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Question Number 174220 by Mastermind last updated on 27/Jul/22

Commented by MJS_new last updated on 27/Jul/22

−21

$$−\mathrm{21} \\ $$

Commented by Mastermind last updated on 27/Jul/22

Show workings!

$$\mathrm{Show}\:\mathrm{workings}! \\ $$$$ \\ $$

Commented by MJS_new last updated on 27/Jul/22

use formula

$$\mathrm{use}\:\mathrm{formula} \\ $$

Commented by Mastermind last updated on 27/Jul/22

If you want to help, help! If we can do it, i won′t posted it here

$$\mathrm{If}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{help},\:\mathrm{help}! \\ $$$$\mathrm{If}\:\mathrm{we}\:\mathrm{can}\:\mathrm{do}\:\mathrm{it},\:\mathrm{i}\:\mathrm{won}'\mathrm{t}\:\mathrm{posted}\:\mathrm{it}\:\mathrm{here} \\ $$

Commented by MJS_new last updated on 27/Jul/22

det [(M) ]= determinant ((M)) determinant ((a,b),(c,d))=ad−bc determinant ((a,b,c),(d,e,f),(g,h,i))=aei+bfg+cdh−(afh+bdi+ceg)= =a(ei−fh)−b(di−fg)+c(dh−eg)= =a determinant ((e,f),(h,i))−b determinant ((d,f),(g,i))+c determinant ((d,e),(g,h)) same principle here: determinant ((a,b,c,d),(e,f,g,h),(i,j,k,l),(m,n,o,p))=a determinant ((f,g,h),(j,k,l),(n,o,p))−b determinant ((e,g,h),(i,k,l),(m,o,p))+c determinant ((e,f,h),(i,j,l),(m,n,p))−d determinant ((e,f,g),(i,j,k),(m,n,o))

$$\mathrm{det}\:\begin{bmatrix}{{M}}\end{bmatrix}=\begin{vmatrix}{{M}}\end{vmatrix} \\ $$$$\begin{vmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{vmatrix}={ad}−{bc} \\ $$$$ \\ $$$$\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{d}}&{{e}}&{{f}}\\{{g}}&{{h}}&{{i}}\end{vmatrix}={aei}+{bfg}+{cdh}−\left({afh}+{bdi}+{ceg}\right)= \\ $$$$\:\:\:\:\:={a}\left({ei}−{fh}\right)−{b}\left({di}−{fg}\right)+{c}\left({dh}−{eg}\right)= \\ $$$$\:\:\:\:\:={a}\begin{vmatrix}{{e}}&{{f}}\\{{h}}&{{i}}\end{vmatrix}−{b}\begin{vmatrix}{{d}}&{{f}}\\{{g}}&{{i}}\end{vmatrix}+{c}\begin{vmatrix}{{d}}&{{e}}\\{{g}}&{{h}}\end{vmatrix} \\ $$$$ \\ $$$$\mathrm{same}\:\mathrm{principle}\:\mathrm{here}: \\ $$$$\begin{vmatrix}{{a}}&{{b}}&{{c}}&{{d}}\\{{e}}&{{f}}&{{g}}&{{h}}\\{{i}}&{{j}}&{{k}}&{{l}}\\{{m}}&{{n}}&{{o}}&{{p}}\end{vmatrix}={a}\begin{vmatrix}{{f}}&{{g}}&{{h}}\\{{j}}&{{k}}&{{l}}\\{{n}}&{{o}}&{{p}}\end{vmatrix}−{b}\begin{vmatrix}{{e}}&{{g}}&{{h}}\\{{i}}&{{k}}&{{l}}\\{{m}}&{{o}}&{{p}}\end{vmatrix}+{c}\begin{vmatrix}{{e}}&{{f}}&{{h}}\\{{i}}&{{j}}&{{l}}\\{{m}}&{{n}}&{{p}}\end{vmatrix}−{d}\begin{vmatrix}{{e}}&{{f}}&{{g}}\\{{i}}&{{j}}&{{k}}\\{{m}}&{{n}}&{{o}}\end{vmatrix} \\ $$

