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Matrices & Determinants — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
Number of real values of xx for which the matrix A=(3−x2224−x1−2−4−1−x)A = \begin{pmatrix}3-x & 2 & 2\\ 2 & 4-x & 1\\ -2 & -4 & -1-x\end{pmatrix} is singular, is:
A11
B22correct
C33
DInfinite
Solution
Step 1: Apply row operations to simplify R1→R1+R3R_1 \to R_1+R_3 and R2→R2+R3R_2 \to R_2+R_3:
(1−x−21−x0−x−x−2−4−1−x)\begin{pmatrix}1-x & -2 & 1-x\\ 0 & -x & -x\\ -2 & -4 & -1-x\end{pmatrix}
 Step 2: Apply a column operation C3→C3−C2C_3 \to C_3 - C_2:
(1−x−23−x0−x0−2−43−x)\begin{pmatrix}1-x & -2 & 3-x\\ 0 & -x & 0\\ -2 & -4 & 3-x\end{pmatrix}
 Step 3: Expand along R2R_2
det⁡(A)=(−x)(−1)2+2∣1−x3−x−23−x∣\det(A) = (-x)(-1)^{2+2}\begin{vmatrix}1-x & 3-x\\ -2 & 3-x\end{vmatrix}
=(−x)[(1−x)(3−x)−(−2)(3−x)]=(−x)(3−x)[(1−x)+2]= (-x)\big[(1-x)(3-x) - (-2)(3-x)\big] = (-x)(3-x)\big[(1-x)+2\big]
=−x(3−x)2= -x(3-x)^2
 Step 4: Solve −x(3−x)2=0-x(3-x)^2 = 0
x=0orx=3x = 0 \quad \text{or} \quad x = 3
 There are 22 real values. Answer: (2)
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