Problem 27
Question
Let the numbers \(2, b, c\) be in an A.P. and \(\mathrm{A}=\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & b & c \\ 4 & b^{2} & c^{2}\end{array}\right] .\) If \(\operatorname{det}(\mathrm{A}) ?[2,16]\), then \(c\) lies in the interval : \(\quad\) [April 08, 2019 (II)] (a) \([2,3)\) (b) \(\left(2+2^{34}, 4\right)\) (c) \([4,6]\) (d) \(\left[3,2+2^{34}\right]\)
Step-by-Step Solution
Verified Answer
Option (d), \([3, 2+2^{34}]\), is the correct interval for \(c\).
1Step 1: Understand Arithmetic Progression
Given the sequence of numbers \(2, b, c\) in an Arithmetic Progression (A.P.), we know that the difference between each consecutive term is constant. Thus, we can express this as \(b = 2 + d\) and \(c = b + d = 2 + 2d\), where \(d\) is the common difference.
2Step 2: Matrix Determinant Expression
We have the matrix \(\mathrm{A}=\begin{bmatrix}1 & 1 & 1 \ 2 & b & c \ 4 & b^2 & c^2\end{bmatrix}\). The determinant of a 3x3 matrix \([a, b, c; d, e, f; g, h, i]\) is given by \(a(ei-fh)-b(di-fg)+c(dh-eg)\). By expanding along the first row, calculate:
3Step 3: Determinant Calculation
The determinant is \(1 \cdot (bc^2 - cb^2) - 1 \cdot (2c^2 - 4c) + 1 \cdot (2b^2 - 4b)\). This simplifies by applying arithmetic progression relations: \(b = 2 + d\), \(c = 2 + 2d\) and plugging these values into \(det(A)\).
4Step 4: Simplify and Solve the Inequality
Substitute \(b\) and \(c\) into the expression and solve the inequality \(det(A)?[2,16]\), finding the condition \(2 \leq det(A) \leq 16\). Calculate and simplify terms until we find the valid range for \(d\), and consequently for \(c\), that satisfies the inequality.
5Step 5: Determine the Range of c
By solving the inequality constraints and back-substituting \(c = 2 + 2d\), we determine the values for \(c\) that fit within this range. Evaluate the result and compare it against the provided options.
Key Concepts
Matrix DeterminantInequality SolvingMatrix Algebra
Matrix Determinant
The concept of the matrix determinant is fundamental in linear algebra, often serving as a measure of the square matrix's properties. For a 3x3 matrix, the determinant can be calculated using a straightforward formula. Given a matrix: \[\mathrm{A} = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix},\]the determinant, denoted as \( \det(A) \), is computed as: \[\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg).\]
This operation involves multiplying specific pairs of elements to capture the matrix's scaling factor. It's noteworthy that if the determinant equals zero, the matrix is singular, meaning it doesn't have an inverse.
In an arithmetic progression exercise, plugging values derived from the A.P. terms into the determinant formula helps check conditions such as constraints given by inequalities, thereby solving the problem step by step.
This operation involves multiplying specific pairs of elements to capture the matrix's scaling factor. It's noteworthy that if the determinant equals zero, the matrix is singular, meaning it doesn't have an inverse.
In an arithmetic progression exercise, plugging values derived from the A.P. terms into the determinant formula helps check conditions such as constraints given by inequalities, thereby solving the problem step by step.
Inequality Solving
Inequality solving is the process of finding the range of values that satisfy an inequality involving mathematical expressions. In the context of matrices and arithmetic progressions, we often derive inequalities from conditions applied to determinants or algebraic expressions.
For instance, as seen in exercises involving matrix properties, we might need to find the determinant of a matrix and determine which values of certain variables meet given constraints – like fitting the determinant within a specified interval. Solving these inequalities typically involves:
For instance, as seen in exercises involving matrix properties, we might need to find the determinant of a matrix and determine which values of certain variables meet given constraints – like fitting the determinant within a specified interval. Solving these inequalities typically involves:
- Substituting the expressions derived from the problem into the inequality.
- Performing algebraic manipulations to simplify and re-arrange terms.
- Progressively narrowing down the range of variable values.
Matrix Algebra
Matrix algebra encompasses operations involving matrices that are crucial in linear algebra, such as addition, multiplication, and finding determinants. These operations enable us to handle problems across various fields, from engineering to physics. In the given arithmetic progression exercise, matrix algebra is utilized to manipulate the matrix and derive its determinant.
Matrices are typically rectangular arrays of numbers, and understanding their manipulation just like regular algebraic variables is essential. Some key operations include:
Matrices are typically rectangular arrays of numbers, and understanding their manipulation just like regular algebraic variables is essential. Some key operations include:
- Matrix Addition: Adding corresponding elements of matrices having the same dimensions.
- Matrix Multiplication: A bit more complex, involving row-by-column multiplication and summation of products.
- Determinant Calculation: As discussed earlier, a special number calculated from the square matrix.
Other exercises in this chapter
Problem 25
A value of \(\theta \in(0, \pi / 3)\), for which \(\left|\begin{array}{ccc}1+\cos ^{2} \theta & \sin ^{2} \theta & 4 \cos 6 \theta \\ \cos ^{2} \theta & 1+\sin
View solution Problem 26
Let \(\alpha\) and \(\beta\) be the roots of the equation \(x^{2}+x+1=0\). Then for \(y^{\prime \prime \prime} 0\) in \(R,\left|\begin{array}{ccc}y+1 & \alpha &
View solution Problem 28
If \(A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]\); then for all \(\theta \in\lef
View solution Problem 29
If \(\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|\) \(=(a+b+c)(x+a+b+c)^{2}, x \neq 0\) and \(a+b+c \ne
View solution