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Matrices and Determinants question

2021 · 1 Sep · Shift 2 · Q26
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Matrices and Determinants question

2021 · 1 Sep · Shift 2 · Q26

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Consider the system of linear equations −-− x + y + 2z = 0 3x −-− ay + 5z = 1 2x −-− 2y −-− az = 7 Let S1 be the set of all a ∈\in∈ R for which the system is inconsistent and S2 be the set of all a ∈\in∈ R for which the system has infinitely many solutions. If n(S1) and n(S2) denote the number of elements in S1 and S2 respectively, then
  1. A
    n(S1) = 2, n(S2) = 2
  2. B
    n(S1) = 1, n(S2) = 0
  3. C
    n(S1) = 2, n(S2) = 0
  4. D
    n(S1) = 0, n(S2) = 2
View written solutionFree

Correct answer: C

  1. Write the system in matrix form

The system is

−x+y+2z=0,-x+y+2z=0,−x+y+2z=0, 3x−ay+5z=1,3x-ay+5z=1,3x−ay+5z=1, 2x−2y−az=7.2x-2y-az=7.2x−2y−az=7.

So the coefficient matrix is

A=(−1123−a52−2−a),A=\begin{pmatrix} -1 & 1 & 2\\ 3 & -a & 5\\ 2 & -2 & -a \end{pmatrix},A=​−132​1−a−2​25−a​​,

and the augmented matrix is

(−11203−a512−2−a7).\left(\begin{array}{ccc|c} -1 & 1 & 2 & 0\\ 3 & -a & 5 & 1\\ 2 & -2 & -a & 7 \end{array}\right).​−132​1−a−2​25−a​017​​.
  1. Find when the system can fail to have a unique solution

A system can be inconsistent or have infinitely many solutions only when

det⁡(A)=0.\det(A)=0.det(A)=0.

Compute the determinant:

det⁡(A)=∣−1123−a52−2−a∣.\det(A)= \begin{vmatrix} -1 & 1 & 2\\ 3 & -a & 5\\ 2 & -2 & -a \end{vmatrix}.det(A)=​−132​1−a−2​25−a​​.

Expanding along the first row,

det⁡(A)=(−1)∣−a5−2−a∣−1∣352−a∣+2∣3−a2−2∣.\det(A)=(-1)\begin{vmatrix}-a & 5\\ -2 & -a\end{vmatrix} -1\begin{vmatrix}3 & 5\\ 2 & -a\end{vmatrix} +2\begin{vmatrix}3 & -a\\ 2 & -2\end{vmatrix}.det(A)=(−1)​−a−2​5−a​​−1​32​5−a​​+2​32​−a−2​​.

Now,

∣−a5−2−a∣=a2+10,\begin{vmatrix}-a & 5\\ -2 & -a\end{vmatrix}=a^2+10,​−a−2​5−a​​=a2+10, ∣352−a∣=−3a−10,\begin{vmatrix}3 & 5\\ 2 & -a\end{vmatrix}=-3a-10,​32​5−a​​=−3a−10, ∣3−a2−2∣=−6+2a.\begin{vmatrix}3 & -a\\ 2 & -2\end{vmatrix}=-6+2a.​32​−a−2​​=−6+2a.

Hence,

det⁡(A)=−(a2+10)−(−3a−10)+2(−6+2a).\det(A)=-(a^2+10)-(-3a-10)+2(-6+2a).det(A)=−(a2+10)−(−3a−10)+2(−6+2a).

Simplifying,

det⁡(A)=−a2+7a−12=−(a−3)(a−4).\det(A)=-a^2+7a-12=-(a-3)(a-4).det(A)=−a2+7a−12=−(a−3)(a−4).

Therefore,

det⁡(A)=0  ⟺  a=3 or a=4.\det(A)=0 \iff a=3 \text{ or } a=4.det(A)=0⟺a=3 or a=4.

So only these two values need to be checked.

  1. Check consistency for a=3a=3a=3

Substitute a=3a=3a=3:

(−11203−3512−2−37).\left(\begin{array}{ccc|c} -1 & 1 & 2 & 0\\ 3 & -3 & 5 & 1\\ 2 & -2 & -3 & 7 \end{array}\right).​−132​1−3−2​25−3​017​​.

Apply row operations.

Take

R2→R2+3R1,R_2 \to R_2+3R_1,R2​→R2​+3R1​, R3→R3+2R1.R_3 \to R_3+2R_1.R3​→R3​+2R1​.

Then

R2=(0,0,11∣1),R_2=(0,0,11\mid 1),R2​=(0,0,11∣1), R3=(0,0,1∣7).R_3=(0,0,1\mid 7).R3​=(0,0,1∣7).

So the matrix becomes

(−1120001110017).\left(\begin{array}{ccc|c} -1 & 1 & 2 & 0\\ 0 & 0 & 11 & 1\\ 0 & 0 & 1 & 7 \end{array}\right).​−100​100​2111​017​​.

Now,

R2→R2−11R3R_2 \to R_2-11R_3R2​→R2​−11R3​

gives

R2=(0,0,0∣−76).R_2=(0,0,0\mid -76).R2​=(0,0,0∣−76).

Thus we get

0=−76,0=-76,0=−76,

which is impossible. Hence the system is inconsistent for a=3a=3a=3.

  1. Check consistency for a=4a=4a=4

Substitute a=4a=4a=4:

(−11203−4512−2−47).\left(\begin{array}{ccc|c} -1 & 1 & 2 & 0\\ 3 & -4 & 5 & 1\\ 2 & -2 & -4 & 7 \end{array}\right).​−132​1−4−2​25−4​017​​.

Again apply

R2→R2+3R1,R_2 \to R_2+3R_1,R2​→R2​+3R1​, R3→R3+2R1.R_3 \to R_3+2R_1.R3​→R3​+2R1​.

Then

R2=(0,−1,11∣1),R_2=(0,-1,11\mid 1),R2​=(0,−1,11∣1), R3=(0,0,0∣7).R_3=(0,0,0\mid 7).R3​=(0,0,0∣7).

So the augmented matrix becomes

(−11200−11110007).\left(\begin{array}{ccc|c} -1 & 1 & 2 & 0\\ 0 & -1 & 11 & 1\\ 0 & 0 & 0 & 7 \end{array}\right).​−100​1−10​2110​017​​.

This gives

0=7,0=7,0=7,

which is impossible. Hence the system is inconsistent for a=4a=4a=4 as well.

  1. Determine S1S_1S1​ and S2S_2S2​
  • Inconsistent values:
S1={3,4}  ⟹  n(S1)=2.S_1=\{3,4\} \implies n(S_1)=2.S1​={3,4}⟹n(S1​)=2.
  • Infinitely many solutions occur when the system is consistent and rank <3<3<3. But for both singular values a=3,4a=3,4a=3,4, the system is inconsistent. Hence
S2=∅  ⟹  n(S2)=0.S_2=\varnothing \implies n(S_2)=0.S2​=∅⟹n(S2​)=0.
  1. Match with the options

Thus,

n(S1)=2,n(S2)=0.n(S_1)=2,\quad n(S_2)=0.n(S1​)=2,n(S2​)=0.

So the correct option is C.

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