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Matrices and Determinants question

2010 · Shift 0 · Q37
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  5. /2010 · Shift 0 · Q37

Matrices and Determinants question

2010 · Shift 0 · Q37

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Consider the system of linear equations; x1+2x2+x3=32x1+3x2+x3=33x1+5x2+2x3=1\begin{matrix} {{x_1} + 2{x_2} + {x_3} = 3} \\ {2{x_1} + 3{x_2} + {x_3} = 3} \\ {3{x_1} + 5{x_2} + 2{x_3} = 1} \\ \end{matrix}x1​+2x2​+x3​=32x1​+3x2​+x3​=33x1​+5x2​+2x3​=1​ The system has :
  1. A
    exactly 333 solutions
  2. B
    a unique solution
  3. C
    no solution
  4. D
    infinitenumber of solutions
View written solutionFree

Correct answer: C

  1. Write the system in matrix form

The given system is

{x1+2x2+x3=32x1+3x2+x3=33x1+5x2+2x3=1\begin{cases} x_1+2x_2+x_3=3 \\ 2x_1+3x_2+x_3=3 \\ 3x_1+5x_2+2x_3=1 \end{cases}⎩⎨⎧​x1​+2x2​+x3​=32x1​+3x2​+x3​=33x1​+5x2​+2x3​=1​

Its coefficient matrix is

A=(121231352)A=\begin{pmatrix} 1&2&1\\ 2&3&1\\ 3&5&2 \end{pmatrix}A=​123​235​112​​

and the augmented matrix is

[A∣B]=(121323133521).[A|B]=\begin{pmatrix} 1&2&1&3\\ 2&3&1&3\\ 3&5&2&1 \end{pmatrix}.[A∣B]=​123​235​112​331​​.
  1. Check whether the coefficient matrix is singular

Compute

det⁡(A)=∣121231352∣.\det(A)= \begin{vmatrix} 1&2&1\\ 2&3&1\\ 3&5&2 \end{vmatrix}.det(A)=​123​235​112​​.

Expanding along the first row,

det⁡(A)=1∣3152∣−2∣2132∣+1∣2335∣.\det(A)=1\begin{vmatrix}3&1\\5&2\end{vmatrix} -2\begin{vmatrix}2&1\\3&2\end{vmatrix} +1\begin{vmatrix}2&3\\3&5\end{vmatrix}.det(A)=1​35​12​​−2​23​12​​+1​23​35​​.

Now,

∣3152∣=6−5=1,\begin{vmatrix}3&1\\5&2\end{vmatrix}=6-5=1,​35​12​​=6−5=1, ∣2132∣=4−3=1,\begin{vmatrix}2&1\\3&2\end{vmatrix}=4-3=1,​23​12​​=4−3=1, ∣2335∣=10−9=1.\begin{vmatrix}2&3\\3&5\end{vmatrix}=10-9=1.​23​35​​=10−9=1.

So,

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

Since det⁡(A)=0\det(A)=0det(A)=0, the system does not have a unique solution. It may have either no solution or infinitely many solutions.

  1. Check consistency using row operations

Start with the augmented matrix:

(121323133521)\begin{pmatrix} 1&2&1&3\\ 2&3&1&3\\ 3&5&2&1 \end{pmatrix}​123​235​112​331​​

Apply:

R2→R2−2R1,R3→R3−3R1R_2 \to R_2-2R_1, \qquad R_3 \to R_3-3R_1R2​→R2​−2R1​,R3​→R3​−3R1​

Then we get

(12130−1−1−30−1−1−8).\begin{pmatrix} 1&2&1&3\\ 0&-1&-1&-3\\ 0&-1&-1&-8 \end{pmatrix}.​100​2−1−1​1−1−1​3−3−8​​.

Now apply:

R3→R3−R2R_3 \to R_3-R_2R3​→R3​−R2​

This gives

(12130−1−1−3000−5).\begin{pmatrix} 1&2&1&3\\ 0&-1&-1&-3\\ 0&0&0&-5 \end{pmatrix}.​100​2−10​1−10​3−3−5​​.

The last row represents

0x1+0x2+0x3=−5,0x_1+0x_2+0x_3=-5,0x1​+0x2​+0x3​=−5,

which is impossible.

Hence the system is inconsistent.

  1. Conclusion

Therefore, the system has no solution.

So the correct option is:

C: no solution\boxed{\text{C: no solution}}C: no solution​
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