Question

There are two possible value of A in the solution of the matrix equation $\begin{bmatrix} 2A + 1 & - 5 \

• 4 & A \end{bmatrix}^{- 1}$$\begin{bmatrix} A - 5 & B \ 2A - 2 & C \end{bmatrix}$=$\begin{bmatrix} 14 & D \ E & F \end{bmatrix}$, where A, B, C, D, E, F are real numbers. The absolute value of the difference of these two solutions, is:

• A. $\frac{8}{3}$
• B. $\frac{11}{3}$
• C. $\frac{1}{3}$
• D. $\frac{19}{3}$

Answer 1

Answer: (D)

$\frac{19}{3}$

Answer 2

$\begin{bmatrix} 2A + 1 & - 5 \

• 4 & A \end{bmatrix}^{- 1}= \begin{bmatrix} A & 5 \ 4 & 2A + 1 \end{bmatrix}$

So $\frac{1}{2A^{2} + A - 20}$$\begin{bmatrix} A - 5 & B \\ 2A - 2 & C \end{bmatrix}$

= $\left[ \begin{array} { c c } 14 & \mathrm { D } \\ \mathrm { E } & \mathrm { F } \end{array} \right]$ $\frac{1}{2A^{2} + A - 20}$ $\begin{bmatrix} A(A - 5) + 5(2A - 2) & AB + 5C \\ 4(A - 5) + (2A + 1)(2A - 2) & 4B + C(2A + 1) \end{bmatrix}$= $\begin{bmatrix} 14 & D \\ E & F \end{bmatrix}$

so $\frac{A^{2} + 5A - 10}{2A^{2} + A - 20}$= 14 Ž A = 3, $\frac{- 10}{3}$