CUET UG 2025 — Mathematics Algebra

Given $A = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$A=[12​31​]

---

Finding $A^2$A2:

$$A^2 = A \times A$$A2=A×A

$$= \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$$=[12​31​]×[12​31​]

$$= \begin{bmatrix} (1)(1) + (3)(2) & (1)(3) + (3)(1) \\ (2)(1) + (1)(2) & (2)(3) + (1)(1) \end{bmatrix}$$=[(1)(1)+(3)(2)(2)(1)+(1)(2)​(1)(3)+(3)(1)(2)(3)+(1)(1)​]

$$= \begin{bmatrix} 7 & 6 \\ 4 & 7 \end{bmatrix}$$=[74​67​]

---

Finding $2A$2A:

$$2A = 2 \times \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$$2A=2×[12​31​]

$$= \begin{bmatrix} 2 & 6 \\ 4 & 2 \end{bmatrix}$$=[24​62​]

---

Finding $A^2 - 2A$A2−2A:

$$A^2 - 2A = \begin{bmatrix} 7 & 6 \\ 4 & 7 \end{bmatrix} - \begin{bmatrix} 2 & 6 \\ 4 & 2 \end{bmatrix}$$A2−2A=[74​67​]−[24​62​]

$$= \begin{bmatrix} 7-2 & 6-6 \\ 4-4 & 7-2 \end{bmatrix}$$=[7−24−4​6−67−2​]

$$= \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$$=[50​05​]

---

Finding the determinant:

$$\text{det}(A^2 - 2A) = \begin{vmatrix} 5 & 0 \\ 0 & 5 \end{vmatrix}$$det(A2−2A)=​50​05​​

$$= (5)(5) - (0)(0)$$=(5)(5)−(0)(0)

$$= 25$$=25

Therefore, $\text{det}(A^2 - 2A) = 25$det(A2−2A)=25