Consider the two matrices \[A\] and \[B\] where \[A = \left[ {\begin{array}{*{20}{c}}1&2\\4&3\end{array}} \right]\] ; \[B = \left[ {\begin{array}{*{20}{c}}5\\{ - 3}\end{array}} \right]\] . Let \[n\left( A \right)\] denote the number of elements in \[A\]. When the two matrices \[X\] and \[Y\] are not conformable for multiplication, then \[n\left( {XY} \right) = 0\]
If \[C = \left( {AB} \right)\left( {B'A} \right)\]; \[D = \left( {B'A} \right)\left( {AB} \right)\] , then find the value of
\[\left[ {\dfrac{{n\left( C \right)\left( {{{\left| D \right|}^2} + n\left( D \right)} \right)}}{{n\left( A \right) - n\left( B \right)}}} \right]\]
Answer
Verified633k+ views
Hint: Here, we need to find the value of the given expression which includes the multiplication of the matrices. We will first find the transpose of the second matrix. Then we will find the number of elements of the required matrices. We will substitute all the values in the given expression to get the required value.
Complete step-by-step answer:
It is given that:
\[A = \left[ {\begin{array}{*{20}{c}}1&2\\4&3\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}5\\{ - 3}\end{array}} \right]\].
We know the number of elements present in any matrix is equal to the product of the number of rows and number of columns of the matrix.
\[n\left( A \right) = 2 \times 2 = 4\] ………. \[\left( 1 \right)\]
\[n\left( B \right) = 2 \times 1 = 2\] ………… \[\left( 2 \right)\]
Now, we will find the transpose of the matrix \[B\] which is denoted by \[B'\] which can be calculated by interchanging their rows and columns.
Therefore,
\[B' = \left[ {\begin{array}{*{20}{c}}5&{ - 3}\end{array}} \right]\]
It is given that \[C = \left( {AB} \right)\left( {B'A} \right)\].
But we need only \[n\left( C \right)\] in the expression, which means the number of elements present in the matrix.
For that, we will calculate the order of the matrix \[C\]
We know that when we multiply two matrices, then the matrix obtained by their product contains a number of rows equal to the number of rows of the first matrix and number of columns equal to the number of columns of the second matrix.
We will use the same concept now.
We know that the order of matrix \[A\] is \[2 \times 2\] and order of matrix \[B\] is \[2 \times 1\]
So, the order of the matrix obtained by their product is equal to \[2 \times 1\].
Therefore, the order of the matrix \[AB\] is equal to \[2 \times 1\].
We know that the order of matrix \[A\] is \[2 \times 2\] and order of matrix \[B'\] is \[1 \times 2\]
So, the order of the matrix obtained by their product i.e. \[B'A\] is equal to \[1 \times 2\].
Therefore, the order of the matrix \[B'A\] is equal to \[1 \times 2\].
Now, we need the order of the matrix \[C\] which is equal to \[\left( {AB} \right)\left( {B'A} \right)\]
As the order of matrix \[AB\] is equal to \[2 \times 1\] and order of matrix \[B'A\] is equal to \[1 \times 2\].
So the order of the matrix obtained by their product i.e. \[\left( {AB} \right)\left( {B'A} \right)\] is equal to \[2 \times 2\].
We know the number of elements present in any matrix is equal to the product of the number of rows and number of columns of the matrix.
Therefore, the number of elements of the matrix \[C\] is equal to \[2 \times 2 = 4\] .
Thus,
\[n\left( C \right) = 4\] ……….. \[\left( 3 \right)\]
Now, we need to find the matrix \[D\] which is equal to \[\left( {B'A} \right)\left( {AB} \right)\]
For that, we will find the product of \[B'A\].
\[B'A = {\left[ {\begin{array}{*{20}{c}}5&{ - 3}\end{array}} \right]_{1 \times 2}} \times {\left[ {\begin{array}{*{20}{c}}1&2\\4&3\end{array}} \right]_{2 \times 2}}\]
On multiplying these two matrices, we get
\[ \Rightarrow B'A = {\left[ {\begin{array}{*{20}{c}}{5 + \left( { - 3} \right) \times 4}&{5 \times 2 + \left( { - 3} \right) \times 3}\end{array}} \right]_{1 \times 2}}\]
On further simplification, we get
\[ \Rightarrow B'A = {\left[ {\begin{array}{*{20}{c}}{ - 7}&1\end{array}} \right]_{1 \times 2}}\]
Now, we will find the product of \[AB\].
\[AB = {\left[ {\begin{array}{*{20}{c}}1&2\\4&3\end{array}} \right]_{2 \times 2}} \times {\left[ {\begin{array}{*{20}{c}}5\\{ - 3}\end{array}} \right]_{2 \times 1}}\]
On multiplying these two matrices, we get
\[ \Rightarrow AB = {\left[ {\begin{array}{*{20}{c}}{5 + 2 \times - 3}\\{4 \times 5 + 3 \times - 3}\end{array}} \right]_{2 \times 1}}\]
On further simplification, we get
\[ \Rightarrow AB = {\left[ {\begin{array}{*{20}{c}}{ - 1}\\{11}\end{array}} \right]_{2 \times 1}}\]
Now, we will find \[\left( {B'A} \right)\left( {AB} \right)\].
