1.Let A=(2a013105b)A = \begin{pmatrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{pmatrix}A=210a3501b. If A3=4A2−A−21IA^{3} = 4A^{2} - A - 21IA3=4A2−A−21I, where III is the identity matrix of order 3×33 \times 33×3, then 2a+3b2a + 3b2a+3b is equal to:a.−9-9−9b.−13-13−13c.−10-10−10d.−12-12−12Login to continueOnly logged in users canattempt or see the solution.
