State True or False for the statements of the following Exercise: If the determinant x+a&p+u&l+f +b&q+v&m+g +c&r+w&n+h splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.

True. Solution: Since, there are two terms in each element of the determinant x+a&p+u&l+f +b&q+v&m+g +c&r+w&n+h . Hence, we can break x+a&p+u&l+f +b&q+v&m+g +c&r+w&n+h as x&p&l +b&q+v&m+g +c&r+w&n+h + a&u&f +b&q+v&m+g +c&r+w&n+h . Again, we can break x&p&l +b&q+v&m+g +c&r+w&n+h + a&u&f +b&q+v&m+g +c&r+w&n+h as = x&p&l &q&m +c&r+w&n+h + x&p&l &v&g +c&r+w&n+h + a&u&f &q&m +c&r+w&n+h + a&u&f &v&g +c&r+w&n+h Similarly, we can split these 4 determinants in 8 determinants by splitting each one in two determinants further. So, given statement is true.

State True or False for the statements of the following Exercise:…

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Question

State True or False for the statements of the following Exercise:
If the determinant $\begin{vmatrix}\text{x}+\text{a}&\text{p}+\text{u}&\text{l}+\text{f}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}$ splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.

✓

Answer

True.
Solution:
Since, there are two terms in each element of the determinant $\begin{vmatrix}\text{x}+\text{a}&\text{p}+\text{u}&\text{l}+\text{f}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}.$
Hence, we can break $\begin{vmatrix}\text{x}+\text{a}&\text{p}+\text{u}&\text{l}+\text{f}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}$ as $\begin{vmatrix}\text{x}&\text{p}&\text{l}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}+\begin{vmatrix}\text{a}&\text{u}&\text{f}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}.$
Again, we can break $\begin{vmatrix}\text{x}&\text{p}&\text{l}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}+\begin{vmatrix}\text{a}&\text{u}&\text{f}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}$ as
$=\begin{vmatrix}\text{x}&\text{p}&\text{l}\\\text{y}&\text{q}&\text{m}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}+\begin{vmatrix}\text{x}&\text{p}&\text{l}\\\text{b}&\text{v}&\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}+\begin{vmatrix}\text{a}&\text{u}&\text{f}\\\text{y}&\text{q}&\text{m}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}+\begin{vmatrix}\text{a}&\text{u}&\text{f}\\\text{b}&\text{v}&\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}$
Similarly, we can split these 4 determinants in 8 determinants by splitting each one in two determinants further. So, given statement is true.

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