Commented by Mastermind last updated on 27/Jul/22

Thanks man

$$\mathrm{Thanks}\:\mathrm{man} \\ $$

Commented by MJS_new last updated on 28/Jul/22

you′re welcome. this principle works for any matrix size n×n

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}.\:\mathrm{this}\:\mathrm{principle}\:\mathrm{works}\:\mathrm{for}\:\mathrm{any} \\ $$$$\mathrm{matrix}\:\mathrm{size}\:{n}×{n} \\ $$

Answered by Rasheed.Sindhi last updated on 27/Jul/22

Using Properties determinant ((2,1,(-2),( 3)),(3,2,(-1),( 2)),(3,3,( 2),(-3)),(0,4,( 3),( 1))) determinant (((-1),(-1),( -1),( 1)),(3,2,(-1),( 2)),(3,3,( 2),(-3)),(0,4,( 3),( 1))) R1−R2 determinant ((0,0,( 0),( 1)),(5,4,( 1),( 2)),(0,0,( -1),(-3)),(1,5,( 4),( 1)))C1+C4_(C2+C4_(C3+C4) ) −(1) determinant ((5,4,( 1)),(0,0,( -1)),(1,5,( 4))) −1(−(−1) ) determinant ((5,4,(-3)),(0,0,( 1)),(1,5,( 2))) =− determinant ((5,4),(1,5)) −(5×5−4×1)=−21

$$\mathrm{Using}\:\mathrm{Properties} \\ $$$$\begin{vmatrix}{\mathrm{2}}&{\mathrm{1}}&{-\mathrm{2}}&{\:\:\mathrm{3}}\\{\mathrm{3}}&{\mathrm{2}}&{-\mathrm{1}}&{\:\:\mathrm{2}}\\{\mathrm{3}}&{\mathrm{3}}&{\:\:\mathrm{2}}&{-\mathrm{3}}\\{\mathrm{0}}&{\mathrm{4}}&{\:\:\mathrm{3}}&{\:\:\mathrm{1}}\end{vmatrix}\: \\ $$$$\begin{vmatrix}{-\mathrm{1}}&{-\mathrm{1}}&{\:\:-\mathrm{1}}&{\:\:\mathrm{1}}\\{\mathrm{3}}&{\mathrm{2}}&{-\mathrm{1}}&{\:\:\mathrm{2}}\\{\mathrm{3}}&{\mathrm{3}}&{\:\:\mathrm{2}}&{-\mathrm{3}}\\{\mathrm{0}}&{\mathrm{4}}&{\:\:\mathrm{3}}&{\:\:\mathrm{1}}\end{vmatrix}\:\mathrm{R1}−\mathrm{R2} \\ $$$$\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\:\:\mathrm{0}}&{\:\:\mathrm{1}}\\{\mathrm{5}}&{\mathrm{4}}&{\:\:\mathrm{1}}&{\:\:\mathrm{2}}\\{\mathrm{0}}&{\mathrm{0}}&{\:\:-\mathrm{1}}&{-\mathrm{3}}\\{\mathrm{1}}&{\mathrm{5}}&{\:\:\mathrm{4}}&{\:\:\mathrm{1}}\end{vmatrix}\underset{\underset{\mathrm{C3}+\mathrm{C4}} {\mathrm{C2}+\mathrm{C4}}} {\mathrm{C1}+\mathrm{C4}} \\ $$$$−\left(\mathrm{1}\right)\begin{vmatrix}{\mathrm{5}}&{\mathrm{4}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}&{\:\:-\mathrm{1}}\\{\mathrm{1}}&{\mathrm{5}}&{\:\:\:\:\mathrm{4}}\end{vmatrix}\: \\ $$$$−\mathrm{1}\left(−\left(−\mathrm{1}\right)\:\right)\begin{vmatrix}{\mathrm{5}}&{\mathrm{4}}&{-\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}&{\:\:\mathrm{1}}\\{\mathrm{1}}&{\mathrm{5}}&{\:\:\mathrm{2}}\end{vmatrix}\:\: \\ $$$$=−\begin{vmatrix}{\mathrm{5}}&{\mathrm{4}}\\{\mathrm{1}}&{\mathrm{5}}\end{vmatrix}\:\: \\ $$$$−\left(\mathrm{5}×\mathrm{5}−\mathrm{4}×\mathrm{1}\right)=−\mathrm{21} \\ $$

Commented by Mastermind last updated on 27/Jul/22

Thank you so much, God bless you

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$

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