Subject: Optional Mathematics
If A and B are two matrices of the same order, then A and B are said to be Conformable or Compatible for addition. If A and B are two matrices of the same order, then they are said to be conformable for subtraction. If A is any matrix and K is any constant or a scalar, then the matrix obtained by multiplying each element of A by K is denoted by KA and it is called the scalar multiple of A by K.
If A and B are two matrices of the same order, then A and B are said to be Conformable or Compatible for addition. The sum of A and B is denoted by A+B and it is obtained by adding corresponding elements of A & B.
The matrix A+B will be of the same order as each of the matrices A and B is
If A= \(\begin{bmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{bmatrix}\) and B= \(\begin{bmatrix} b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\end{bmatrix}\)
then A+B =\(\begin{bmatrix} a_{11}+b_{11}&a_{12}+b_{12}&a_{13}+b_{13}\\a_{21}+_b{21}&a_{22}+b_{22}&a_{23}+b_{23}\end{bmatrix}\)
Foe example, if A= \(\begin{bmatrix} 1&2&3\\4&-3&7\end{bmatrix}\) and B= \(\begin{bmatrix} -3&4&7\\8&9&-3\end{bmatrix}\)
Then,
A+B =\(\begin{bmatrix} 1&2&3\\4&-3&7\end{bmatrix}\) + \(\begin{bmatrix} -3&4&7\\8&9&-3\end{bmatrix}\)
= \(\begin{bmatrix} 1-3&2+4&3+7\\4+8&-3+9&7-3\end{bmatrix}\)
= \(\begin{bmatrix} -2&6&10\\12&6&4\end{bmatrix}\)
If A and B are two matrices of the same order, then they are said to be conformable for subtraction. The difference of the matrix B from A is denoted by A-B and it is obtained by subtracting the elements of B from the corresponding elements of A.
The order of the matrix A-B is same as the order of A or B.
If A = \(\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22} &a_{23}\end{bmatrix}\) and B = \(\begin{bmatrix}b_{11}&b_{12}&b_{13}
\\b_{21}&b_{22}&b_{23}\end{bmatrix}\)
then A-B = \(\begin{bmatrix} a_{11}-b_{11}&a_{12}-b_{12}&a_{13}-b_{13}\\a_{21}-b_{21}&a_{22}-b_{22}&a_{23}-b_{23}\end{bmatrix}\)
Foe example, if A =\(\begin{bmatrix} 1&2&3\\4&-3&7\end{bmatrix}\) and B= \(\begin{bmatrix} -3&4&7\\8&9&-3\end{bmatrix}\)
Then
A-B=\(\begin{bmatrix} 1&2&3\\4&-3&7\end{bmatrix}\) and B= \(\begin{bmatrix} -3&4&7\\8&9&-3\end{bmatrix}\)
=\(\begin{bmatrix} 1+3&2-1&3-7\\4-8&-3-9&7+3\end{bmatrix}\)
=\(\begin{bmatrix} 4&-2&-4\\-4&-12&10\end{bmatrix}\)
B-A =\(\begin{bmatrix} -3&4&7\\8&9&-3\end{bmatrix}\) -\(\begin{bmatrix} 1&2&3\\4&-3&7\end{bmatrix}\)
=\(\begin{bmatrix} -3-1&4-2&7-3\\8-4&9+3&-3-7\end{bmatrix}\)
=\(\begin{bmatrix} -4&2&4\\4&-12&-10\end{bmatrix}\)
If A is any matrix and K is any constant or a scalar, then the matrix obtained by multiplying each element of A by K is denoted by KA and it is called the scalar multiple of A by K.
Addition of matrices satisfy the following properties:
Closure property
If A and B are two matrices of the same order, then their A+B is also a matrix of the same order as that of A or B.
If A=\(\begin{bmatrix} 2&4\\3&5\end{bmatrix}\) and B=\(\begin{bmatrix} 6&5\\2&1\end{bmatrix}\)
Now,
A+B = \(\begin{bmatrix} 2&4\\3&5\end{bmatrix}\) + \(\begin{bmatrix} 6&5\\2&1\end{bmatrix}\)
= \(\begin{bmatrix} 8&9\\5&6\end{bmatrix}\) which is again is again a2x2 matrix. Hence the closer property is satisfied.
