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Question: For two \(3 \times 3\) matrices A and B , let A + B = 2B’ and 3A + 2B = \({I_3}\) where B’ is the tr...

For two 3×33 \times 3 matrices A and B , let A + B = 2B’ and 3A + 2B = I3{I_3} where B’ is the transpose of B and I3{I_3}is 3×33 \times 3 identity matrix . Then
A) 5A+10B=2I35A + 10B = 2{I_3}
B) 3A+6B=2I33A + 6B = 2{I_3}
C) 10A+5B=3I310A + 5B = 3{I_3}
D) B+2A=I3B + 2A = {I_3}

Explanation

Solution

Taking transposes on the given equations we get 3A′+2B′=I33{A ^{'}} + 2{B ^{'}} = {I_3} and A′=2B−B′{A ^{'}} = 2B - {B ^{'}} and simplifying further we get that A = B and I3{I_3}= 5A and using these values we can get the equation which satisfies these equations.

Complete step by step solution:
We are given that A + B = 2B’
Applying transpose on both sides
⇒(A+B)′=(2B′)′\Rightarrow {\left( {A + B} \right) ^{'}} = {\left( {2{B ^{'}}} \right) ^{'}}
We know that (A+B)′=A′+B′{\left( {A + B} \right) ^{'}} = {A ^{'}} + {B ^{'}} and (A′)′=A{\left( {{A ^{'}}} \right) ^{'}} = A
From this we get ,
⇒A′+B′=2B ⇒A′=2B−B′  \Rightarrow {A ^{'}} + {B ^{'}} = 2B \\\ \Rightarrow {A ^{'}} = 2B - {B ^{'}} \\\
We are given that 3A + 2B = I3{I_3}
Applying transpose on both sides we get
⇒(3A + 2B)′=(I3)′ ⇒3A′+2B′=I3  \Rightarrow {\left( {3A{\text{ }} + {\text{ }}2B} \right) ^{'}} = {\left( {{I_3}} \right) ^{'}} \\\ \Rightarrow 3{A ^{'}} + 2{B ^{'}} = {I_3} \\\
Substitute the value of A′{A ^{'}}in the above equation
⇒3(2B−B′)+2B′=I3 ⇒6B−3B′+2B′=I3 ⇒6B−B′=I3  \Rightarrow 3(2B - {B ^{'}}) + 2{B ^{'}} = {I_3} \\\ \Rightarrow 6B - 3{B ^{'}} + 2{B ^{'}} = {I_3} \\\ \Rightarrow 6B - {B ^{'}} = {I_3} \\\
We are given that A + B = 2B’
From this B′=A+B2{B ^{'}} = \dfrac{{A + B}}{2}
Substitute this in the previous equation
⇒6B−(A+B2)=I3 ⇒12B−A−B=2I3  \Rightarrow 6B - \left( {\dfrac{{A + B}}{2}} \right) = {I_3} \\\ \Rightarrow 12B - A - B = 2{I_3} \\\
Substitute the value of I3{I_3}from the given equation 3A + 2B = I3{I_3}
⇒12B−A−B=2(3A+2B) ⇒11B−A=6A+4B ⇒7A=7B ⇒A=B  \Rightarrow 12B - A - B = 2(3A + 2B) \\\ \Rightarrow 11B - A = 6A + 4B \\\ \Rightarrow 7A = 7B \\\ \Rightarrow A = B \\\
Substituting in 3A + 2B = I3{I_3}we get
⇒3A+2A=I3 ⇒5A=I3  \Rightarrow 3A + 2A = {I_3} \\\ \Rightarrow 5A = {I_3} \\\
Now we have A = B and I3{I_3}= 5A
We have asked 5A +10B
⇒5A+10B=5A+10A=15A=3×5A=3I3\Rightarrow 5A + 10B = 5A + 10A = 15A = 3 \times 5A = 3{I_3}

Therefore the correct option is C.

Note:

  1. In mathematics, a matrix (plural: matrices) is a rectangular table of cells of numbers, with rows and columns.
  2. Every square dimension set of a matrix has a special counterpart called an "identity matrix". The identity matrix has nothing but zeroes except on the main diagonal, where there are all ones.
  3. An inverse matrix is a matrix that, when multiplied by another matrix, equals the identity matrix.