Q.Construct a $2 \times 2$ matrix where

We are given a $3\times3$ matrix $A=[a_{ij}]$ with entries defined piecewise:

The condition $i=j$ gives the diagonal entries; $i>j$ means entries below the diagonal are zero; $i<j$ means entries above the diagonal are all $1$. So the matrix is upper‑triangular (all entries below the main diagonal are zero). For such a matrix, the determinant is simply the product of the diagonal entries — a key fact that saves us from any heavy computation.

Let’s build the matrix step by step.

1. Diagonal entries ($i=j$):

   For $i=1$: $a_{11}=1-2\cdot1 = -1$.

   For $i=2$: $a_{22}=2-2\cdot2 = -2$.

   For $i=3$: $a_{33}=3-2\cdot3 = -3$.

2. Below the diagonal ($i>j$):

   These are positions $(2,1)$, $(3,1)$, $(3,2)$. By definition they are $0$.

3. Above the diagonal ($i<j$):

   These are positions $(1,2)$, $(1,3)$, $(2,3)$. By definition they are $1$.

Thus the matrix is:

4. Determinant of an upper‑triangular matrix: The determinant is the product of the entries on the main diagonal. No expansion or row operations are needed. So

5. Compute the product: $(-1)(-2)=2$, then $2 \times (-3) = -6$.

The core idea here is matrix inverses. When you see an equation of the form $PA = I$, where $P$ is a square matrix and $I$ is the identity matrix, you are looking at the definition of the inverse: $A$ must be $P^{-1}$. The problem is therefore asking: which of the given matrices is the inverse of $\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}$?

Let’s work through it step by step.

1. Identify the given matrix and the equation.

   We have $P A = I$, where $P = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}$ and $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.

   This is a $2 \times 2$ matrix equation. Since $P$ is square, the only matrix $A$ that satisfies $PA = I$ is the inverse $P^{-1}$.

2. Recall the formula for the inverse of a $2 \times 2$ matrix.

   For a matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, its inverse is

provided the determinant $ad - bc \neq 0$.

3. Compute the determinant of $P$. Here $a = 2$, $b = 1$, $c = 3$, $d = 2$.

Since the determinant is $1$, the inverse is simply the matrix of cofactors without any scaling factor.

4. Write the inverse using the formula.

5. Match with the options. Option (B) is exactly $\begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix}$. The other options are either the original matrix (A), a sign-changed version (C), or a matrix with the wrong sign on one entry (D).

Step 1 — The concept: inverse of a product

If $A$ and $B$ are invertible square matrices of the same order, then

Why: an inverse is defined by the property $X X^{-1}=I$. Check the candidate:

Since inverses are unique, $B^{-1}A^{-1}$ is $(AB)^{-1}$.

Step 2 — Specialise to $B=A$

(Here the order-reversal is invisible because both factors are the same matrix, but the law is still what licenses the step.)

Step 3 — Verify by direct multiplication

So $(A^{-1})^2$ satisfies the defining property of $(A^2)^{-1}$.

Step 4 — Why the other options fail

• (B) $A^2$ would require $A^2A^2 = I$, i.e. $A^4=I$ — true only for very special matrices, not for a general $3\times3$ matrix.
• (C) $(-A)^2 = A^2$ — identical to (B) (the two minus signs cancel), so equally wrong.
• (D) $(-A^2)^2 = A^4$ — would need $A^6=I$; again not general.

ANSWER: A

Step 1 — Use the determinant property.

For any square matrices, $|AB|=|A||B|$. Applying it twice,

This is far cheaper than actually cubing the matrix — that is the point of the question.

Step 2 — Solve for $|A|$.

(The real cube root is unique, so there is no $\pm$ ambiguity here — a cube, unlike a square, preserves sign.)

Step 3 — Compute $|A|$ from the entries.

Step 4 — Equate and solve.

Check: for $k=3$, $|A|=9-4=5$ and $|A^{3}|=5^{3}=125$ ✓. For $k=-3$, $|A|=9-4=5$ as well (since $k$ appears squared) and again $|A^{3}|=125$ ✓. Both signs work, which is why the answer must be $\pm3$ and not just one of them — options (C) and (D), single negative values, ignore this, and (A) comes from mistakenly solving $k^{2}=4$.