Commutative property
If A and B are two matrices of the same order, then A+B=B+A.
Let, A = \(\begin{bmatrix} 2&4\\6&8\end{bmatrix}\) + \(\begin{bmatrix} 7&-2\\8&4\end{bmatrix}\)
Then A+B =\(\begin{bmatrix} 2&4\\6&8\end{bmatrix}\) +\(\begin{bmatrix} 7&-2\\8&4\end{bmatrix}\)
=\(\begin{bmatrix} 9&2\\14&12\end{bmatrix}\)
B+A =\(\begin{bmatrix} 7&-2\\8&4\end{bmatrix}\) +\(\begin{bmatrix} 2&4\\6&8\end{bmatrix}\)
=\(\begin{bmatrix}9&2\\14&12\end{bmatrix}\)
∴ A+B=B+A
Hence the commutative property is satisfied.
Associative property
If A, B and C three matrices of the same order, then (A+B) + C = A +(B+C)
Existence of identity element
If A is any matrix, then three exists a null matrix 0 of the same order such that A+0 = 0+A=A.
Existence of additive inverse
If A is a matrix of any order, then there exists another matrix -A of same order such that A+(-A) =(-A) +A = 0, then additive identity.
If A and B are the matrices of the same order and K is a scalar, then
K(A+B) = KA+KB.
If A is a matrix and C,K are any two scalars, then (C+K) A=CA+KA
If C,K are any two scalars and A is a matrix, then C(KA) = (CK)A.
To solve the matrix equation A + X =B where A and B are two given matrices of the same order and X is unknown, we proceed in a manner similar to numbers.
Here, A + X = B
Adding the matrix (-A)to both sides, we get
(-A)+ A+ X = (-A) + B
or, (-A+A) +X = B-A
or, 0 + X = B-A
or, X = B-A, which is the required solution.
Let A be a matrix. Then the new matrix obtained by interchanging the corresponding rows and column of A is called the transpose of A. It is denoted by A' or At.
If A = \(\begin{bmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{bmatrix}\)
Then At= \(\begin{bmatrix} a_{11}&a_{12}\\a_{13}&a_{21}\\a_{22}&a_{23}\end{bmatrix}\)
Here, the order of A' is 2 x 3 and that of At is 3 x 2. Hence, if the order of matrix A is mxn, then the order of At will be nxm. If A' is a square, matrix of order n, then At is also a square matrix of order n. If A' is a row matrix, then At is a column matrix.
Properties of transpose
Find the values of x and y if
\(\begin{bmatrix} x+y \\ x-y \\ \end{bmatrix}\) = \(\begin{bmatrix} 4 \\ 2 \\ \end{bmatrix}\)
Soln:
(a) here given,\(\begin{bmatrix} x+y \\ x-y \\ \end{bmatrix}\)=\(\begin{bmatrix} 4 \\ 2 \\ \end{bmatrix}\)so,
x+y=4.........(i)
and x-y=2 ........(ii)
Adding two equation, we get,
2x=6 ∴ x=\(\frac{6}{2}\)=3
Putting x=3 in equation (i), we get, x
3+y=4 or, y=4-3 or, y=1
∴x=3 and y=1 Ans.
Find the values of x and y if
\(\begin{bmatrix} x & x+y \\ 2 & 3\\ \end{bmatrix}\)=\(\begin{bmatrix} 4 & 5 \\ 5 & 3 \\ \end{bmatrix}\)
here given,
\(\begin{bmatrix} x & x+y \\ 2 & 3\\ \end{bmatrix}\)=\(\begin{bmatrix} 4 & 5 \\ 5 & 3 \\ \end{bmatrix}\)
x=4 .........(i)
and x+y=5 ..........(ii)
putting x=4 in equation (ii) we get, 4+y=5,or,y=5-4=1
∴ x=4,y=1. Ans.
If A =\(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{bmatrix}\) and B= \(\begin {bmatrix} -1 & 0 & 3 \\ 1 & 0 & 5 \\ \end{bmatrix}\), Verify commutative law of addition.
