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Matrices and Determinants question

2021 · 18 Mar · Shift 1 · Q37
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Matrices and Determinants question

2021 · 18 Mar · Shift 1 · Q37

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A+2B=[1206−33−531]A + 2B = \left[ {\begin{matrix} 1 & 2 & 0 \\ 6 & { - 3} & 3 \\ { - 5} & 3 & 1 \\ \end{matrix} } \right]A+2B=​16−5​2−33​031​​ and 2A−B=[2−152−16012]2A - B = \left[ {\begin{matrix} 2 & { - 1} & 5 \\ 2 & { - 1} & 6 \\ 0 & 1 & 2 \\ \end{matrix} } \right]2A−B=​220​−1−11​562​​. If Tr(A) denotes the sum of all diagonal elements of the matrix A, then Tr(A) −-− Tr(B) has value equal to
  1. A
    1
  2. B
    2
  3. C
    0
  4. D
    3
View written solutionFree

Correct answer: B

  1. Let M1=A+2B=[1206−33−531],M2=2A−B=[2−152−16012].M_1=A+2B=\begin{bmatrix}1&2&0\\6&-3&3\\-5&3&1\end{bmatrix},\qquad M_2=2A-B=\begin{bmatrix}2&-1&5\\2&-1&6\\0&1&2\end{bmatrix}.M1​=A+2B=​16−5​2−33​031​​,M2​=2A−B=​220​−1−11​562​​.

We need to find Tr⁡(A)−Tr⁡(B)=Tr⁡(A−B).\operatorname{Tr}(A)-\operatorname{Tr}(B)=\operatorname{Tr}(A-B).Tr(A)−Tr(B)=Tr(A−B).

  1. Use linearity of trace: Tr⁡(A+2B)=Tr⁡(A)+2Tr⁡(B),\operatorname{Tr}(A+2B)=\operatorname{Tr}(A)+2\operatorname{Tr}(B),Tr(A+2B)=Tr(A)+2Tr(B), Tr⁡(2A−B)=2Tr⁡(A)−Tr⁡(B).\operatorname{Tr}(2A-B)=2\operatorname{Tr}(A)-\operatorname{Tr}(B).Tr(2A−B)=2Tr(A)−Tr(B).

So first compute the traces of the given matrices.

  1. Trace of M1M_1M1​: Tr⁡(M1)=1+(−3)+1=−1.\operatorname{Tr}(M_1)=1+(-3)+1=-1.Tr(M1​)=1+(−3)+1=−1. Hence, Tr⁡(A)+2Tr⁡(B)=−1.\operatorname{Tr}(A)+2\operatorname{Tr}(B)=-1. Tr(A)+2Tr(B)=−1.

  2. Trace of M2M_2M2​: Tr⁡(M2)=2+(−1)+2=3.\operatorname{Tr}(M_2)=2+(-1)+2=3.Tr(M2​)=2+(−1)+2=3. Hence, 2Tr⁡(A)−Tr⁡(B)=3.2\operatorname{Tr}(A)-\operatorname{Tr}(B)=3. 2Tr(A)−Tr(B)=3.

  3. Let x=Tr⁡(A),y=Tr⁡(B).x=\operatorname{Tr}(A),\qquad y=\operatorname{Tr}(B).x=Tr(A),y=Tr(B). Then we have the system x+2y=−1,x+2y=-1,x+2y=−1, 2x−y=3.2x-y=3.2x−y=3.

  4. Solve the system: From the first equation, x=−1−2y.x=-1-2y.x=−1−2y. Substitute into the second: 2(−1−2y)−y=32(-1-2y)-y=32(−1−2y)−y=3 −2−4y−y=3-2-4y-y=3−2−4y−y=3 −5y=5-5y=5−5y=5 y=−1.y=-1.y=−1. Then x=−1−2(−1)=1.x=-1-2(-1)=1.x=−1−2(−1)=1.

  5. Therefore, Tr⁡(A)−Tr⁡(B)=x−y=1−(−1)=2.\operatorname{Tr}(A)-\operatorname{Tr}(B)=x-y=1-(-1)=2.Tr(A)−Tr(B)=x−y=1−(−1)=2.

  6. Checking options:

  • A: 111 ❌
  • B: 222 ✅
  • C: 000 ❌
  • D: 333 ❌

So the correct answer is Option B.

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