ANSWER: B

Step 1 — What is being asked.

$AB=I$ means $B$ is the inverse of $A$, so $B=A^{-1}$. For a $2\times2$ matrix,

Step 2 — Apply it. Write $t=\tan\dfrac{\alpha}{2}$, so

Determinant:

Adjugate:

Therefore

Step 3 — Recognise the bracket.

Transposing $A$ swaps the off-diagonal entries:

which is exactly the matrix in the bracket. (This is the skew-symmetric-off-diagonal structure of a rotation-type matrix: its adjugate equals its transpose.) So

Note it is not $\cos^2\frac{\alpha}{2}A$, since $A$ itself has $+t$ in the top-right; the sign flip matters.

ANSWER: D

Step 1 — Why we may expand binomially.

Matrix multiplication is generally non-commutative, so $(X+Y)^3$ is not the ordinary binomial expansion in general. But here $X=I$, the identity, and $IA=AI=A$ — the identity commutes with every matrix. That restores the binomial theorem:

Step 2 — Reduce the powers of $A$ using idempotency.

Given $A^2=A$. Then

So every positive power of $A$ equals $A$: $A^2=A^3=\dots=A$.

Step 3 — Substitute back.

Step 4 — Collect like terms.

Verification with a concrete idempotent. Take $A=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ (indeed $A^2=A$). Then $I+A=\begin{bmatrix}2&0\\0&1\end{bmatrix}$ and $(I+A)^3=\begin{bmatrix}8&0\\0&1\end{bmatrix}$. Meanwhile $7A+I=\begin{bmatrix}7&0\\0&0\end{bmatrix}+\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}8&0\\0&1\end{bmatrix}$ ✓. (Note $7A=\begin{bmatrix}7&0\\0&0\end{bmatrix}\neq(I+A)^3$, so option (B) fails; and $7A-I$ gives $\begin{bmatrix}6&0\\0&-1\end{bmatrix}$ ✗.)

ANSWER: C

Step 1 — Compute $A^2$.

Entry by entry:

• $(1,1): (2)(2) + (-2)(-2) = 4 + 4 = 8$
• $(1,2): (2)(-2) + (-2)(2) = -4 - 4 = -8$
• $(2,1): (-2)(2) + (2)(-2) = -4 - 4 = -8$
• $(2,2): (-2)(-2) + (2)(2) = 4 + 4 = 8$

So $A$ satisfies $A^2 = 4A$ — the key structural fact. (This is the matrix analogue of an "idempotent up to a scalar": $A = 2P$ where $P = \begin{bmatrix}1&-1\\-1&1\end{bmatrix}$ and $P^2 = 2P$.)

Step 2 — Iterate to get the general power.

The pattern of exponents is $2, 4, 6, \dots$ for $n = 2, 3, 4, \dots$

Step 3 — Prove by induction / write the formula.

Claim: $A^n = 2^{2(n-1)}A$.

• Base $n=1$: $2^{0}A = A$. ✓
• Step: if $A^n = 2^{2(n-1)}A$, then

Since $A^n = 2^k A$, we read off

Check with $n=2$: $k = 2(1) = 2 \Rightarrow A^2 = 2^2A = 4A$ ✓ (matches Step 1). Option (C), $k = n-1$, would give $A^2 = 2A$ — false.

ANSWER: D

Step 1 — Expand the $2\times2$ determinant.

Step 2 — Expand the $3\times3$ determinant (along the first row):

(Cross-check by the standard factorisation: this determinant equals $(x-1)^{2}(x+2) = (x^{2}-2x+1)(x+2) = x^{3} - 3x + 2$. ✓)

Step 3 — Differentiate $B$ with respect to $x$.

Step 4 — Express the answer in terms of $A$.

Reject the rest: $-3B = -3(x^3-3x+2)$ is a cubic, not a quadratic; $3B+1$ likewise; $1-3A = 1-3x^{2}+3 = 4-3x^{2} \neq 3x^{2}-3$.

ANSWER: A