
For two $3 \times 3$ matrices A and B , let A + B = 2B’ and 3A + 2B = ${I_3}$ where B’ is the transpose of B and ${I_3}$is $3 \times 3$ identity matrix . Then
A) $5A + 10B = 2{I_3}$
B) $3A + 6B = 2{I_3}$
C) $10A + 5B = 3{I_3}$
D) $B + 2A = {I_3}$
Answer
483.9k+ views
Hint:
Taking transposes on the given equations we get $3{A ^{'}} + 2{B ^{'}} = {I_3}$ and ${A ^{'}} = 2B - {B ^{'}}$ and simplifying further we get that A = B and ${I_3}$= 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
$ \Rightarrow {\left( {A + B} \right) ^{'}} = {\left( {2{B ^{'}}} \right) ^{'}}$
We know that ${\left( {A + B} \right) ^{'}} = {A ^{'}} + {B ^{'}}$ and ${\left( {{A ^{'}}} \right) ^{'}} = A$
From this we get ,
$
\Rightarrow {A ^{'}} + {B ^{'}} = 2B \\
\Rightarrow {A ^{'}} = 2B - {B ^{'}} \\
$
We are given that 3A + 2B = ${I_3}$
Applying transpose on both sides we get
$
\Rightarrow {\left( {3A{\text{ }} + {\text{ }}2B} \right) ^{'}} = {\left( {{I_3}} \right) ^{'}} \\
\Rightarrow 3{A ^{'}} + 2{B ^{'}} = {I_3} \\
$
Substitute the value of ${A ^{'}}$in the above equation
$
\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 ${B ^{'}} = \dfrac{{A + B}}{2}$
Substitute this in the previous equation
$
\Rightarrow 6B - \left( {\dfrac{{A + B}}{2}} \right) = {I_3} \\
\Rightarrow 12B - A - B = 2{I_3} \\
$
Substitute the value of ${I_3}$from the given equation 3A + 2B = ${I_3}$
$
\Rightarrow 12B - A - B = 2(3A + 2B) \\
\Rightarrow 11B - A = 6A + 4B \\
\Rightarrow 7A = 7B \\
\Rightarrow A = B \\
$
Substituting in 3A + 2B = ${I_3}$we get
$
\Rightarrow 3A + 2A = {I_3} \\
\Rightarrow 5A = {I_3} \\
$
Now we have A = B and ${I_3}$= 5A
We have asked 5A +10B
$ \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.
Taking transposes on the given equations we get $3{A ^{'}} + 2{B ^{'}} = {I_3}$ and ${A ^{'}} = 2B - {B ^{'}}$ and simplifying further we get that A = B and ${I_3}$= 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
$ \Rightarrow {\left( {A + B} \right) ^{'}} = {\left( {2{B ^{'}}} \right) ^{'}}$
We know that ${\left( {A + B} \right) ^{'}} = {A ^{'}} + {B ^{'}}$ and ${\left( {{A ^{'}}} \right) ^{'}} = A$
From this we get ,
$
\Rightarrow {A ^{'}} + {B ^{'}} = 2B \\
\Rightarrow {A ^{'}} = 2B - {B ^{'}} \\
$
We are given that 3A + 2B = ${I_3}$
Applying transpose on both sides we get
$
\Rightarrow {\left( {3A{\text{ }} + {\text{ }}2B} \right) ^{'}} = {\left( {{I_3}} \right) ^{'}} \\
\Rightarrow 3{A ^{'}} + 2{B ^{'}} = {I_3} \\
$
Substitute the value of ${A ^{'}}$in the above equation
$
\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 ${B ^{'}} = \dfrac{{A + B}}{2}$
Substitute this in the previous equation
$
\Rightarrow 6B - \left( {\dfrac{{A + B}}{2}} \right) = {I_3} \\
\Rightarrow 12B - A - B = 2{I_3} \\
$
Substitute the value of ${I_3}$from the given equation 3A + 2B = ${I_3}$
$
\Rightarrow 12B - A - B = 2(3A + 2B) \\
\Rightarrow 11B - A = 6A + 4B \\
\Rightarrow 7A = 7B \\
\Rightarrow A = B \\
$
Substituting in 3A + 2B = ${I_3}$we get
$
\Rightarrow 3A + 2A = {I_3} \\
\Rightarrow 5A = {I_3} \\
$
Now we have A = B and ${I_3}$= 5A
We have asked 5A +10B
$ \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.
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