 114879 by bobhans last updated on 21/Sep/20
If the product of the matrices    (((1   1)),((0   1)) ) (((1    2)),((0    1)) ) (((1     3)),((0     1)) )... (((1     k)),((0     1)) )= (((1    378)),((0        1)) )  then k =
$${If}\:{the}\:{product}\:{of}\:{the}\:{matrices}\: \\ $$$$\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{3}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{pmatrix}…\begin{pmatrix}{\mathrm{1}\:\:\:\:\:{k}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{378}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${then}\:{k}\:=\: \\ $$
Answered by john santu last updated on 21/Sep/20
(i) (((1   1)),((0   1)) ) (((1    2)),((0    1)) )= (((1   3)),((0   1)) )  (ii) (((1    1)),((0    1)) ) (((1     2)),((0     1)) ) (((1   3)),((0   1)) )= (((1    3)),((0   1)) ) (((1  3)),((0   1)) )= (((1   6)),((0   1)) )  (iii) (((1    6)),((0   1)) ) (((1    4)),((0    1)) )= (((1     10)),((0       1)) )  (iv) (((1    10)),((0      1)) ) (((1     5)),((0     1)) )= (((1     15)),((0       1)) )  therefore    (((1   1)),((0   1)) ) (((1    2)),((0    1)) ) (((1    3)),((0    1)) )... (((1     k)),((0    1)) )= (((1    378)),((0       1)) )   (((1     1+2+3+...+k)),((0                   1)) )= (((1     378)),((0        1)) )  ⇔ 1+2+3+...+k = 378       ((k(k+1))/2) = 378 ⇒k^2 +k−756=0      (k+27)(k−28) = 0     ⇔ k = 28 .
$$\left({i}\right)\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{3}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\left({ii}\right)\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{3}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{3}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\mathrm{3}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{6}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\left({iii}\right)\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{6}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{4}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{10}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\left({iv}\right)\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{10}}\\{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{5}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{15}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${therefore}\: \\ $$$$\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{3}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix}…\begin{pmatrix}{\mathrm{1}\:\:\:\:\:{k}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{378}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{k}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{378}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\Leftrightarrow\:\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{k}\:=\:\mathrm{378}\: \\ $$$$\:\:\:\:\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}\:=\:\mathrm{378}\:\Rightarrow{k}^{\mathrm{2}} +{k}−\mathrm{756}=\mathrm{0} \\ $$$$\:\:\:\:\left({k}+\mathrm{27}\right)\left({k}−\mathrm{28}\right)\:=\:\mathrm{0}\: \\ $$$$\:\:\Leftrightarrow\:{k}\:=\:\mathrm{28}\:. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *