Find $x$ in each of the following.
a. $(x-2)\binom{1}{4}=(10)$
b. $\left(\begin{array}{ll}2 & x\end{array}\right)\binom{-3}{2}=(4)$
c. $\left(\begin{array}{ll}x & 1\end{array}\right)\binom{4}{x}=(-30)$
d. $(x-3)\binom{x}{-1}=(12)$
e. $\left(\begin{array}{ll}2 & x\end{array}\right)\binom{x}{1}=(9)$
Answer (1)
Part a has no solution because it leads to a contradiction.
Part b involves matrix multiplication, leading to the equation − 6 + 2 x = 4 , which solves to x = 5 .
Part c involves matrix multiplication, leading to the equation 5 x = − 30 , which solves to x = − 6 .
Part d has no solution because it leads to a contradiction.
Part e involves matrix multiplication, leading to the equation 3 x = 9 , which solves to x = 3 .
The final answers are: a. No solution, b. x = 5 , c. x = − 6 , d. No solution, e. x = 3 .
Explanation
Problem Analysis We are given a set of equations involving binomial coefficients and variables, and our goal is to find the value of x in each equation. We will analyze each equation separately, keeping in mind the properties of binomial coefficients and matrix multiplication. The binomial coefficient ( k n ) is defined as k ! ( n − k )! n ! if n and k are non-negative integers and k ≤ n , and 0 otherwise. For matrix multiplication, we multiply corresponding elements and sum the results.
Solving for x in part a a. ( x − 2 ) ( 4 1 ) = ( 10 ) . Since 1"> 4 > 1 , ( 4 1 ) = 0 . Therefore, the equation becomes ( x − 2 ) ( 0 ) = 10 , which simplifies to 0 = 10 . This is a contradiction, so there is no solution for x in this case.
Solving for x in part b b. ( 2 x ) ( 2 − 3 ) = ( 4 ) . This is a matrix multiplication problem. The equation can be written as 2 ∗ ( − 3 ) + x ∗ 2 = 4 , which simplifies to − 6 + 2 x = 4 . Adding 6 to both sides gives 2 x = 10 , and dividing by 2 gives x = 5 .
Solving for x in part c c. ( x 1 ) ( x 4 ) = ( − 30 ) . This is a matrix multiplication problem. The equation can be written as x ∗ 4 + 1 ∗ x = − 30 , which simplifies to 4 x + x = − 30 , or 5 x = − 30 . Dividing by 5 gives x = − 6 .
Solving for x in part d d. ( x − 3 ) ( − 1 x ) = ( 12 ) . Since the bottom number of the binomial coefficient is negative, ( − 1 x ) = 0 . Therefore, the equation becomes ( x − 3 ) ( 0 ) = 12 , which simplifies to 0 = 12 . This is a contradiction, so there is no solution for x in this case.
Solving for x in part e e. ( 2 x ) ( 1 x ) = ( 9 ) . This is a matrix multiplication problem. The equation can be written as 2 ∗ x + x ∗ 1 = 9 , which simplifies to 2 x + x = 9 , or 3 x = 9 . Dividing by 3 gives x = 3 .
Final Answer Therefore, the solutions are: a. No solution b. x = 5 c. x = − 6 d. No solution e. x = 3
Examples
Understanding binomial coefficients and matrix operations is crucial in various fields such as computer science, statistics, and physics. For instance, in computer science, these concepts are used in algorithm design and analysis. In statistics, binomial coefficients are fundamental in probability calculations, such as determining the likelihood of a certain number of successes in a series of trials. Matrix operations are used in image processing and data analysis to transform and manipulate data efficiently. These mathematical tools provide a foundation for solving complex problems in diverse real-world applications.