\[\left( {B'A} \right)\left( {AB} \right) = {\left[ {\begin{array}{*{20}{c}}{ - 7}&1\end{array}} \right]_{1 \times 2}} \times {\left[ {\begin{array}{*{20}{c}}{ - 1}\\{11}\end{array}} \right]_{2 \times 1}}\]
On multiplying these two matrices, we get
\[ \Rightarrow \left( {B'A} \right)\left( {AB} \right) = {\left[ {\left( { - 7} \right) \times \left( { - 1} \right) + 1 \times 11} \right]_{1 \times 1}}\]
On further simplification, we get
\[ \Rightarrow D = \left( {B'A} \right)\left( {AB} \right) = {\left[ {18} \right]_{1 \times 1}}\]
Therefore,
\[ \Rightarrow n\left( D \right) = 1 \times 1 = 1\] …….. \[\left( 4 \right)\]
Now, we will find the determinant of matrix \[D\] i.e. \[\left| D \right|\] but we also know that the determinant of the matrix of order \[1 \times 1\] is equal to its element.
Therefore,
\[\left| D \right| = 18\]
Now, we will calculate the square of it.
\[ \Rightarrow {\left| D \right|^2} = {\left( {18} \right)^2}\]
On further simplification, we get
\[ \Rightarrow {\left| D \right|^2} = 324\] …….. \[\left( 5 \right)\]
Now, we will calculate the value of the given expression i.e. \[\left[ {\dfrac{{n\left( C \right)\left( {{{\left| D \right|}^2} + n\left( D \right)} \right)}}{{n\left( A \right) - n\left( B \right)}}} \right]\]
Now, we will substitute all the values calculated in equation 1, equation 2, equation 3, equation 4 and equation 5.
\[\left[ {\dfrac{{n\left( C \right)\left( {{{\left| D \right|}^2} + n\left( D \right)} \right)}}{{n\left( A \right) - n\left( B \right)}}} \right] = \dfrac{{4 \times \left( {324 + 1} \right)}}{{4 - 2}}\]
On simplifying the terms, we get
\[ \Rightarrow \left[ {\dfrac{{n\left( C \right)\left( {{{\left| D \right|}^2} + n\left( D \right)} \right)}}{{n\left( A \right) - n\left( B \right)}}} \right] = 2 \times 325 = 650\]
Hence, the required value of the given expression is equal to 650.
Note: Matrix is defined as a set of numbers which are arranged in a particular row and columns. We need to keep in mind that when we multiply two matrices, then the matrix obtained by their product contains a number of rows equal to the number of rows of the first matrix and number of columns equal to the number of columns of the second matrix. Multiplication of matrices is not commutative.
Complete step-by-step answer:
It is given that:
\[A = \left[ {\begin{array}{*{20}{c}}1&2\\4&3\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}5\\{ - 3}\end{array}} \right]\].
We know the number of elements present in any matrix is equal to the product of the number of rows and number of columns of the matrix.
\[n\left( A \right) = 2 \times 2 = 4\] ………. \[\left( 1 \right)\]
\[n\left( B \right) = 2 \times 1 = 2\] ………… \[\left( 2 \right)\]
Now, we will find the transpose of the matrix \[B\] which is denoted by \[B'\] which can be calculated by interchanging their rows and columns.
Therefore,
\[B' = \left[ {\begin{array}{*{20}{c}}5&{ - 3}\end{array}} \right]\]
It is given that \[C = \left( {AB} \right)\left( {B'A} \right)\].
But we need only \[n\left( C \right)\] in the expression, which means the number of elements present in the matrix.
For that, we will calculate the order of the matrix \[C\]
We know that when we multiply two matrices, then the matrix obtained by their product contains a number of rows equal to the number of rows of the first matrix and number of columns equal to the number of columns of the second matrix.
We will use the same concept now.
We know that the order of matrix \[A\] is \[2 \times 2\] and order of matrix \[B\] is \[2 \times 1\]
So, the order of the matrix obtained by their product is equal to \[2 \times 1\].
Therefore, the order of the matrix \[AB\] is equal to \[2 \times 1\].
We know that the order of matrix \[A\] is \[2 \times 2\] and order of matrix \[B'\] is \[1 \times 2\]
So, the order of the matrix obtained by their product i.e. \[B'A\] is equal to \[1 \times 2\].
Therefore, the order of the matrix \[B'A\] is equal to \[1 \times 2\].
Now, we need the order of the matrix \[C\] which is equal to \[\left( {AB} \right)\left( {B'A} \right)\]
As the order of matrix \[AB\] is equal to \[2 \times 1\] and order of matrix \[B'A\] is equal to \[1 \times 2\].
So the order of the matrix obtained by their product i.e. \[\left( {AB} \right)\left( {B'A} \right)\] is equal to \[2 \times 2\].
We know the number of elements present in any matrix is equal to the product of the number of rows and number of columns of the matrix.
