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Matrices and Determinants question

2021 · 25 Feb · Shift 1 · Q42
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Matrices and Determinants question

2021 · 25 Feb · Shift 1 · Q42

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
If the system of equations kx + y + 2z = 1 3x −-− y −-− 2z = 2 −-− 2x −-− 2y −-− 4z = 3 has infinitely many solutions, then k is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 21

  1. Write the system in standard form

The equations are:

kx+y+2z=13x−y−2z=2−2x−2y−4z=3\begin{aligned} kx+y+2z&=1 \\ 3x-y-2z&=2 \\ -2x-2y-4z&=3 \end{aligned}kx+y+2z3x−y−2z−2x−2y−4z​=1=2=3​

So the coefficient matrix is

A=(k123−1−2−2−2−4)A=\begin{pmatrix} k&1&2\\ 3&-1&-2\\ -2&-2&-4 \end{pmatrix}A=​k3−2​1−1−2​2−2−4​​

and the augmented matrix is

[A∣B]=(k1213−1−22−2−2−43).[A|B]=\begin{pmatrix} k&1&2&1\\ 3&-1&-2&2\\ -2&-2&-4&3 \end{pmatrix}.[A∣B]=​k3−2​1−1−2​2−2−4​123​​.
  1. Condition for infinitely many solutions

A system has infinitely many solutions when

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)=∣k123−1−2−2−2−4∣\det(A)=\begin{vmatrix} k&1&2\\ 3&-1&-2\\ -2&-2&-4 \end{vmatrix}det(A)=​k3−2​1−1−2​2−2−4​​

Expand along the first row:

det⁡(A)=k∣−1−2−2−4∣−1∣3−2−2−4∣+2∣3−1−2−2∣\det(A)=k\begin{vmatrix}-1&-2\\-2&-4\end{vmatrix}-1\begin{vmatrix}3&-2\\-2&-4\end{vmatrix}+2\begin{vmatrix}3&-1\\-2&-2\end{vmatrix}det(A)=k​−1−2​−2−4​​−1​3−2​−2−4​​+2​3−2​−1−2​​

Now,

∣−1−2−2−4∣=(−1)(−4)−(−2)(−2)=4−4=0\begin{vmatrix}-1&-2\\-2&-4\end{vmatrix}=(-1)(-4)-(-2)(-2)=4-4=0​−1−2​−2−4​​=(−1)(−4)−(−2)(−2)=4−4=0 ∣3−2−2−4∣=3(−4)−(−2)(−2)=−12−4=−16\begin{vmatrix}3&-2\\-2&-4\end{vmatrix}=3(-4)-(-2)(-2)=-12-4=-16​3−2​−2−4​​=3(−4)−(−2)(−2)=−12−4=−16 ∣3−1−2−2∣=3(−2)−(−1)(−2)=−6−2=−8\begin{vmatrix}3&-1\\-2&-2\end{vmatrix}=3(-2)-(-1)(-2)=-6-2=-8​3−2​−1−2​​=3(−2)−(−1)(−2)=−6−2=−8

Hence,

det⁡(A)=k(0)−1(−16)+2(−8)=16−16=0.\det(A)=k(0)-1(-16)+2(-8)=16-16=0.det(A)=k(0)−1(−16)+2(−8)=16−16=0.

So det⁡(A)=0\det(A)=0det(A)=0 for all kkk.

  1. Check consistency using row operations

Start with the augmented matrix:

(k1213−1−22−2−2−43)\begin{pmatrix} k&1&2&1\\ 3&-1&-2&2\\ -2&-2&-4&3 \end{pmatrix}​k3−2​1−1−2​2−2−4​123​​

Notice from the coefficients of the 2nd and 3rd equations:

(−2,−2,−4)(-2,-2,-4)(−2,−2,−4)

is not a scalar multiple of

(3,−1,−2),(3,-1,-2),(3,−1,−2),

and also not generally of the first row unless k=−1k=-1k=−1.

Let us test the special value k=−1k=-1k=−1. Then the first row becomes

(−1,1,2∣1).(-1,1,2|1).(−1,1,2∣1).

Multiplying by 222,

2R1=(−2,2,4∣2),2R_1=(-2,2,4|2),2R1​=(−2,2,4∣2),

which is still not equal to R3=(−2,−2,−4∣3)R_3=(-2,-2,-4|3)R3​=(−2,−2,−4∣3). So even then, rows are not dependent in the augmented sense.

A cleaner way is to compute a 2×22\times 22×2 minor of AAA:

∣3−1−2−2∣=3(−2)−(−1)(−2)=−6−2=−8≠0.\begin{vmatrix} 3&-1\\ -2&-2 \end{vmatrix}=3(-2)-(-1)(-2)=-6-2=-8\neq 0.​3−2​−1−2​​=3(−2)−(−1)(−2)=−6−2=−8=0.

Therefore,

rank⁡(A)≥2.\operatorname{rank}(A)\ge 2.rank(A)≥2.

Now check a 3×33\times 33×3 minor of the augmented matrix using columns 1,2,4:

∣k113−12−2−23∣\begin{vmatrix} k&1&1\\ 3&-1&2\\ -2&-2&3 \end{vmatrix}​k3−2​1−1−2​123​​

Compute it:

=k∣−12−23∣−1∣32−23∣+1∣3−1−2−2∣= k\begin{vmatrix}-1&2\\-2&3\end{vmatrix}-1\begin{vmatrix}3&2\\-2&3\end{vmatrix}+1\begin{vmatrix}3&-1\\-2&-2\end{vmatrix}=k​−1−2​23​​−1​3−2​23​​+1​3−2​−1−2​​ =k((−1)(3)−2(−2))−(3⋅3−2(−2))+(3(−2)−(−1)(−2))= k((-1)(3)-2(-2))-(3\cdot 3-2(-2))+(3(-2)-(-1)(-2))=k((−1)(3)−2(−2))−(3⋅3−2(−2))+(3(−2)−(−1)(−2)) =k(−3+4)−(9+4)+(−6−2)= k(-3+4)-(9+4)+(-6-2)=k(−3+4)−(9+4)+(−6−2) =k−13−8=k−21.= k-13-8=k-21.=k−13−8=k−21.

For infinitely many solutions, we need rank of augmented matrix also to be less than 3, so this determinant must be zero:

k−21=0k-21=0k−21=0

Thus,

k=21.k=21.k=21.
  1. Verify ranks for k=21k=21k=21

For k=21k=21k=21, we already know det⁡(A)=0\det(A)=0det(A)=0, so rank⁡(A)<3\operatorname{rank}(A)<3rank(A)<3. Also the above 3×33\times 33×3 minor of augmented matrix becomes zero. Since a 2×22\times 22×2 minor of AAA is nonzero, rank is at least 2. Hence

rank⁡(A)=rank⁡([A∣B])=2<3.\operatorname{rank}(A)=\operatorname{rank}([A|B])=2<3.rank(A)=rank([A∣B])=2<3.

So the system has infinitely many solutions.

Therefore, the required value is

21.\boxed{21}.21​.
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