- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.21 Miscellaneous review exercises on matrices
3. $\quad$ If $A = \begin{bmatrix} 1 & 2 \\ 5 & 4 \end{bmatrix},$ find a nonsingular matrix $P$ such that $P^{-1}AP = \begin{bmatrix} 6 & 0 \\ 0 & -1 \end{bmatrix}.$
Solution. $\quad$ We wish to find a $2 \times 2$ matrix $P = \begin{bmatrix}p & q \\ r & s \end{bmatrix},$ whose inverse is given by $P^{-1} = \frac{1}{ps - qr}\begin{bmatrix}s & -q \\ -r & p \end{bmatrix}$ such that \begin{align*} P^{-1}AP = \begin{bmatrix} 6 & 0 \\ 0 & -1 \end{bmatrix} \end{align*} In other words, for $ps - qr \neq 0,$ we wish to find $A, P,$ and $P^{-1}$ such that \begin{align*} P^{-1}AP &= \frac{1}{ps - qr} \begin{bmatrix} s & -q \\ -r & p \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 5 & 4 \end{bmatrix} \begin{bmatrix} p & q \\ r & s \end{bmatrix} \\ &= \begin{bmatrix} s - 5q & 2s - 4q \\ 5p - r & 4p - 2r \end{bmatrix} \begin{bmatrix} p & q \\ r & s \end{bmatrix} \\ &= \begin{bmatrix} p(s-5q) + r(2s - 4q) & q(s - 5q) + s(2s - 4q) \\ p(5p - r) + r(4p - 2r) & q(5p - r) + s(4p - 2r) \end{bmatrix} \\ &= \begin{bmatrix} 6 & 0 \\ 0 & -1 \end{bmatrix} \end{align*} From this, we find that \begin{align*} \\ 2s^2 - 3qs - 5q^2 &= (2s - 5q)(s + q) \\ &= 0 \\ \\ 5p^2 + 3pr - 2r^2 &= (5p - 2r)(p + r) \\ &= 0 \end{align*} \begin{align*} ps - 5pq + 2rs - 4qr &= 6(ps - qr) \\ 2r(q + s) &= 5p(q + s) \\ \\ 5pq - qr + 4ps - 2rs &= -1(ps - qr) \\ 5p(q + s) &= 2r(q + s) \end{align*} This system of equations is satisfied when \begin{align*} p &= 2, \quad q = -1, \quad r = 5, \quad s = 1 \end{align*} \begin{align*} P &= \begin{bmatrix}2 & -1 \\ 5 & 1 \end{bmatrix}. \quad \blacksquare \end{align*}