Let p, q, r be three real numbers satisfying | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Performing the matrix multiplication $[p \, q \, r] \begin{bmatrix} 2 & p & q \ -3 & q & -p+r \ 12 & r & -q+3r \end{bmatrix} = [5 \, b \, c]$ gi

Vedclass · Accepted Answer

$\frac{25}{49}$

Vedclass · Answer

(B) Performing the matrix multiplication $[p \, q \, r] \begin{bmatrix} 2 & p & q \ -3 & q & -p+r \ 12 & r & -q+3r \end{bmatrix} = [5 \, b \, c]$ gives the following equations:$1$) $2p - 3q + 12r = 5$$2$) $b = p^2 + q^2 + r^2$$3$) $c = pq - qp + qr - qr + 3r^2 = 3r^2$Adding $b$ and $c$,we get $b + c = p^2 + q^2 + 4r^2$.We need to minimize $p^2 + q^2 + 4r^2$ subject to the constraint $2p - 3q + 12r = 5$.Using the Cauchy-Schwarz inequality in the form $(\vec{u} \cdot \vec{v})^2 \le |\vec{u}|^2 |\vec{v}|^2$,let $\vec{u} = (2, -3, 6)$ and $\vec{v} = (p, q, 2r)$.Then $(\vec{u} \cdot \vec{v}) = 2p - 3q + 12r = 5$.$|\vec{u}|^2 = 2^2 + (-3)^2 + 6^2 = 4 + 9 + 36 = 49$.$|\vec{v}|^2 = p^2 + q^2 + (2r)^2 = p^2 + q^2 + 4r^2$.Thus,$5^2 \le 49(p^2 + q^2 + 4r^2)$.$25 \le 49(b+c) \implies b+c \ge \frac{25}{49}$.

Let $p, q, r$ be three real numbers satisfying $[p \, q \, r] \begin{bmatrix} 2 & p & q \\ -3 & q & -p+r \\ 12 & r & -q+3r \end{bmatrix} = [5 \, b \, c]$. Then the minimum value of $(b+c)$ is:

Similar Questions

Let $\omega$ be a complex cube root of unity with $\omega \neq 1$ and $P = [p_{ij}]$ be an $n \times n$ matrix with $p_{ij} = \omega^{i+j}$. Then $P^2 \neq 0$ when $n =$

For $M=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$ and for any $n \in N$,the matrix $M^{n+1}-M^n=$

If $A=\left[\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$,then $A+A^3+A^4+A^5+3 I=$

Let $A = \begin{bmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{bmatrix}$,where $a, b \in \mathbb{R}$. If for some $n \in \mathbb{N}$,$A^n = \begin{bmatrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{bmatrix}$,then $n + a + b$ is equal to:

Let matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 5 & 4 \\ 0 & 3 & 2 \end{bmatrix}$ and $A^3 - 8A^2 + \alpha A + \beta I = O$,then the ordered pair $(\alpha, \beta)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module