Let \(\beta\) be a real number Consider the
\(matrix\ A=\begin{pmatrix}\beta & 0 & 1 \\2 & 1 & -2 \\3 & 1 & -2\end{pmatrix}If A^7-(\beta-1) A^6-\beta A^5\) is a singular matrix, then the value of \(9 \beta\) is _____
The given equation is:
\(|A|^5 |A^2 - (\beta - 1) A - \beta| = 0\)
From the equation, we know that if \( |A| \neq 0 \), we can simplify the equation to:
\(A^2 - (\beta - 1) A - \beta = 0\)
Now, factor the expression:
\(\Rightarrow |A + 1| |A - \beta| = 0\)
Since \( |A + 1| \neq 0 \) (we assume \( A \neq -1 \)), we get:
\(|A - \beta| = 0\)
This implies:
\(A = \beta\)
Now, solving for \( \beta \), we get:
\(\beta = \frac{1}{3} \Rightarrow 9\beta = 3\)
Thus, the value of \( \beta \) is \( \frac{1}{3} \), and \( 9\beta = 3 \).
Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $ D $ of a tube. The measured value of $ D $ is:
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
A matrix is a rectangular array of numbers, variables, symbols, or expressions that are defined for the operations like subtraction, addition, and multiplications. The size of a matrix is determined by the number of rows and columns in the matrix.