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Matrices and Determinants question

2021 · 31 Aug · Shift 2 · Q40
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  5. /2021 · 31 Aug · Shift 2 · Q40

Matrices and Determinants question

2021 · 31 Aug · Shift 2 · Q40

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
The number of elements in the set {A=(ab0d):a,b,d∈{−1,0,1} and (I−A)3=I−A3}\left\{ {A = \left( {\begin{matrix} a & b \\ 0 & d \\ \end{matrix} } \right):a,b,d \in \{ - 1,0,1\} \,and\,{{(I - A)}^3} = I - {A^3}} \right\}{A=(a0​bd​):a,b,d∈{−1,0,1}and(I−A)3=I−A3}, where I is 2 ×\times× 2 identity matrix, is :
Numerical answer
View written solutionFree

Correct answer: 8

  1. We are given
A=(ab0d),a,b,d∈{−1,0,1}A=\begin{pmatrix}a&b\\0&d\end{pmatrix},\qquad a,b,d\in\{-1,0,1\}A=(a0​bd​),a,b,d∈{−1,0,1}

and we need to count how many such matrices satisfy

(I−A)3=I−A3.(I-A)^3=I-A^3.(I−A)3=I−A3.
  1. Let
B=I−A=(1−a−b01−d).B=I-A=\begin{pmatrix}1-a&-b\\0&1-d\end{pmatrix}.B=I−A=(1−a0​−b1−d​).

Since both AAA and I−AI-AI−A are upper triangular, we can use the formula: for

M=(xy0z),M=\begin{pmatrix}x&y\\0&z\end{pmatrix},M=(x0​yz​),

we have

M3=(x3y(x2+xz+z2)0z3).M^3=\begin{pmatrix}x^3&y(x^2+xz+z^2)\\0&z^3\end{pmatrix}.M3=(x30​y(x2+xz+z2)z3​).

So,

(I−A)3=((1−a)3−b((1−a)2+(1−a)(1−d)+(1−d)2)0(1−d)3).(I-A)^3=\begin{pmatrix}(1-a)^3&-b\big((1-a)^2+(1-a)(1-d)+(1-d)^2\big)\\0&(1-d)^3\end{pmatrix}.(I−A)3=((1−a)30​−b((1−a)2+(1−a)(1−d)+(1−d)2)(1−d)3​).

Also,

A3=(a3b(a2+ad+d2)0d3),A^3=\begin{pmatrix}a^3&b(a^2+ad+d^2)\\0&d^3\end{pmatrix},A3=(a30​b(a2+ad+d2)d3​),

therefore

I−A3=(1−a3−b(a2+ad+d2)01−d3).I-A^3=\begin{pmatrix}1-a^3&-b(a^2+ad+d^2)\\0&1-d^3\end{pmatrix}.I−A3=(1−a30​−b(a2+ad+d2)1−d3​).
  1. Equating corresponding entries of (I−A)3=I−A3(I-A)^3=I-A^3(I−A)3=I−A3:
  • From the (1,1)(1,1)(1,1) entry,
(1−a)3=1−a3.(1-a)^3=1-a^3.(1−a)3=1−a3.

Expand:

1−3a+3a2−a3=1−a31-3a+3a^2-a^3=1-a^31−3a+3a2−a3=1−a3 −3a+3a2=0-3a+3a^2=0−3a+3a2=0 3a(a−1)=0.3a(a-1)=0.3a(a−1)=0.

Hence

a=0 or a=1.a=0\ \text{or}\ a=1.a=0 or a=1.
  • From the (2,2)(2,2)(2,2) entry,
(1−d)3=1−d3(1-d)^3=1-d^3(1−d)3=1−d3

which similarly gives

d=0 or d=1.d=0\ \text{or}\ d=1.d=0 or d=1.

So a,d∈{0,1}a,d\in\{0,1\}a,d∈{0,1}.

  1. Now compare the (1,2)(1,2)(1,2) entries:
−b((1−a)2+(1−a)(1−d)+(1−d)2)=−b(a2+ad+d2).-b\big((1-a)^2+(1-a)(1-d)+(1-d)^2\big)=-b(a^2+ad+d^2).−b((1−a)2+(1−a)(1−d)+(1−d)2)=−b(a2+ad+d2).

Multiply by −1-1−1:

b((1−a)2+(1−a)(1−d)+(1−d)2)=b(a2+ad+d2).b\big((1-a)^2+(1-a)(1-d)+(1-d)^2\big)=b(a^2+ad+d^2).b((1−a)2+(1−a)(1−d)+(1−d)2)=b(a2+ad+d2).

Thus either b=0b=0b=0, or

(1−a)2+(1−a)(1−d)+(1−d)2=a2+ad+d2.(1-a)^2+(1-a)(1-d)+(1-d)^2=a^2+ad+d^2.(1−a)2+(1−a)(1−d)+(1−d)2=a2+ad+d2.

Let us simplify the left side:

(1−a)2+(1−a)(1−d)+(1−d)2(1-a)^2+(1-a)(1-d)+(1-d)^2(1−a)2+(1−a)(1−d)+(1−d)2 =(1−2a+a2)+(1−a−d+ad)+(1−2d+d2)=(1-2a+a^2)+(1-a-d+ad)+(1-2d+d^2)=(1−2a+a2)+(1−a−d+ad)+(1−2d+d2) =3−3a−3d+a2+ad+d2.=3-3a-3d+a^2+ad+d^2.=3−3a−3d+a2+ad+d2.

So equality with a2+ad+d2a^2+ad+d^2a2+ad+d2 gives

3−3a−3d=03-3a-3d=03−3a−3d=0 a+d=1.a+d=1.a+d=1.

Therefore, for b≠0b\neq 0b=0, we must have

a+d=1.a+d=1.a+d=1.
  1. Now count all possibilities.

Since a,d∈{0,1}a,d\in\{0,1\}a,d∈{0,1}, the possible pairs are:

(0,0),(0,1),(1,0),(1,1).(0,0),(0,1),(1,0),(1,1).(0,0),(0,1),(1,0),(1,1).

For each pair:

  • If b=0b=0b=0, condition is automatically satisfied. So all 444 pairs give valid matrices.

  • If b=1b=1b=1 or b=−1b=-1b=−1, then we need a+d=1a+d=1a+d=1. This happens for exactly two pairs:

(a,d)=(0,1),(1,0).(a,d)=(0,1),(1,0).(a,d)=(0,1),(1,0).

For each such pair, there are 222 choices of bbb (111 or −1-1−1), giving

2×2=42\times 2=42×2=4

matrices.

Total number of matrices:

4+4=8.4+4=8.4+4=8.
  1. Hence the required number of elements is
8.\boxed{8}.8​.
  1. Comparison with stored answer: Stored correct answer = 888. This matches our derived answer.
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