Q.Construct a $2 \times 2$ matrix, $A = [a_{ij}]$, whose elements are given by:

Concept and Intuition

This is a direct construction problem: build the matrix entry-by-entry from the given rule and match against the options, watching the fractions carefully (many wrong options differ only by a dropped $\frac13$ or a sign/arithmetic slip in a single entry).

Step-by-Step Solution

1. $i=1$: $a_{11}=\frac13|1-5|=\frac43$, $a_{12}=\frac13|1-10|=\frac93=3$, $a_{13}=\frac13|1-15|=\frac{14}3$.
2. $i=2$: $a_{21}=\frac13|2-5|=1$, $a_{22}=\frac13|2-10|=\frac83$, $a_{23}=\frac13|2-15|=\frac{13}3$.
3. $i=3$: $a_{31}=\frac13|3-5|=\frac23$, $a_{32}=\frac13|3-10|=\frac73$, $a_{33}=\frac13|3-15|=\frac{12}3=4$. …

Concept and Intuition

A cofactor $C_{ij}=(-1)^{i+j}M_{ij}$ where $M_{ij}$ is the minor obtained by deleting row $i$ and column $j$. Being told three cofactor values turns the problem into three simple linear equations for $x,y,z$ — once those are known, $AB$ is pure computation.

Step-by-Step Solution

1. Matrix: $A=\begin{bmatrix}x&y&0\\-3&1&2\\1&-2&z\end{bmatrix}$.
2. Cofactor of $z$ (entry $(3,3)$): delete row 3, column 3 $\Rightarrow$ minor $=\begin{vmatrix}x&y\\-3&1\end{vmatrix}=x+3y$. Sign $(-1)^{3+3}=+1$. Given $=9$:

3. Cofactor of "1" in row 3 — the entry "1" sits at position $(3,1)$. Delete row 3, column 1 $\Rightarrow$ minor $=\begin{vmatrix}y&0\\1&2\end{vmatrix}=2y$. Sign $(-1)^{3+1}=+1$. Given $=4$:

Substituting into step 2: $x+6=9\Rightarrow x=3$.

4. Cofactor of $x$ (entry $(1,1)$): delete row 1, column 1 $\Rightarrow$ minor $=\begin{vmatrix}1&2\\-2&z\end{vmatrix}=z+4$. Sign $(-1)^{1+1}=+1$. Given $=3$:

5. So $A=\begin{bmatrix}3&2&0\\-3&1&2\\1&-2&-1\end{bmatrix}$, $B=\begin{bmatrix}1&-2&-2\\2&0&1\\2&1&0\end{bmatrix}$.
6. Multiply row by row:
   • Row 1: $(3\cdot1+2\cdot2+0\cdot2,\ 3\cdot(-2)+2\cdot0+0\cdot1,\ 3\cdot(-2)+2\cdot1+0\cdot0) = (7,-6,-4)$.
   • Row 2: $(-3\cdot1+1\cdot2+2\cdot2,\ -3\cdot(-2)+1\cdot0+2\cdot1,\ -3\cdot(-2)+1\cdot1+2\cdot0) = (3,8,7)$. …

Concept and Intuition

Rather than trying to count $B$ and $C$ directly (hard — most $0/1$ matrices have determinant $0$, and enumerating the nonzero-determinant ones by brute force is tedious), look for a structure-preserving bijection between them. Swapping two rows of any matrix is a classical determinant-sign-flipping operation, and critically it never changes what the entries are (only their row positions) — so a $0/1$ matrix stays a $0/1$ matrix after the swap.

Step-by-Step Solution

1. Let $M \in B$, so $M$ is a $3\times3$ matrix with entries in $\{0,1\}$ and $\det(M)=1$.
2. Define $\sigma(M)$ = the matrix obtained by swapping (say) rows 1 and 2 of $M$.
3. Swapping two rows of any matrix multiplies its determinant by $-1$: $\det(\sigma(M)) = -\det(M) = -1$. So $\sigma(M)\in C$.
4. Since $\sigma(M)$ only rearranges $M$'s entries (no entry is changed, only relocated), $\sigma(M)$ is still a valid $0/1$ matrix.
5. $\sigma$ is an involution: swapping rows 1,2 twice returns the original matrix, $\sigma(\sigma(M))=M$. So $\sigma:B\to C$ is a bijection (it has an inverse, namely itself).
6. A bijection between two finite sets means they have exactly the same cardinality: $|B|=|C|$. …