Question

Let ai, 02, 03, 04, 05 be real numbers. 2 7 1 1 Compute det ((a :) (1 :) ( .) (:) (* .)) 1 1 1 Determine all values of x E R such that matrix 4 0 1 3 C 2 25 A= is invertable. 2 0 0 1 4 0 0 0

Answer 1

We find out diterminant value of the given matrix product.

We find out all the value of x such that the given matrix is invertible.
** (*)(***)(**)(**)(*)) (91.) der (920) 17. Joten ) der og -196|1975|1976|1981197.1 = det :) (0-1) (0-1)(0-1 (0-1) (0-1) Eu (-1 (-0 (-0 (-1) (-)5 5 = -1 Therefore determinauit value is -1 Giron matrice iş Az 13 25 XO 이 OO 10 det (A) 4 0 1 3 na 2 5 xo O O 4 D 0 Expand along now a.

40 3 xx 5 XO 1 now 1. = as x 1 xa Expand along ** [*1*51-01 $1+3 3-4 -4 [4(x-91-0 +3(0-x)] -4 [ ^^-34] The matrie A is inverstible if det (A) to When det (A) = 0 , Then 40-3X= 0 *(4-3x)=0 neo => x= Therefore for all values of xf 1R-20,938 hech the the matrix A is investible. All the values are HER-20,4/3} →nco 4-3x zo 4/3