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Matrices and Determinants question

2004 · Shift 0 · Q80
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  5. /2004 · Shift 0 · Q80

Matrices and Determinants question

2004 · Shift 0 · Q80

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A=(00−10−10−100)A = \left( {\begin{matrix} 0 & 0 & { - 1} \\ 0 & { - 1} & 0 \\ { - 1} & 0 & 0 \\ \end{matrix} } \right)A=​00−1​0−10​−100​​. The only correct statement about the matrix AAA is
  1. A
    A2=1{A^2} = 1A2=1
  2. B
    A=(−1)I,A=(-1)I,A=(−1)I, where III is a unit matrix
  3. C
    A−1{A^{ - 1}}A−1 does not exist
  4. D
    AAA is a zero matrix
View written solutionFree

Correct answer: A

  1. Given matrix
A=(00−10−10−100)A=\begin{pmatrix} 0&0&-1\\ 0&-1&0\\ -1&0&0 \end{pmatrix}A=​00−1​0−10​−100​​

We check each option.

  1. Compute A2A^2A2

Let

A=(00−10−10−100)A=\begin{pmatrix} 0&0&-1\\ 0&-1&0\\ -1&0&0 \end{pmatrix}A=​00−1​0−10​−100​​

Then

A2=A⋅AA^2=A\cdot AA2=A⋅A

Now multiply row by column:

  • First row with columns:

    (1,1):0⋅0+0⋅0+(−1)⋅(−1)=1(1,1): 0\cdot 0+0\cdot 0+(-1)\cdot(-1)=1(1,1):0⋅0+0⋅0+(−1)⋅(−1)=1 (1,2):0⋅0+0⋅(−1)+(−1)⋅0=0(1,2): 0\cdot 0+0\cdot(-1)+(-1)\cdot 0=0(1,2):0⋅0+0⋅(−1)+(−1)⋅0=0 (1,3):0⋅(−1)+0⋅0+(−1)⋅0=0(1,3): 0\cdot(-1)+0\cdot 0+(-1)\cdot 0=0(1,3):0⋅(−1)+0⋅0+(−1)⋅0=0

  • Second row with columns:

    (2,1):0⋅0+(−1)⋅0+0⋅(−1)=0(2,1): 0\cdot 0+(-1)\cdot 0+0\cdot(-1)=0(2,1):0⋅0+(−1)⋅0+0⋅(−1)=0 (2,2):0⋅0+(−1)⋅(−1)+0⋅0=1(2,2): 0\cdot 0+(-1)\cdot(-1)+0\cdot 0=1(2,2):0⋅0+(−1)⋅(−1)+0⋅0=1 (2,3):0⋅(−1)+(−1)⋅0+0⋅0=0(2,3): 0\cdot(-1)+(-1)\cdot 0+0\cdot 0=0(2,3):0⋅(−1)+(−1)⋅0+0⋅0=0

  • Third row with columns:

    (3,1):(−1)⋅0+0⋅0+0⋅(−1)=0(3,1): (-1)\cdot 0+0\cdot 0+0\cdot(-1)=0(3,1):(−1)⋅0+0⋅0+0⋅(−1)=0 (3,2):(−1)⋅0+0⋅(−1)+0⋅0=0(3,2): (-1)\cdot 0+0\cdot(-1)+0\cdot 0=0(3,2):(−1)⋅0+0⋅(−1)+0⋅0=0 (3,3):(−1)⋅(−1)+0⋅0+0⋅0=1(3,3): (-1)\cdot(-1)+0\cdot 0+0\cdot 0=1(3,3):(−1)⋅(−1)+0⋅0+0⋅0=1

Thus,

A2=(100010001)=IA^2=\begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}=IA2=​100​010​001​​=I

So option A is true.

  1. Check option B: A=(−1)IA=(-1)IA=(−1)I

Now,

(−1)I=(−1000−1000−1)(-1)I=\begin{pmatrix} -1&0&0\\ 0&-1&0\\ 0&0&-1 \end{pmatrix}(−1)I=​−100​0−10​00−1​​

But

A=(00−10−10−100)A=\begin{pmatrix} 0&0&-1\\ 0&-1&0\\ -1&0&0 \end{pmatrix}A=​00−1​0−10​−100​​

These are clearly not equal.

So B is false.

  1. Check option C: A−1A^{-1}A−1 does not exist

Since

A2=IA^2=IA2=I

we have

A⋅A=IA\cdot A=IA⋅A=I

Hence,

A−1=AA^{-1}=AA−1=A

So inverse exists.

Therefore C is false.

  1. Check option D: AAA is a zero matrix

A zero matrix has all entries equal to 000. But here entries like −1-1−1 are present.

So D is false.

  1. Conclusion

The only correct statement is:

A2=I\boxed{A^2=I}A2=I​

So the correct option is A.

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