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

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(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:

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