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Matrices and Determinants question

2019 · 11 Jan · Shift 1 · Q33
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Matrices and Determinants question

2019 · 11 Jan · Shift 1 · Q33

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If the system of linear equations 2x + 2y + 3z = a 3x – y + 5z = b x – 3y + 2z = c where a, b, c are non zero real numbers, has more one solution, then :
  1. A
    b – c – a = 0
  2. B
    a + b + c = 0
  3. C
    b – c + a = 0
  4. D
    b + c – a = 0
View written solutionFree

Correct answer: A

  1. Write the system in matrix form:
{2x+2y+3z=a3x−y+5z=bx−3y+2z=c\begin{cases} 2x+2y+3z=a\\ 3x-y+5z=b\\ x-3y+2z=c \end{cases}⎩⎨⎧​2x+2y+3z=a3x−y+5z=bx−3y+2z=c​

Coefficient matrix is

A=(2233−151−32)A=\begin{pmatrix} 2&2&3\\ 3&-1&5\\ 1&-3&2 \end{pmatrix}A=​231​2−1−3​352​​

For the system to have more than one solution, it must have infinitely many solutions. Hence:

  • det⁡(A)=0\det(A)=0det(A)=0,
  • and the equations must be consistent.
  1. Compute det⁡(A)\det(A)det(A):
det⁡(A)=2∣−15−32∣−2∣3512∣+3∣3−11−3∣\det(A)= 2\begin{vmatrix}-1&5\\-3&2\end{vmatrix} -2\begin{vmatrix}3&5\\1&2\end{vmatrix} +3\begin{vmatrix}3&-1\\1&-3\end{vmatrix}det(A)=2​−1−3​52​​−2​31​52​​+3​31​−1−3​​

Now,

∣−15−32∣=(−1)(2)−5(−3)=−2+15=13\begin{vmatrix}-1&5\\-3&2\end{vmatrix}=(-1)(2)-5(-3)=-2+15=13​−1−3​52​​=(−1)(2)−5(−3)=−2+15=13 ∣3512∣=3⋅2−5⋅1=6−5=1\begin{vmatrix}3&5\\1&2\end{vmatrix}=3\cdot 2-5\cdot 1=6-5=1​31​52​​=3⋅2−5⋅1=6−5=1 ∣3−11−3∣=3(−3)−(−1)(1)=−9+1=−8\begin{vmatrix}3&-1\\1&-3\end{vmatrix}=3(-3)-(-1)(1)=-9+1=-8​31​−1−3​​=3(−3)−(−1)(1)=−9+1=−8

Therefore,

det⁡(A)=2(13)−2(1)+3(−8)=26−2−24=0\det(A)=2(13)-2(1)+3(-8)=26-2-24=0det(A)=2(13)−2(1)+3(−8)=26−2−24=0

So the coefficient matrix is singular, and infinite solutions are possible only if the constants satisfy the same linear dependence.

  1. Find the dependence among the rows.

Let

λ1(2,2,3)+λ2(3,−1,5)+λ3(1,−3,2)=(0,0,0)\lambda_1(2,2,3)+\lambda_2(3,-1,5)+\lambda_3(1,-3,2)=(0,0,0)λ1​(2,2,3)+λ2​(3,−1,5)+λ3​(1,−3,2)=(0,0,0)

This gives

2λ1+3λ2+λ3=02\lambda_1+3\lambda_2+\lambda_3=02λ1​+3λ2​+λ3​=0 2λ1−λ2−3λ3=02\lambda_1-\lambda_2-3\lambda_3=02λ1​−λ2​−3λ3​=0 3λ1+5λ2+2λ3=03\lambda_1+5\lambda_2+2\lambda_3=03λ1​+5λ2​+2λ3​=0

Try solving the first two:

From

2λ1+3λ2+λ3=0...(1)2\lambda_1+3\lambda_2+\lambda_3=0 \quad ...(1)2λ1​+3λ2​+λ3​=0...(1) 2λ1−λ2−3λ3=0...(2)2\lambda_1-\lambda_2-3\lambda_3=0 \quad ...(2)2λ1​−λ2​−3λ3​=0...(2)

Subtract (2) from (1):

4λ2+4λ3=0⇒λ2=−λ34\lambda_2+4\lambda_3=0 \Rightarrow \lambda_2=-\lambda_34λ2​+4λ3​=0⇒λ2​=−λ3​

Put into (1):

2λ1+3(−λ3)+λ3=02\lambda_1+3(-\lambda_3)+\lambda_3=02λ1​+3(−λ3​)+λ3​=0 2λ1−2λ3=0⇒λ1=λ32\lambda_1-2\lambda_3=0 \Rightarrow \lambda_1=\lambda_32λ1​−2λ3​=0⇒λ1​=λ3​

So one relation is

λ1=1,λ2=−1,λ3=1\lambda_1=1,\quad \lambda_2=-1,\quad \lambda_3=1λ1​=1,λ2​=−1,λ3​=1

Hence,

R1−R2+R3=0R_1-R_2+R_3=0R1​−R2​+R3​=0

Check:

(2,2,3)−(3,−1,5)+(1,−3,2)=(0,0,0)(2,2,3)-(3,-1,5)+(1,-3,2)=(0,0,0)(2,2,3)−(3,−1,5)+(1,−3,2)=(0,0,0)

Correct.

  1. For consistency, the constants must satisfy the same relation:
a−b+c=0a-b+c=0a−b+c=0

Rearranging,

a+c−b=0⇒b−c−a=0a+c-b=0 \Rightarrow b-c-a=0a+c−b=0⇒b−c−a=0

since

a−b+c=0  ⟺  a+c=b  ⟺  b−c−a=0a-b+c=0 \iff a+c=b \iff b-c-a=0a−b+c=0⟺a+c=b⟺b−c−a=0
  1. Compare with options:
  • A: b−c−a=0b-c-a=0b−c−a=0 ✅
  • B: a+b+c=0a+b+c=0a+b+c=0 ❌
  • C: b−c+a=0b-c+a=0b−c+a=0 ❌
  • D: b+c−a=0b+c-a=0b+c−a=0 ❌

Therefore, the correct option is A.

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