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
610.5k+ 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.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Find the value of the expression given below sin 30circ class 11 maths CBSE

Difference between physical and chemical change class 11 chemistry CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

