If A=[ beginmatrix 2 -1 1 2 endmatrix ], then A2+2A-3I=

Question

If A=[ beginmatrix 2 -1 1 2 endmatrix ], then A2+2A-3I=  (A)  [ beginmatrix 4 -6 6 4 endmatrix ]  (B)  0  (C)  [ beginmatrix -6 2 -2 6 endmatrix ]  (D)  5I

Tardigrade Admin · Accepted Answer

Given, A=[ beginmatrix 2 -1 1 2 endmatrix ] Now, A2=[ beginmatrix 2 -1 1 2 endmatrix ][ beginmatrix 2 -1 1 2 endmatrix ]=[ beginmatrix 4-1 -2-2 2+2 -1+4 endmatrix ] ⇒ A2=[ beginmatrix 3 -4 4 3 endmatrix ] Now, A2+2A-3l=[ beginmatrix 3 -4 4 3 endmatrix ]+2[ beginmatrix 2 -1 1 2 endmatrix ]-3[ beginmatrix 1 0 0 1 endmatrix ] =[ beginmatrix 3+4-3-4-2-0 4+2-0 3+4-3 endmatrix ] = [ beginmatrix 4 -6 6 4 endmatrix ]

Q. If $A = [21 - 1 2],$ then $A^{2} + 2 A - 3 I =$

$[46 - 6 4]$

$0$

$[- 6 - 2 26]$

$5 I$

Given, $A = [21 - 1 2]$ Now, $A^{2} = [21 - 1 2] [21 - 1 2] = [4 - 1 2 + 2 - 2 - 2 - 1 + 4]$
$\Rightarrow$ $A^{2} = [34 - 4 3]$
Now, $A^{2} + 2 A - 3 l = [34 - 4 3] + 2 [21 - 1 2] - 3 [1001]$
$= [3 + 4 - 3 - 4 - 2 - 0 4 + 2 - 03 + 4 - 3] = [46 - 6 4]$

Q. If A=[21​−12​], then A2+2A−3I=

[46​−64​]

0

[−6−2​26​]

5I

Given, A=[21​−12​] Now, A2=[21​−12​][21​−12​]=[4−12+2​−2−2−1+4​] ⇒ A2=[34​−43​] Now, A2+2A−3l=[34​−43​]+2[21​−12​]−3[10​01​] =[3+4−3−4−2−04+2−03+4−3​]=[46​−64​]

Q. If $A = [21 - 1 2],$ then $A^{2} + 2 A - 3 I =$

$[46 - 6 4]$

$0$

$[- 6 - 2 26]$

$5 I$

Given, $A = [21 - 1 2]$ Now, $A^{2} = [21 - 1 2] [21 - 1 2] = [4 - 1 2 + 2 - 2 - 2 - 1 + 4]$
$\Rightarrow$ $A^{2} = [34 - 4 3]$
Now, $A^{2} + 2 A - 3 l = [34 - 4 3] + 2 [21 - 1 2] - 3 [1001]$
$= [3 + 4 - 3 - 4 - 2 - 0 4 + 2 - 03 + 4 - 3] = [46 - 6 4]$