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Matrices and Determinants question

2017 · 9 Apr · Shift 1 · Q27
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Matrices and Determinants question

2017 · 9 Apr · Shift 1 · Q27

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
For two 3 × 3 matrices A and B, let A + B = 2BT and 3A + 2B = I3, where BT is the transpose of B and I3 is 3 × 3 identity matrix. Then :
  1. A
    5A + 10B = 2I3
  2. B
    10A + 5B = 3I3
  3. C
    B + 2A = I3
  4. D
    3A + 6B = 2I3
View written solutionFree

Correct answer: B

  1. We are given A+B=2BT ,3A+2B=I3.A+B=2B^T \, , \qquad 3A+2B=I_3.A+B=2BT,3A+2B=I3​.

We need to determine which option must be true.

  1. First, use the first equation: A+B=2BT.A+B=2B^T.A+B=2BT. Transpose both sides: (A+B)T=(2BT)T.(A+B)^T=(2B^T)^T.(A+B)T=(2BT)T. So, AT+BT=2B.A^T+B^T=2B.AT+BT=2B.

But from the original equation, A=2BT−B.A=2B^T-B.A=2BT−B. Taking transpose, AT=2B−BT.A^T=2B-B^T.AT=2B−BT.

Now substitute into AT+BT=2B:A^T+B^T=2B:AT+BT=2B: (2B−BT)+BT=2B,(2B-B^T)+B^T=2B,(2B−BT)+BT=2B, which is an identity. So the first relation gives A=2BT−B.A=2B^T-B.A=2BT−B.

  1. Substitute this into the second equation: 3A+2B=I3.3A+2B=I_3.3A+2B=I3​. Using A=2BT−BA=2B^T-BA=2BT−B, 3(2BT−B)+2B=I3,3(2B^T-B)+2B=I_3,3(2BT−B)+2B=I3​, 6BT−3B+2B=I3,6B^T-3B+2B=I_3,6BT−3B+2B=I3​, 6B^T-B=I_3. \tag{1}

Transpose (1): 6B-B^T=I_3. \tag{2}

  1. Solve equations (1) and (2): (1):  6BT−B=I3(1): \; 6B^T-B=I_3(1):6BT−B=I3​ (2):  6B−BT=I3(2): \; 6B-B^T=I_3(2):6B−BT=I3​

Multiply (2) by 6: 36B−6BT=6I3.36B-6B^T=6I_3.36B−6BT=6I3​. Add with (1): (36B−6BT)+(6BT−B)=6I3+I3,(36B-6B^T)+(6B^T-B)=6I_3+I_3,(36B−6BT)+(6BT−B)=6I3​+I3​, 35B=7I3,35B=7I_3,35B=7I3​, B=15I3.B=\frac{1}{5}I_3.B=51​I3​.

Then from 3A+2B=I3,3A+2B=I_3,3A+2B=I3​, 3A+25I3=I3,3A+\frac{2}{5}I_3=I_3,3A+52​I3​=I3​, 3A=35I3,3A=\frac{3}{5}I_3,3A=53​I3​, A=15I3.A=\frac{1}{5}I_3.A=51​I3​.

So, A=B=15I3.A=B=\frac{1}{5}I_3.A=B=51​I3​.

  1. Now check each option.

Option A: 5A+10B=5⋅15I3+10⋅15I3=I3+2I3=3I3≠2I3.5A+10B=5\cdot \frac{1}{5}I_3+10\cdot \frac{1}{5}I_3=I_3+2I_3=3I_3 \neq 2I_3.5A+10B=5⋅51​I3​+10⋅51​I3​=I3​+2I3​=3I3​=2I3​. So A is false.

Option B: 10A+5B=10⋅15I3+5⋅15I3=2I3+I3=3I3.10A+5B=10\cdot \frac{1}{5}I_3+5\cdot \frac{1}{5}I_3=2I_3+I_3=3I_3.10A+5B=10⋅51​I3​+5⋅51​I3​=2I3​+I3​=3I3​. So B is true.

Option C: B+2A=15I3+2⋅15I3=35I3≠I3.B+2A=\frac{1}{5}I_3+2\cdot \frac{1}{5}I_3=\frac{3}{5}I_3 \neq I_3.B+2A=51​I3​+2⋅51​I3​=53​I3​=I3​. So C is false.

Option D: 3A+6B=3⋅15I3+6⋅15I3=95I3≠2I3.3A+6B=3\cdot \frac{1}{5}I_3+6\cdot \frac{1}{5}I_3=\frac{9}{5}I_3 \neq 2I_3.3A+6B=3⋅51​I3​+6⋅51​I3​=59​I3​=2I3​. So D is false.

  1. Therefore the only correct option is B.\boxed{\text{B}}.B​.
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