Let S = { A = ( 0 1 c 1 a d 1 b e ) : a , b , c , d , e { 0 , 1 } and | A | { - 1 , 1 } } , where | A | denotes the determinant of A . Then the number of elements in S is ______________. [2024]

Question

Let $S = {A = (\begin{matrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{matrix}) : a, b, c, d, e \in {0, 1} and | A | \in {- 1, 1}},$ where $| A |$ denotes the determinant of $A$ .

Then the number of elements in $S$ is ______________. [2024]

Smart Achievers · Accepted Answer

(16)

$| A |$ $= - (e - d) + c (b - a) = \pm 1$

Case (i): $c = 0 \Rightarrow e - d = \pm 1$

$\Rightarrow (e, d) = (1, 0), (0, 1) \to 2 ways$

$b$ and $S$ can each take $2$ ways

$\Rightarrow Total = 1 \times 2 \times 2 \times 2 = 8 ways$

Case (ii): $c = 1 \Rightarrow d - e + b - a = \pm 1$

$\begin{matrix} c & a & b & d & e \\ 1 & 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \end{matrix}}$ $\Rightarrow 4 \times 2 = 8 ways$

$∴$ $Total = 16 ways$

Solution

Q.
Let $S = {A = (\begin{matrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{matrix}) : a, b, c, d, e \in {0, 1} and | A | \in {- 1, 1}},$ where $| A |$ denotes the determinant of $A$ .

Then the number of elements in $S$ is ______________. [2024]

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Solution

Q. Let S={A=(01c1ad1be): a,b,c,d,e∈{0,1} and |A|∈{-1,1}}, where |A| denotes the determinant of A. Then the number of elements in S is ______________. [2024]

Ans. (16) |A|=-(e-d)+c(b-a)=±1 Case (i): c=0⇒e-d=±1 ⇒(e,d)=(1,0),(0,1)→2 ways b and a can each take 2 ways ⇒Total=1×2×2×2=8 ways Case (ii): c=1⇒d-e+b-a=±1 cabde11110111011001010001}⇒4×2=8 ways ∴ Total=16 ways

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Q.
Let $S = {A = (\begin{matrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{matrix}) : a, b, c, d, e \in {0, 1} and | A | \in {- 1, 1}},$ where $| A |$ denotes the determinant of $A$ .

Then the number of elements in $S$ is ______________. [2024]