Q.In the matrix $A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}$, write:

Step 1 — The rule being tested.

If $A$ is $m\times n$ and $B$ is $p\times q$, then $AB$ exists only if $n=p$, and then

i.e. the product inherits the rows of the first matrix and the columns of the second.

Step 2 — Apply it to the inner dimension.

We are told $B$ is $3\times 4$, so $p=3$. For $AB$ to be defined, the number of columns of $A$ must be $3$. Write $A$ as $m\times 3$.

Step 3 — Apply it to the outer dimensions.

But $AB$ is an identity matrix, and every identity matrix $I_k$ is square. Therefore

Step 4 — Conclude.

A quick sanity check: $A_{4\times3}B_{3\times4}=(AB)_{4\times4}$ — square, so it can indeed be $I_4$. (Note $BA$ would then be $3\times3$; a non-square $A$ can have a one-sided inverse only.)

ANSWER: C

The key here is that the equations involve both a matrix and its transpose. You cannot directly add or subtract matrices of different orders — $A$ and $B$ are $3 \times 2$ matrices (since $B'$ is $2 \times 3$ in the first equation, and $B$ is $3 \times 2$ in the second). The trick is to take the transpose of one equation to make the unknown matrices align, then solve the resulting system.

Let $A$ and $B$ both be $3 \times 2$ matrices. Then $A'$ and $B'$ are $2 \times 3$.

1. Write the given equations clearly.

   Equation (1): $3A + 4B' = \begin{bmatrix} 7 & -10 & 17 \\ 0 & 6 & 31 \end{bmatrix}$ (a $2 \times 3$ matrix).

   Equation (2): $2B - 3A' = \begin{bmatrix} -1 & 18 \\ 4 & 0 \\ -5 & -7 \end{bmatrix}$ (a $3 \times 2$ matrix).

2. Take the transpose of equation (2).

   Since $(2B - 3A')' = 2B' - 3A$, we get:

Now both equation (1) and this new equation involve $A$ and $B'$, both $2 \times 3$ matrices.

3. Solve the system for $B'$. We have:

where $P = \begin{bmatrix} 7 & -10 & 17 \\ 0 & 6 & 31 \end{bmatrix}$ and $Q = \begin{bmatrix} -1 & 4 & -5 \\ 18 & 0 & -7 \end{bmatrix}$.

Add the two equations to eliminate $A$:

So:

4. Transpose back to get $B$. Since $B$ is the transpose of $B'$:

Step 0 — A note on the printed matrices.

As printed, $A$ shows only two columns while $B$ is $3\times2$, so $AB$ would be undefined — a column of $A$ was lost in typesetting. The missing entries are recoverable and self-consistent: taking

reproduces every printed entry of $A$ and makes the product conformable ($2\times3$ times $3\times2$ ⇒ $2\times2$, matching the $2\times2$ options). All four resulting entries agree with the option block, so the reconstruction is over-determined and consistent.

Step 1 — Multiply $AB$ (row of $A$ · column of $B$).

Step 2 — Why the order of dimensions matters.

$A_{2\times3}B_{3\times2}$ is defined because the inner dimensions match (3 = 3); the product inherits the outer dimensions, $2\times2$. That is exactly the shape of every option.

Step 3 — Take the transpose.

The transpose $M'$ is obtained by writing the rows of $M$ as columns: $(M')_{ij}=M_{ji}$.

The main-diagonal entries $(-3$ and $7)$ stay put; the off-diagonal entries $-2$ and $10$ swap places.

Step 4 — Spot the trap.

Option (A) $\begin{bmatrix}-3&-2\\10&7\end{bmatrix}$ is $AB$ itself — offered for the student who forgets the dash. The question asks for $(AB)'$.

ANSWER: B

Step 1 — Write down $A$ and its transpose $A'$.

(The transpose is obtained by interchanging rows and columns.)

Step 2 — Multiply.

Entry by entry:

• $(1,1) = \cos\alpha\cos\alpha + \sin\alpha\sin\alpha = \cos^2\alpha + \sin^2\alpha = 1$
• $(1,2) = \cos\alpha(-\sin\alpha) + \sin\alpha\cos\alpha = -\sin\alpha\cos\alpha + \sin\alpha\cos\alpha = 0$
• $(2,1) = (-\sin\alpha)\cos\alpha + \cos\alpha\sin\alpha = 0$
• $(2,2) = (-\sin\alpha)(-\sin\alpha) + \cos\alpha\cos\alpha = \sin^2\alpha + \cos^2\alpha = 1$

Step 3 — The concept behind it.

A matrix satisfying $AA' = A'A = I$ is called orthogonal, and its transpose is therefore its inverse: $A^{-1} = A'$. Geometrically $A$ here is a rotation of the plane through angle $\alpha$; rotations preserve lengths and angles, and undoing a rotation by $\alpha$ is just rotating by $-\alpha$ — which is exactly $A'$. So $AA' = I$ is the algebraic statement of "rotate, then rotate back."

A quick confirmation: $\det A = \cos^2\alpha + \sin^2\alpha = 1 \ne 0$, so $A$ is invertible — consistent with $A' = A^{-1}$ and $AA' = I$ (and it rules out the zero matrix, option B).

ANSWER: D

A=[[0,1],[0,0]] satisfies A²=0 (nilpotent). Since I and A commute, binomial theorem applies: (aI+bA)^n = Σ_{k=0}^n C(n,k) a^(n-k)(bA)^k. Since A^k=0 for k≥2, only k=0 and k=1 terms survive: = a^n I + n·a^(n-1)·b·A. (Note options A and B in the source are identical text — both represent this same correct formula; I'm reporting 'A' since it's listed first.)

Step 1 — Determinant of the general matrix

For a $2\times 2$ matrix $\begin{bmatrix} a & b \\ c & d\end{bmatrix}$ the determinant is $ad-bc$. Here $a=d=1-n$ and $b=c=n$, so

Use the difference of squares $p^{2}-q^{2}=(p-q)(p+q)$ with $p=1-n,\ q=n$:

Sanity check: $|A_1|=(0)(0)-(1)(1)=-1$, and $1-2(1)=-1$ ✓.

Step 2 — Recognise the series

This is an AP with first term $a=-1$, common difference $d=-2$, and $N=2021$ terms.

Step 3 — Sum it

Split the sum:

using $\sum_{n=1}^{N} n=\dfrac{N(N+1)}{2}$. Then

(Equivalently: the sum of the first $N$ negative odd numbers is $-N^2$, so with $N=2021$ we get $-(2021)^2$ directly.)

ANSWER: A

Step 1 — Recall the definition

A matrix of order $1 \times n$ (one row, $n$ columns, $n \geq 1$) is called a row matrix. The number of columns can be one or more; what makes it a row matrix is that it has exactly one row.

Step 2 — Match to the options

"One element" only describes a $1\times1$ matrix (a special case, not the general definition). "One column with one or more rows" describes a column matrix, not a row matrix. "One row and one column" again only describes the $1\times 1$ case. The correct general definition is "one row with one or more columns."