CUET UG 2025 — Mathematics Algebra

The value of the determinant needs to be found where $x + y + z = 0$x+y+z=0.

Adding all elements in Row 1:

$$3x + (-x+y) + (-x+z)$$3x+(−x+y)+(−x+z)

$$= 3x - x + y - x + z$$=3x−x+y−x+z

$$= x + y + z$$=x+y+z

$$= 0$$=0

---

Adding all elements in Row 2:

$$(x-y) + 3y + (z-y)$$(x−y)+3y+(z−y)

$$= x - y + 3y + z - y$$=x−y+3y+z−y

$$= x + y + z$$=x+y+z

$$= 0$$=0

---

Adding all elements in Row 3:

$$(x-z) + (y-z) + 3z$$(x−z)+(y−z)+3z

$$= x - z + y - z + 3z$$=x−z+y−z+3z

$$= x + y + z$$=x+y+z

$$= 0$$=0

---

Each row sums to zero, which means:

$$C_1 + C_2 + C_3 = \begin{pmatrix}0\\0\\0\end{pmatrix}$$C1​+C2​+C3​=​000​​

This shows that one column can be written as a linear combination of the other two, making the columns linearly dependent.

When columns of a matrix are linearly dependent, the determinant equals zero.

---

The value of the determinant is $0$0