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Matrices and Determinants question

2021 · 20 Jul · Shift 2 · Q37
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Matrices and Determinants question

2021 · 20 Jul · Shift 2 · Q37

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The value of k ∈\in∈ R, for which the following system of linear equations 3x −-− y + 4z = 3, x + 2y −-− 3z =−-− 2 6x + 5y + kz = −-− 3, has infinitely many solutions, is :
  1. A
    3
  2. B
    −-− 5
  3. C
    5
  4. D
    −-− 3
View written solutionFree

Correct answer: B

  1. Write the system in matrix form:
{3x−y+4z=3x+2y−3z=−26x+5y+kz=−3\begin{cases} 3x-y+4z=3 \\ x+2y-3z=-2 \\ 6x+5y+kz=-3 \end{cases}⎩⎨⎧​3x−y+4z=3x+2y−3z=−26x+5y+kz=−3​

The coefficient matrix is

A=(3−1412−365k)A=\begin{pmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & k \end{pmatrix}A=​316​−125​4−3k​​

and the augmented matrix is

(3−14312−3−265k−3).\left(\begin{array}{ccc|c} 3 & -1 & 4 & 3 \\ 1 & 2 & -3 & -2 \\ 6 & 5 & k & -3 \end{array}\right).​316​−125​4−3k​3−2−3​​.

For infinitely many solutions, we need:

rank⁡(A)=rank⁡(A∣B)<3.\operatorname{rank}(A)=\operatorname{rank}(A|B)<3.rank(A)=rank(A∣B)<3.

So first, the determinant of the coefficient matrix must be zero.

  1. Compute det⁡(A)\det(A)det(A):
det⁡(A)=∣3−1412−365k∣\det(A)=\begin{vmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & k \end{vmatrix}det(A)=​316​−125​4−3k​​

Expanding along the first row,

det⁡(A)=3∣2−35k∣−(−1)∣1−36k∣+4∣1265∣\det(A)=3\begin{vmatrix}2 & -3 \\ 5 & k\end{vmatrix}-(-1)\begin{vmatrix}1 & -3 \\ 6 & k\end{vmatrix}+4\begin{vmatrix}1 & 2 \\ 6 & 5\end{vmatrix}det(A)=3​25​−3k​​−(−1)​16​−3k​​+4​16​25​​ =3(2k+15)+(k+18)+4(5−12)=3(2k+15)+(k+18)+4(5-12)=3(2k+15)+(k+18)+4(5−12) =6k+45+k+18−28=6k+45+k+18-28=6k+45+k+18−28 =7k+35=7(k+5).=7k+35=7(k+5).=7k+35=7(k+5).

For infinitely many solutions,

det⁡(A)=0⇒7(k+5)=0⇒k=−5.\det(A)=0 \Rightarrow 7(k+5)=0 \Rightarrow k=-5.det(A)=0⇒7(k+5)=0⇒k=−5.
  1. Now verify consistency when k=−5k=-5k=−5.

Substitute k=−5k=-5k=−5 in the third equation:

6x+5y−5z=−3.6x+5y-5z=-3.6x+5y−5z=−3.

Now check whether the third equation is a linear combination of the first two.

Let

a(3x−y+4z)+b(x+2y−3z)=6x+5y−5z.a(3x-y+4z)+b(x+2y-3z)=6x+5y-5z.a(3x−y+4z)+b(x+2y−3z)=6x+5y−5z.

Comparing coefficients:

3a+b=6,3a+b=6,3a+b=6, −a+2b=5.-a+2b=5.−a+2b=5.

Solving:

From 3a+b=63a+b=63a+b=6, we get

b=6−3a.b=6-3a.b=6−3a.

Substitute into the second:

−a+2(6−3a)=5-a+2(6-3a)=5−a+2(6−3a)=5 −a+12−6a=5-a+12-6a=5−a+12−6a=5 −7a=−7⇒a=1,-7a=-7 \Rightarrow a=1,−7a=−7⇒a=1,

so

b=6−3=3.b=6-3=3.b=6−3=3.

Check the zzz-coefficient:

4(1)−3(3)=4−9=−5,4(1)-3(3)=4-9=-5,4(1)−3(3)=4−9=−5,

which matches.

Check constants:

1⋅3+3⋅(−2)=3−6=−3,1\cdot 3+3\cdot(-2)=3-6=-3,1⋅3+3⋅(−2)=3−6=−3,

which also matches.

Hence,

Eq. 3=Eq. 1+3 Eq. 2.\text{Eq. 3} = \text{Eq. 1} + 3\,\text{Eq. 2}.Eq. 3=Eq. 1+3Eq. 2.

So the system is consistent and dependent, giving infinitely many solutions.

  1. Therefore, the required value is
k=−5.\boxed{k=-5}.k=−5​.

Thus the correct option is B.

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