Q.Examine the consistency of the following system of equations: $x + 2y = 2$ $2x + 3y = 3$

Step 1 — The key property.

From the fundamental identity

take determinants of both sides:

(the last step uses $|kI_n| = k^n$). Dividing by $|A| \ne 0$:

Step 2 — Specialise to $n = 2$.

Here $B = \operatorname{adj}A$ is $2\times2$, so $A$ is $2\times2$ too, giving $n = 2$:

(A nice special fact: for a $2\times2$ matrix, the adjoint has the same determinant as the matrix.)

Step 3 — Compute $|B|$ directly from its entries.

Step 4 — Equate and solve.

Step 5 — Sanity check.

With $\alpha = 5$, $B = \begin{bmatrix}1 & 3\\ 1 & 5\end{bmatrix}$, $|B| = 5-3 = 2 = |A|$. $\checkmark$ Consistent. (Indeed $A = \operatorname{adj}(B)\!/\!...$ — one can recover $A = \begin{bmatrix}5 & -3\\ -1 & 1\end{bmatrix}$, whose determinant is $5-3=2$. $\checkmark$)

Trap check: Option (A) 4 arises if you wrongly set $|B| = |A|^2 = 4$, i.e. mis-remembering the exponent as $n$ instead of $n-1$.

ANSWER: B

We are given $A M = B$, where

Since $A M = B$, isolate $A$ by right-multiplying with $M^{-1}$:

Find $M^{-1}$. The determinant is

So

Multiply.

Entry by entry:

• $(1,1): (-4)(-\tfrac{1}{7}) + (1)(\tfrac{3}{7}) = \tfrac{4}{7} + \tfrac{3}{7} = 1$
• $(1,2): (-4)(\tfrac{2}{7}) + (1)(\tfrac{1}{7}) = -\tfrac{8}{7} + \tfrac{1}{7} = -1$
• $(2,1): (7)(-\tfrac{1}{7}) + (7)(\tfrac{3}{7}) = -1 + 3 = 2$
• $(2,2): (7)(\tfrac{2}{7}) + (7)(\tfrac{1}{7}) = 2 + 1 = 3$

Hence

We have

Multiply the second equation by 2:

Subtract the first equation:

(Check: $A=\begin{bmatrix}-11&-2&-1\\3&-14&4\end{bmatrix}$ then gives $A+2B=\begin{bmatrix}5&0&3\\1&6&2\end{bmatrix}$, confirming the result.)

We are given a product of three matrices that equals the $1 \times 1$ matrix $[0]$. That means the entire expression is just a number — zero. The trick is to multiply step by step, keeping careful track of dimensions: the first is $1 \times 3$, the second is $3 \times 3$, the third is $3 \times 1$, so the result is $1 \times 1$.

Let’s denote:

We have $A (B C) = 0$ or $(A B) C = 0$ — matrix multiplication is associative, so we can choose whichever order is simpler.

1. First multiply $B$ and $C$ (the $3 \times 3$ times the $3 \times 1$):

2. Now multiply $A$ by that result (a $1 \times 3$ times a $3 \times 1$):

3. Simplify the expression:

4. Set equal to zero (since the product is $[0]$):

5. Solve the quadratic:

So:

Thus the values of $x$ are $-2$ and $-14$.

Step 1 — Given.

Step 2 — Show $A^2=A$.

Since $BA=A$, replace the second $A$:

Matrix multiplication is associative, so

Hence $A^2=A$.

Step 3 — Show $B^2=B$ (same trick).

Since $AB=B$, replace the second $B$:

Hence $B^2=B$.

Step 4 — Add.

Why not (A)/(B)/(C)? $AB=B$ and $BA=A$, so $2AB=2B$, $AB=B$, $2BA=2A$ — none of these equals $A+B$ in general.

ANSWER: D

Step 1 — Why we may expand binomially.

$I$ commutes with every matrix ($AI = IA = A$), so $I$ and $A$ commute and the ordinary binomial expansion is valid:

Step 2 — Use idempotency to reduce the powers.

Given $A^{2} = A$. Then

So both $A^{2}$ and $A^{3}$ equal $A$.

Step 3 — Substitute and simplify.

Step 4 — A neat cross-check.

If $A$ is idempotent then so is $(I-A)$:

Since $(I-A)$ is itself idempotent, every positive power of it equals itself — so $(I-A)^{3} = (I-A)$ immediately, confirming Step 3. ✓

ANSWER: A

We are given two matrix equations involving $A$ and $B$ (both are $2 \times 3$ matrices, as we can infer from the dimensions of the given matrices). The trick is to treat these as simultaneous equations in the unknown matrices, but we must handle the transpose operation carefully.

1. Understand the dimensions and the unknowns

The first equation:

The right-hand side is $2 \times 3$, so $3A$ and $4B^t$ must also be $2 \times 3$. Hence $A$ is $2 \times 3$ and $B^t$ is $2 \times 3$, meaning $B$ is $3 \times 2$.

The second equation:

The right-hand side is $3 \times 2$, so $2B$ and $3A^t$ are $3 \times 2$. Since $A$ is $2 \times 3$, $A^t$ is $3 \times 2$, consistent.

So we have two matrix equations:

2. Eliminate $A$ by taking a transpose

If we transpose the second equation, we get:

Since $(B)^t = B^t$ and $(A^t)^t = A$, this becomes:

Now $N^t$ is a $2 \times 3$ matrix (transpose of the given $3 \times 2$ matrix):

So we now have two equations in the unknowns $A$ and $B^t$ (both $2 \times 3$):

3. Solve the linear system for $B^t$

Add the two equations to eliminate $A$:

Now compute $M + N^t$:

Thus:

Divide by 6:

4. Find $(5B)^t$

Recall that $(5B)^t = 5 B^t$. So:

5. Match with the options

The matrix we got is:

This matches option (D).

Step 1 — Square the matrix.

Step 2 — Why this makes all powers easy.

Because $A^2 = 2A$ (a scalar times $A$), multiplying again by $A$ just pulls out another factor of $2$:

Step 3 — Induction.

Claim: $A^{n} = 2^{\,n-1}A$ for $n \ge 1$. True for $n=1$ ($A^1 = 2^0A$). If $A^{k}=2^{\,k-1}A$, then

So the formula holds for all $n$.

Step 4 — Put $n = 10$.

Check by trace: $A$ has eigenvalues $2$ and $0$, so $A^{10}$ has eigenvalues $2^{10}$ and $0$; $2^{9}A$ indeed has eigenvalues $2^9 \cdot 2 = 2^{10}$ and $0$. ✓

ANSWER: B