For the matrix A=1 1 01 2 12 1 0, which of the following is correct.

A2=A⋅A=1 1 01 2 12 1 0∣1 1 01 2 12 1 0=2 3 15 6 23 4 1,A3=A2A=231562341110121210=793151969124A3−3⋅A2=793151969124−693151869123=100010001=I⇒ A3−3A2−I=0

For the matrix A=1 1 01 2 12 1 0, which of the following is correct.

A2=A⋅A=1 1 01 2 12 1 0∣1 1 01 2 12 1 0=2 3 15 6 23 4 1,A3=A2A=231562341110121210=793151969124A3−3⋅A2=793151969124−693151869123=100010001=I⇒ A3−3A2−I=0

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

For the matrix $A = [\begin{array}{lll} 1 1 0 \\ 1 2 1 \\ 2 1 0 \end{array}],$ which of the following is correct.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\begin{aligned} A^{3} + 3 A^{2} - I = 0 \end{aligned}$

$\begin{aligned} A^{3} - 3 A^{2} - I = 0 \end{aligned}$

$A^{3} + 2 A^{2} - I = 0$

$A^{3} - A^{2} + I = 0$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$A^{2} = A \cdot A = [\begin{array}{lll} 1 1 0 \\ 1 2 1 \\ 2 1 0 \end{array} ∣] [\begin{array}{lll} 1 1 0 \\ 1 2 1 \\ 2 1 0 \end{array}] = [\begin{array}{lll} 2 3 1 \\ 5 6 2 \\ 3 4 1 \end{array}],$

$\begin{array}{l} A^{3} = A^{2} A = [\begin{array}{lll} 2 & 3 & 1 \\ 5 & 6 & 2 \\ 3 & 4 & 1 \end{array}] [\begin{array}{lll} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 2 & 1 & 0 \end{array}] = [\begin{array}{ccc} 7 & 9 & 3 \\ 15 & 19 & 6 \\ 9 & 12 & 4 \end{array}] \\ A^{3} - 3 \cdot A^{2} = [\begin{array}{ccc} 7 & 9 & 3 \\ 15 & 19 & 6 \\ 9 & 12 & 4 \end{array}] - [\begin{array}{ccc} 6 & 9 & 3 \\ 15 & 18 & 6 \\ 9 & 12 & 3 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = I \\ \Rightarrow A^{3} - 3 A^{2} - I = 0 \end{array}$