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In-a-third-order-determinant-each-element-of-the-first-column-consists-of-sum-of-2-terms-each-element-of-the-second-column-consists-of-sum-of-3-terms-and-each-element-of-the-third-column-consists-of




Question Number 15158 by arnabpapu550@gmail.com last updated on 08/Jun/17
In a third order determinant, each  element of the first column consists of  sum of 2 terms, each element of the  second column consists of sum of 3 terms  and each element of the third column  consists of sum of 4 terms. Then it can  be decomposed into n determinants,  where n has the value  Ans= 24
$$\mathrm{In}\:\mathrm{a}\:\mathrm{third}\:\mathrm{order}\:\mathrm{determinant},\:\mathrm{each} \\ $$$$\mathrm{element}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{column}\:\mathrm{consists}\:\mathrm{of} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{2}\:\mathrm{terms},\:\mathrm{each}\:\mathrm{element}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{second}\:\mathrm{column}\:\mathrm{consists}\:\mathrm{of}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{3}\:\mathrm{terms} \\ $$$$\mathrm{and}\:\mathrm{each}\:\mathrm{element}\:\mathrm{of}\:\mathrm{the}\:\mathrm{third}\:\mathrm{column} \\ $$$$\mathrm{consists}\:\mathrm{of}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{4}\:\mathrm{terms}.\:\mathrm{Then}\:\mathrm{it}\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{decomposed}\:\mathrm{into}\:{n}\:\mathrm{determinants}, \\ $$$$\mathrm{where}\:{n}\:\mathrm{has}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{Ans}=\:\mathrm{24} \\ $$
Commented by 1kanika# last updated on 07/Jun/17
n+1 may be
$$\mathrm{n}+\mathrm{1}\:\mathrm{may}\:\mathrm{be} \\ $$
Commented by arnabpapu550@gmail.com last updated on 08/Jun/17
But its answer is 24
$$\mathrm{But}\:\mathrm{its}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{24} \\ $$
Commented by prakash jain last updated on 08/Jun/17
2×3×4  e.g     [((a+b),(c+d)),((e+f),(g+h)) ]   =  [(a,(c+d)),(e,(g+h)) ]+ [(b,(c+d)),(f,(g+h)) ]   =  [(a,c),(e,g) ]+ [(a,d),(e,h) ]+ [(b,c),(f,g) ]+ [(b,d),(f,h) ]
$$\mathrm{2}×\mathrm{3}×\mathrm{4} \\ $$$${e}.{g} \\ $$$$\:\:\begin{bmatrix}{{a}+{b}}&{{c}+{d}}\\{{e}+{f}}&{{g}+{h}}\end{bmatrix} \\ $$$$\:=\:\begin{bmatrix}{{a}}&{{c}+{d}}\\{{e}}&{{g}+{h}}\end{bmatrix}+\begin{bmatrix}{{b}}&{{c}+{d}}\\{{f}}&{{g}+{h}}\end{bmatrix} \\ $$$$\:=\:\begin{bmatrix}{{a}}&{{c}}\\{{e}}&{{g}}\end{bmatrix}+\begin{bmatrix}{{a}}&{{d}}\\{{e}}&{{h}}\end{bmatrix}+\begin{bmatrix}{{b}}&{{c}}\\{{f}}&{{g}}\end{bmatrix}+\begin{bmatrix}{{b}}&{{d}}\\{{f}}&{{h}}\end{bmatrix} \\ $$

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