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Matrices and Determinants question

2021 · 25 Feb · Shift 2 · Q39
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  5. /2021 · 25 Feb · Shift 2 · Q39

Matrices and Determinants question

2021 · 25 Feb · Shift 2 · Q39

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The following system of linear equations 2x + 3y + 2z = 9 3x + 2y + 2z = 9 x −-− y + 4z = 8
  1. A
    does not have any solution
  2. B
    has a solution (α\alphaα, β\betaβ, γ\gammaγ) satisfying α\alphaα+β\betaβ 2 + γ\gammaγ 3 = 12
  3. C
    has a unique solution
  4. D
    has infinitely many solutions
View written solutionFree

Correct answer: C

  1. Write the system

Given:

{2x+3y+2z=93x+2y+2z=9x−y+4z=8\begin{cases} 2x+3y+2z=9 \\ 3x+2y+2z=9 \\ x-y+4z=8 \end{cases}⎩⎨⎧​2x+3y+2z=93x+2y+2z=9x−y+4z=8​

We need to determine whether the system has no solution, a unique solution, or infinitely many solutions. If needed, we can also check option B.

  1. Form the coefficient matrix

The coefficient matrix is

A=(2323221−14)A=\begin{pmatrix} 2 & 3 & 2 \\ 3 & 2 & 2 \\ 1 & -1 & 4 \end{pmatrix}A=​231​32−1​224​​

If det⁡(A)≠0\det(A) \neq 0det(A)=0, then the system has a unique solution.

  1. Compute the determinant
det⁡(A)=∣2323221−14∣\det(A)= \begin{vmatrix} 2 & 3 & 2 \\ 3 & 2 & 2 \\ 1 & -1 & 4 \end{vmatrix}det(A)=​231​32−1​224​​

Expand along the first row:

det⁡(A)=2∣22−14∣−3∣3214∣+2∣321−1∣\det(A)=2\begin{vmatrix}2 & 2 \\ -1 & 4\end{vmatrix} -3\begin{vmatrix}3 & 2 \\ 1 & 4\end{vmatrix} +2\begin{vmatrix}3 & 2 \\ 1 & -1\end{vmatrix}det(A)=2​2−1​24​​−3​31​24​​+2​31​2−1​​

Now evaluate each minor:

∣22−14∣=2⋅4−2⋅(−1)=8+2=10\begin{vmatrix}2 & 2 \\ -1 & 4\end{vmatrix}=2\cdot 4-2\cdot(-1)=8+2=10​2−1​24​​=2⋅4−2⋅(−1)=8+2=10 ∣3214∣=3⋅4−2⋅1=12−2=10\begin{vmatrix}3 & 2 \\ 1 & 4\end{vmatrix}=3\cdot 4-2\cdot 1=12-2=10​31​24​​=3⋅4−2⋅1=12−2=10 ∣321−1∣=3⋅(−1)−2⋅1=−3−2=−5\begin{vmatrix}3 & 2 \\ 1 & -1\end{vmatrix}=3\cdot(-1)-2\cdot 1=-3-2=-5​31​2−1​​=3⋅(−1)−2⋅1=−3−2=−5

So,

det⁡(A)=2(10)−3(10)+2(−5)=20−30−10=−20\det(A)=2(10)-3(10)+2(-5)=20-30-10=-20det(A)=2(10)−3(10)+2(−5)=20−30−10=−20

Since

det⁡(A)=−20≠0,\det(A)=-20\neq 0,det(A)=−20=0,

the system has a unique solution.

Thus, option C is correct.

  1. Optional check by solving the system

Let us solve to verify option B as well.

Given:

2x+3y+2z=9...(1)2x+3y+2z=9 \quad ...(1)2x+3y+2z=9...(1) 3x+2y+2z=9...(2)3x+2y+2z=9 \quad ...(2)3x+2y+2z=9...(2) x−y+4z=8...(3)x-y+4z=8 \quad ...(3)x−y+4z=8...(3)

Subtract (1) from (2):

(3x+2y+2z)−(2x+3y+2z)=9−9(3x+2y+2z)-(2x+3y+2z)=9-9(3x+2y+2z)−(2x+3y+2z)=9−9 x−y=0x-y=0x−y=0 x=y...(4)x=y \quad ...(4)x=y...(4)

Using (3):

x−y+4z=8x-y+4z=8x−y+4z=8

From (4), x−y=0x-y=0x−y=0, so

4z=8⇒z=24z=8 \Rightarrow z=24z=8⇒z=2

Now put z=2z=2z=2 and x=yx=yx=y into (1):

2x+3x+2(2)=92x+3x+2(2)=92x+3x+2(2)=9 5x+4=95x+4=95x+4=9 5x=5⇒x=15x=5 \Rightarrow x=15x=5⇒x=1

So,

y=1y=1y=1

Hence the unique solution is

(x,y,z)=(1,1,2)(x,y,z)=(1,1,2)(x,y,z)=(1,1,2)

Now check option B:

α+β2+γ3=1+12+23=1+1+8=10\alpha+\beta^2+\gamma^3=1+1^2+2^3=1+1+8=10α+β2+γ3=1+12+23=1+1+8=10

This is not 12. So option B is false.

  1. Evaluate all options
  • A: does not have any solution — False
  • B: has a solution (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) satisfying α+β2+γ3=12\alpha+\beta^2+\gamma^3=12α+β2+γ3=12 — False
  • C: has a unique solution — True
  • D: has infinitely many solutions — False
  1. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

They agree.

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