- 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
2.16 Exercises
13. $\quad$ If $A = \begin{bmatrix} 2 & -1 \\ -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 7 & 6 \\ 9 & 8 \end{bmatrix},$ find 2 x 2 matrices $C$ and $D$ such that $AC = B$ and $DA = B.$
Solution. $\quad$ Let $C = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ and $D = \begin{bmatrix}p & q \\ r & s \end{bmatrix}.$ We have: \begin{align*} AC &= \begin{bmatrix} 2 & - 1 \\ -2 & 3 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \\ &= \begin{bmatrix} 2a - c & 2b - d \\ -2a + 3c & -2b + 3d \end{bmatrix} \\ &= \begin{bmatrix} 7 & 6 \\ 9 & 8 \end{bmatrix} \\ \\ DA &= \begin{bmatrix} p & q \\ r & s \end{bmatrix} \begin{bmatrix} 2 & - 1 \\ -2 & 3 \end{bmatrix} \\ &= \begin{bmatrix} 2p - 2q & -p + 3q \\ 2r - 2s & -r + 3s \end{bmatrix} \\ &= \begin{bmatrix} 7 & 6 \\ 9 & 8 \end{bmatrix} \end{align*} This gives us the following two systems of equations: \begin{align*} 2a - c &= 7 \\ 2b - d &= 6 \\ -2a + 3c &= 9 \\ -2b + 3d &= 8 \\ \\ 2p - 2q &= 7 \\ -p + 3q &= 6 \\ 2r - 2s &= 9 \\ -r + 3s &= 8 \end{align*} From this, we get: \begin{align*} a &= \frac{15}{2} \quad b = \frac{13}{2} \quad c = 8 \quad d = 7 \\ \\ p &= \frac{33}{4} \quad q = \frac{19}{4} \quad r = \frac{43}{4} \quad s = \frac{25}{4} \end{align*} Thus, $C$ and $D$ satisfying $AC = B$ and $DA = B$ are the $2 \times 2$ matrices: \begin{align*} C &= \begin{bmatrix}\frac{15}{2} & \frac{13}{2} \\ 8 & 7 \end{bmatrix} \quad D = \begin{bmatrix}\frac{33}{4} & \frac{19}{4} \\ \frac{43}{4} & \frac{25}{4}\end{bmatrix} \quad \blacksquare \end{align*}