Soln:
Here given,
A =\(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{bmatrix}\) and \(\begin {bmatrix} -1 & 0 & 3 \\ 1 & 0 & 5 \\ \end{bmatrix}\)
We have, commutative law of addition A+B=B+A
Now, A+B=\(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{bmatrix}\)+\(\begin {bmatrix} -1 & 0 & 3 \\ 1 & 0 & 5 \\ \end{bmatrix}\)
=\(\begin{bmatrix} 1+ (-2) & 2+0 & 3+3 \\ 4+1 & 5+0 & 6+5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 0 & 2 & 6 \\ 5 & 5 & 11 \\ \end{bmatrix}\)....(i)
Again, B+A=\(\begin {bmatrix} -1 & 0 & 3 \\ 1 & 0 & 5 \\ \end{bmatrix}\)+\(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{bmatrix}\)=\(\begin{bmatrix} -1 + 1 & 0+2 & 3+3 \\ 1+4 & 0+5 & 5+6 \\ \end{bmatrix}\)
=\(\begin{bmatrix} 0 & 2 & 6 \\ 5 & 5 & 11 \\ \end{bmatrix}\)........(ii)
So, from (i) and (ii) A+B=B+A proved
If \(\begin{bmatrix} x \\ -3 \\ \end{bmatrix}\)+\(\begin{bmatrix} 5 \\ y \\ \end{bmatrix}\)=\(\begin{bmatrix} 3 \\ 8 \\ \end{bmatrix}\), Find the value of x and y.
Soln:
here,
\(\begin{bmatrix} x \\ -3 \\ \end{bmatrix}\)+\(\begin{bmatrix} 5 \\ y \\ \end{bmatrix}\)=\(\begin{bmatrix} 3 \\ 8 \\ \end{bmatrix}\)
or, \(\begin{bmatrix} x+5 \\ -3+y \\ \end{bmatrix}\)=\(\begin{bmatrix} 3 \\ 8 \\ \end{bmatrix}\)
or, x+5=3 and -3+y=8
or, x=3-5 and y=8+3
∴ x=-2 and y=11 Ans.
If \(\begin{bmatrix} 4 & 4 \\ m & 0 \\ \end{bmatrix}\)= \(\begin{bmatrix} n & p \\ 3 & 4 \\ \end{bmatrix}\)+\(\begin{bmatrix} 2 & 3 \\ 3 & r \\ \end{bmatrix}\), then find the value of m,n,p and r.
soln:
here given,
\(\begin{bmatrix} 4 & 4 \\ m & 0 \\ \end{bmatrix}\)= \(\begin{bmatrix} n & p \\ 3 & 4 \\ \end{bmatrix}\)+\(\begin{bmatrix} 2 & 3 \\ 3 & r \\ \end{bmatrix}\)
or,\(\begin{bmatrix} 4 & 4 \\ m & 0 \\ \end{bmatrix}\)=\(\begin{bmatrix} n+2 & p+4 \\ 3+3 & 4+r \\ \end{bmatrix}\)
or, \(\begin{bmatrix} 4 & 4 \\ m & 0 \\ \end{bmatrix}\)=\(\begin{bmatrix} n+2 & p+4 \\ 6 & 4+r \\ \end{bmatrix}\)
here, n+2=4..........(i) p+4=4..........(ii) m=9........(iii)
4+r=0..........(iv)
∴n=4-2=2, p=4-4=0, m=6 and r=-4
∴n=2, p=0, m=6 and r=-4 Ans.
If \(\begin{bmatrix} 3x-2 & 5y+4 \\ 2 & 4+2z \\ \end{bmatrix}\)=\(\begin{bmatrix} x+2 & y-4 \\ 2 & z-2 \\ \end{bmatrix}\),the find the value of x,y and z.
soln:
here given, \(\begin{bmatrix} 3x-2 & 5y+4 \\ 2 & 4+2z \\ \end{bmatrix}\)=\(\begin{bmatrix} x+2 & y-4 \\ 2 & z-2 \\ \end{bmatrix}\)
or, 3x-2=x+2, 5y+4=y-4 and 4+2z=z-2
or, 3x-x=2+2 , 5y-y=-4-4 and 2z-z=-2-4
or, 2x=4, 4y=-8 and z=-6
or,x=\(\frac{4}{2}\), y=\(\frac{-8}{4}\) and z=-6
∴x=2, y=-2 and z=-6 Ans.
