For what value of $n$ is the determinant
$[\begin{vmatrix}C\binom{9}{4} & C\binom{9}{3} & C\binom{10}{n-2} \\C\binom{11}{6} & C\binom{11}{5} & C\binom{12}{n} \\C\binom{m}{7} & C\binom{m}{6} & C\binom{m+1}{n+1} \end{vmatrix} = 0]$
for every $m > n$?
Determinant with Binomial Coefficients
The problem asks for the value of '$n$' that makes a specific determinant equal to zero for all '$m$' greater than '$n$'. The determinant involves binomial coefficients, denoted as $C(p, k)$ or $\binom{p}{k}$.
The determinant in question is:
$ \begin{vmatrix} \binom{9}{4} & \binom{9}{3} & \binom{10}{n-2} \\ \binom{11}{6} & \binom{11}{5} & \binom{12}{n} \\ \binom{m}{7} & \binom{m}{6} & \binom{m+1}{n+1} \end{vmatrix} = 0 $A fundamental property of determinants is that they are zero if and only if the columns (or rows) of the matrix are linearly dependent. This means one column can be expressed as a linear combination of the others. Let the columns be $C_1$, $C_2$, and $C_3$. We are looking for a condition where $C_3 = aC_1 + bC_2$ for some constants $a$ and $b$.
Binomial Coefficient Identity
To analyze the determinant, we can use key properties of binomial coefficients. Pascal's Identity is particularly relevant here. It states: $ \binom{p}{k} + \binom{p}{k-1} = \binom{p+1}{k} $ This identity shows that the sum of two adjacent binomial coefficients in one row equals a binomial coefficient in the next row.
Determinant Structure Analysis
Let's examine the structure of the given determinant by looking at the indices in each column:
- Column 1: $\binom{9}{4}$, $\binom{11}{6}$, $\binom{m}{7}$
- Column 2: $\binom{9}{3}$, $\binom{11}{5}$, $\binom{m}{6}$
- Column 3: $\binom{10}{n-2}$, $\binom{12}{n}$, $\binom{m+1}{n+1}$
We hypothesize that the determinant might be zero if the third column is the sum of the first two columns, i.e., $C_3 = C_1 + C_2$. This implies that for each row, the element in the third column should equal the sum of the elements in the first and second columns. Let's check this using Pascal's Identity.
Verifying n=6
The options provided are $n=4, 5, 6, 7$. Let's test the value $n=6$. If $n=6$, the third column becomes: $ \begin{pmatrix} \binom{10}{6-2} \\ \binom{12}{6} \\ \binom{m+1}{6+1} \end{pmatrix} = \begin{pmatrix} \binom{10}{4} \\ \binom{12}{6} \\ \binom{m+1}{7} \end{pmatrix} $ Now, let's see if this column is the sum of the first two columns ($C_1 + C_2$) for $n=6$:
- Row 1 Check: Is $\binom{10}{4} = \binom{9}{4} + \binom{9}{3}$? Yes, according to Pascal's Identity, $\binom{9}{4} + \binom{9}{3} = \binom{9+1}{4} = \binom{10}{4}$.
- Row 2 Check: Is $\binom{12}{6} = \binom{11}{6} + \binom{11}{5}$? Yes, by Pascal's Identity, $\binom{11}{6} + \binom{11}{5} = \binom{11+1}{6} = \binom{12}{6}$.
- Row 3 Check: Is $\binom{m+1}{7} = \binom{m}{7} + \binom{m}{6}$? Yes, Pascal's Identity states $\binom{m}{7} + \binom{m}{6} = \binom{m+1}{7}$. This holds true for $m \ge 7$. The problem states $m > n$, which means $m > 6$. Therefore, $m \ge 7$, satisfying the condition for Pascal's Identity.
Since $n=6$ makes the third column the sum of the first two columns ($C_3 = C_1 + C_2$), the columns are linearly dependent. This condition ensures the determinant is equal to 0 for all $m > 6$.
Final Answer Determination
The value $n=6$ satisfies the condition that the determinant is zero for every $m > n$, because it leads to the linear dependence of the columns based on Pascal's Identity.
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