Q.Solve the system of equations $2x + 5y = 1$, $3x + 2y = 7$.

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 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).

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 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 — The concept.

By definition, $A^{-1}$ is the unique matrix with $A\,A^{-1} = I$. So if we can manipulate the given equation into the shape

that something must be $A^{-1}$. (This also silently proves $A$ is invertible, which is why the trick works.)

Step 2 — Rearrange the given relation.

Step 3 — Factor out $A$ on the right.

Matrix multiplication distributes over subtraction, and $A\cdot I = A$, so

Hence

Step 4 — Divide by the scalar and read off the inverse.

Comparing with $A A^{-1} = I$:

Step 5 — Verify (always check on an inverse question).

Multiply back:

using $A^2 = 5A - 7I$ from the original equation.

Why the others fail: (B) $\frac17(A-5I)$ is the negative of the correct answer. (C) $7(5I-A)$ has the scalar inverted ($7$ instead of $\frac17$). (D) uses the wrong constants entirely.

ANSWER: A

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 row vector $A = \begin{bmatrix} x & 1 & -2 \end{bmatrix}$ and a $3\times3$ matrix $B$. The product $A B A^t$ is a $1\times1$ matrix (a scalar), and we are told it equals $[-20]$. Our job is to find $x$.

Concept & Intuition

The expression $A B A^t$ is a quadratic form in the entries of $A$. Since $A$ is $1\times3$ and $B$ is $3\times3$, $A B$ is a $1\times3$ row vector; multiplying that by $A^t$ (a $3\times1$ column) yields a single number. We can compute step by step: first multiply $A$ by $B$, then dot the result with $A^t$. This will give a quadratic equation in $x$.

Step-by-step solution

1. Compute $A B$ $A$ is $1\times3$, $B$ is $3\times3$, so $AB$ is $1\times3$.

Multiply:

• First column: $x\cdot1 + 1\cdot4 + (-2)\cdot7 = x + 4 - 14 = x - 10$
• Second column: $x\cdot2 + 1\cdot5 + (-2)\cdot8 = 2x + 5 - 16 = 2x - 11$
• Third column: $x\cdot3 + 1\cdot6 + (-2)\cdot9 = 3x + 6 - 18 = 3x - 12$ So