NIMCET 2007 — Mathematics PYQ

Let's evaluate each option step-by-step to determine which statement is correct.

Checking Option (a): Does $A^{-1}$ exist?

An inverse matrix $A^{-1}$ exists if and only if the determinant of matrix $A$ is non-zero ($\det (A)\ne 0$). Let's find the determinant of $A$:

$$\det (A)=\left|\begin{array}{ccc}0& amp;0& amp;-1\\ 0& amp;-1& amp;0\\ -1& amp;0& amp;0\end{array}\right|$$

Expanding along the first row:

$$\det (A)=0-0+(-1)\cdot \left|\begin{array}{cc}0& amp;-1\\ -1& amp;0\end{array}\right|$$

$$\det (A)=-1\cdot \left((0\cdot 0)-(-1\cdot -1)\right)$$

$$\det (A)=-1\cdot (0-1)=1$$

Since $\det (A)=1\ne 0$, $A^{-1}$ does exist. Therefore, statement (a) is incorrect.

Checking Option (b): Is $A=(-1)I$?

The identity matrix $I$ of order $3\times 3$ is:

$$I=\left(\begin{array}{ccc}1& amp;0& amp;0\\ 0& amp;1& amp;0\\ 0& amp;0& amp;1\end{array}\right)⟹(-1)I=\left(\begin{array}{ccc}-1& amp;0& amp;0\\ 0& amp;-1& amp;0\\ 0& amp;0& amp;-1\end{array}\right)$$

Comparing this with our matrix $A$:

$$A=\left(\begin{array}{ccc}0& amp;0& amp;-1\\ 0& amp;-1& amp;0\\ -1& amp;0& amp;0\end{array}\right)\ne \left(\begin{array}{ccc}-1& amp;0& amp;0\\ 0& amp;-1& amp;0\\ 0& amp;0& amp;-1\end{array}\right)$$

Therefore, statement (b) is incorrect.

Checking Option (c): Is $A$ a zero matrix?

A zero matrix requires all elements to be $0$. Since matrix $A$ has non-zero values (like $-1$), it is clearly not a zero matrix. Therefore, statement (c) is incorrect.

Checking Option (d): Does $A^2=I$?

Let's calculate $A^2$ by multiplying matrix $A$ with itself:

$$A^2=A\cdot A=\left(\begin{array}{ccc}0& amp;0& amp;-1\\ 0& amp;-1& amp;0\\ -1& amp;0& amp;0\end{array}\right)\left(\begin{array}{ccc}0& amp;0& amp;-1\\ 0& amp;-1& amp;0\\ -1& amp;0& amp;0\end{array}\right)$$

Let's compute the rows of the resulting matrix:

• Row 1:

  • Element $(1,1)=(0)(0)+(0)(0)+(-1)(-1)=1$

  • Element $(1,2)=(0)(0)+(0)(-1)+(-1)(0)=0$

  • Element $(1,3)=(0)(-1)+(0)(0)+(-1)(0)=0$

• Row 2:

  • Element $(2,1)=(0)(0)+(-1)(0)+(0)(-1)=0$

  • Element $(2,2)=(0)(0)+(-1)(-1)+(0)(0)=1$

  • Element $(2,3)=(0)(-1)+(-1)(0)+(0)(0)=0$

• Row 3:

  • Element $(3,1)=(-1)(0)+(0)(0)+(0)(-1)=0$

  • Element $(3,2)=(-1)(0)+(0)(-1)+(0)(0)=0$

  • Element $(3,3)=(-1)(-1)+(0)(0)+(0)(0)=1$

Putting it all together:

$$A^2=\left(\begin{array}{ccc}1& amp;0& amp;0\\ 0& amp;1& amp;0\\ 0& amp;0& amp;1\end{array}\right)=I$$

Since $A^2=I$, this statement is perfectly true. (A matrix that equals its own inverse is also known as an involutory matrix).

Final Answer

The correct option is (d) $A^2=I$.