Therefore, the number of elements of the matrix \[C\] is equal to \[2 \times 2 = 4\] .
Thus,
\[n\left( C \right) = 4\] ……….. \[\left( 3 \right)\]
Now, we need to find the matrix \[D\] which is equal to \[\left( {B'A} \right)\left( {AB} \right)\]
For that, we will find the product of \[B'A\].
\[B'A = {\left[ {\begin{array}{*{20}{c}}5&{ - 3}\end{array}} \right]_{1 \times 2}} \times {\left[ {\begin{array}{*{20}{c}}1&2\\4&3\end{array}} \right]_{2 \times 2}}\]
On multiplying these two matrices, we get
\[ \Rightarrow B'A = {\left[ {\begin{array}{*{20}{c}}{5 + \left( { - 3} \right) \times 4}&{5 \times 2 + \left( { - 3} \right) \times 3}\end{array}} \right]_{1 \times 2}}\]
On further simplification, we get
\[ \Rightarrow B'A = {\left[ {\begin{array}{*{20}{c}}{ - 7}&1\end{array}} \right]_{1 \times 2}}\]
Now, we will find the product of \[AB\].
\[AB = {\left[ {\begin{array}{*{20}{c}}1&2\\4&3\end{array}} \right]_{2 \times 2}} \times {\left[ {\begin{array}{*{20}{c}}5\\{ - 3}\end{array}} \right]_{2 \times 1}}\]
On multiplying these two matrices, we get
\[ \Rightarrow AB = {\left[ {\begin{array}{*{20}{c}}{5 + 2 \times - 3}\\{4 \times 5 + 3 \times - 3}\end{array}} \right]_{2 \times 1}}\]
On further simplification, we get
\[ \Rightarrow AB = {\left[ {\begin{array}{*{20}{c}}{ - 1}\\{11}\end{array}} \right]_{2 \times 1}}\]
Now, we will find \[\left( {B'A} \right)\left( {AB} \right)\].
\[\left( {B'A} \right)\left( {AB} \right) = {\left[ {\begin{array}{*{20}{c}}{ - 7}&1\end{array}} \right]_{1 \times 2}} \times {\left[ {\begin{array}{*{20}{c}}{ - 1}\\{11}\end{array}} \right]_{2 \times 1}}\]
On multiplying these two matrices, we get
\[ \Rightarrow \left( {B'A} \right)\left( {AB} \right) = {\left[ {\left( { - 7} \right) \times \left( { - 1} \right) + 1 \times 11} \right]_{1 \times 1}}\]
On further simplification, we get
\[ \Rightarrow D = \left( {B'A} \right)\left( {AB} \right) = {\left[ {18} \right]_{1 \times 1}}\]
Therefore,
\[ \Rightarrow n\left( D \right) = 1 \times 1 = 1\] …….. \[\left( 4 \right)\]
Now, we will find the determinant of matrix \[D\] i.e. \[\left| D \right|\] but we also know that the determinant of the matrix of order \[1 \times 1\] is equal to its element.
Therefore,
\[\left| D \right| = 18\]
Now, we will calculate the square of it.
\[ \Rightarrow {\left| D \right|^2} = {\left( {18} \right)^2}\]
On further simplification, we get
\[ \Rightarrow {\left| D \right|^2} = 324\] …….. \[\left( 5 \right)\]
Now, we will calculate the value of the given expression i.e. \[\left[ {\dfrac{{n\left( C \right)\left( {{{\left| D \right|}^2} + n\left( D \right)} \right)}}{{n\left( A \right) - n\left( B \right)}}} \right]\]
Now, we will substitute all the values calculated in equation 1, equation 2, equation 3, equation 4 and equation 5.
\[\left[ {\dfrac{{n\left( C \right)\left( {{{\left| D \right|}^2} + n\left( D \right)} \right)}}{{n\left( A \right) - n\left( B \right)}}} \right] = \dfrac{{4 \times \left( {324 + 1} \right)}}{{4 - 2}}\]
On simplifying the terms, we get
\[ \Rightarrow \left[ {\dfrac{{n\left( C \right)\left( {{{\left| D \right|}^2} + n\left( D \right)} \right)}}{{n\left( A \right) - n\left( B \right)}}} \right] = 2 \times 325 = 650\]
Hence, the required value of the given expression is equal to 650.
Note: Matrix is defined as a set of numbers which are arranged in a particular row and columns. We need to keep in mind that when we multiply two matrices, then the matrix obtained by their product contains a number of rows equal to the number of rows of the first matrix and number of columns equal to the number of columns of the second matrix. Multiplication of matrices is not commutative.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE
A car covers the first half distance between two places class 11 physics CBSE
The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE
Find the value of cos 135 class 11 maths CBSE
A mass M is held in place by an applied force F and class 11 physics CBSE
A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE
Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Find the value of the expression given below sin 30circ class 11 maths CBSE
What do you mean by retardation What is its SI uni class 11 physics CBSE
Draw a diagram of nephron and explain its structur class 11 biology CBSE
10 examples of friction in our daily life
Difference between physical and chemical change class 11 chemistry CBSE