IGDTUW 2026 — Mathematics PYQ

Let the general $2\times 2$ matrix be represented as:

$$A=\left[\begin{array}{cc}a& amp;b\\ c& amp;d\end{array}\right]$$

The determinant of this matrix is given by the formula:

$$\Delta =\det (A)=ad-bc$$

According to the question, the entries $a,b,c,$ and $d$ belong to the set $S=\{-1,0,1\}$.

Step 1: Find all possible values for the products $ad$ and $bc$

Since $a,d\in \{-1,0,1\}$, the possible values for the product $ad$ are obtained by multiplying any two elements from the set:

• $(-1)\times (-1)=1$

• $(-1)\times 1=-1$

• Any element multiplied by $0=0$

Thus, the possible values for the product $ad$ are:

$$ad\in \{-1,0,1\}$$

Similarly, since $b,c\in \{-1,0,1\}$, the possible values for the product $bc$ are:

$$bc\in \{-1,0,1\}$$

Step 2: Calculate all possible combinations for $ad-bc$

Let's find all distinct outcomes for $\Delta =ad-bc$ by substituting values from the sets calculated above:

• When $ad=1$:

  • $\Delta =1-(1)=0$

  • $\Delta =1-(0)=1$

  • $\Delta =1-(-1)=2$

• When $ad=0$:

  • $\Delta =0-(1)=-1$

  • $\Delta =0-(0)=0$

  • $\Delta =0-(-1)=1$

• When $ad=-1$:

  • $\Delta =-1-(1)=-2$

  • $\Delta =-1-(0)=-1$

  • $\Delta =-1-(-1)=0$

Step 3: List the distinct values

Gathering all unique values from the calculations above, we get the set of distinct values for the determinant:

$$\Delta \in \{-2,-1,0,1,2\}$$

Counting these values, there are exactly 5 distinct outcomes.

Correct Answer:

(c) 5