If A=\(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}\) then
(a) Find A' and (A')'
(b) What type of relation is between A and (A')'?
(c) If B=\(\begin{bmatrix} 2 & 3 &-4 \\ -5 & 0 & 6 \\ \end{bmatrix}\) then is the relation between B and (B')' same as relation A and (A')' in (b)?
soln:
here given,
A=\(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}\)
(a) here, A'=\(\begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \\ \end{bmatrix}\)=\(\begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \\ \end{bmatrix}\) .Ans.
and (A')'=\(\begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \\ \end{bmatrix}\)= A=\(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}\)Ans.
(b) A and (A')' are equal.
i.e. (A')'=A
(c) here, B=\(\begin{bmatrix} 2 & 3 & -4 \\ -5 & 0 & 6 \\ \end{bmatrix}\)
∴B'=\(\begin{bmatrix} 2 & 3 & -4 \\ -5 & 0 & 6 \\ \end{bmatrix}\)'=\(\begin{bmatrix} 2 & -5 \\ 3 & 0 \\ -4 & 6 \\ \end{bmatrix}\).Ans.
and (B')'=\(\begin{bmatrix} 2 & -5 \\ 3 & 0 \\ -4 & 6 \\ \end{bmatrix}\)'=\(\begin{bmatrix} 2 & 3 & -4 \\ -5 & 0 & 6 \\ \end{bmatrix}\).Ans.
∴ (B')'=B
here, the relation of B and (B')' is same as A and (A')' in (b)
If A=\(\begin{bmatrix} 2 & 3 \\ -4 & -5 \\ \end{bmatrix}\) and B=\(\begin{bmatrix} 0 & -3 \\ 2 & 5 \\ \end{bmatrix}\) then
(a) Find A'+B' (b) Find B'+A'
(c) Can we write A' + B' = B' + A'
soln:
here given,
A=\(\begin{bmatrix} 2 & 3 \\ -4 & -5 \\ \end{bmatrix}\) and B=\(\begin{bmatrix} 0 & -3 \\ 2 & 5 \\ \end{bmatrix}\)
Here, A'=\(\begin{bmatrix} 2 & 3 \\ -4 & -5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 2 & -4 \\ 3 & -5 \\ \end{bmatrix}\)
and, B'=\(\begin{bmatrix} 0 & -3 \\ 2 & 5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 0 & 2 \\ -3 & 5 \\ \end{bmatrix}\)
(a) Now, A'+B'=\(\begin{bmatrix} 2 & -4 \\ 3 & -5 \\ \end{bmatrix}\)+\(\begin{bmatrix} 0 & 2 \\ -3 & 5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 2+0-4+2 \\ 3-3-5+5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 2-2 \\ 0 & 0 \\ \end{bmatrix}\)Ans.
(b) B'+A'=\(\begin{bmatrix} 0 & 2 \\ -3 & 5 \\ \end{bmatrix}\)+\(\begin{bmatrix} 2 & -4 \\ 3 & -5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 0+2 & 2-4 \\ -3+3 & 5-5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 2 & -2 \\ 0 & 0 \\ \end{bmatrix}\).Ans.
(c) From (a) and (b) , we get,
A'+B'=B'+A'
Find the transpose of the given matrix.
P=[1 0 2]
Here,
Given,
P=[1 0 2]
∴Transpose of P (P')=[1 0 2]'=\(\begin{bmatrix} 1 \\ 0 \\ 2 \\ \end{bmatrix}\)Ans.
Find the transpose of the given matrix.
Q=\(\begin{bmatrix} 3 & 4 & 0 \\ 5 & 6 & -2 \\ \end{bmatrix}\)
Here,
Given,
Q=\(\begin{bmatrix} 3 & 4 & 0 \\ 5 & 6 & -2 \\ \end{bmatrix}\)
∴Transpose of Q (Q')=Q=\(\begin{bmatrix} 3 & 4 & 0 \\ 5 & 6 & -2 \\ \end{bmatrix}\)=\(\begin{bmatrix} 3 & 5 \\ 4 & 6 \\ 0 & -2 \\ \end{bmatrix}\)Ans.
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