Matrix A rectangular array of variables or constants
Matrix - A rectangular array of variables or constants in horizontal rows and vertical columns enclosed in brackets. Element - Each value in a matrix; either a number or a constant. Dimension - Number of rows by number of columns of a matrix. **A matrix is named by its dimensions.
Examples: Find the dimensions of each matrix. Dimensions: 3 x 2 Dimensions: 4 x 1 Dimensions: 2 x 4
Different types of Matrices • Column Matrix - A matrix with only one column. • Row Matrix - A matrix with only one row. • Square Matrix - A matrix that has the same number of rows and columns.
NAME DESCRIPTION Row matrix A matrix with only 1 row Column matrix A matrix with only I column Square matrix A matrix with same number of rows and columns Zero matrix A matrix with all zero entries EXAMPLE
Equal Matrices - Two matrices that have the same dimensions and each element of one matrix is equal to the corresponding element of the other matrix. *The definition of equal matrices can be used to find values when elements of the matrices are algebraic expressions.
EQUAL MATRICES: Matrices having equal corresponding entries. For Example:
If A and B are both m × n matrices then the sum of A and B, denoted A + B, is a matrix obtained by adding corresponding elements of AA and B. B. add these add these
If A is an m × n matrix and s is a scalar, then we let k. A denote the matrix obtained by multiplying every element of A by k. This procedure is called scalar multiplication. PROPERTIES OF SCALAR MULTIPLICATION
The m × n zero matrix, denoted 0, is the m × n matrix whose elements are all zeros. 2× 2 1× 3
For example:
The multiplication of matrices is easier shown than put into words. You multiply the rows of the first matrix with the columns of the second adding products Find AB First we multiply across the first row and down the first column adding products. We put the answer in the first row, first column of the answer.
Find AB Notice the sizes of A and B and the size of the product AB. Now we multiply across the first second rowrow andand down the second first We multiplied across first row and down first column second and we’lland putwe’ll the answer put the in answer the second first in the row, second firstrow, so we put the answer in the first row, first column. second
To multiply matrices A and B look at their dimensions MUST BE SAME SIZE OF PRODUCT If the number of columns of A does not equal the number of rows of B then the product AB is undefined.
Now let’s look at the product BA. acrossthird second third row row across first second row as as aswe we godown we as we go go go down first third first second third column: column: second column: first column: 3 2 can multiply size of answ Completely different than AB! er 2 3
PROPERTIES OF MATRIX MULTIPLICATION Is it possible for AB = BA ? , yes it is possible.
What is AI? Multiplying a matrix by the identity gives the matrix back again. What is IA? an n n matrix with ones on the main diagonal and zeros elsewhere
Can we find a matrix to multiply the first matrix by to get the identity? ? Let A be an n n matrix. If there exists a matrix B such that AB = BA = I then we call this matrix the inverse of A and denote it A-1.
For example: FIND (a. ) AB and (b. ) BA
The inverse of the matrix
Find the inverse of Solution:
If A has an inverse we say that A is nonsingular. A-1 does not exist we say A is singular. If To matrix A, A, a a To find the inverse of of a a matrix we we put the matrix line identity matrix. We We then perform row line and then the identity operations toturnititintothe theidentity. We We operations on on matrix A A to carry right hand side carry the row operations across and the right will turn into the inverse. r 2 r 1 r 2 2 r 1+r 2
Check this answer by multiplying. We should get the identity matrix if we’ve found the inverse.
We can use A-1 to solve a system of equations To see how, we can re-write a system of equations as matrices. coefficient matrix variable matrix constant matrix
left multiply both sides by the inverse of A This is just the identity but the identity times a matrix just gives us back the matrix so we This then gives us a formula have: for finding the variable matrix: Multiply A inverse by the constants.
find the inverse -2 r 1+r 2 r 1 -3 r 2 -r 2 x y This is the answer to the system
Determinant - A square array of numbers or variables enclosed between parallel vertical bars. **To find a determinant you must have a SQUARE MATRIX!!** Finding a 2 x 2 determinant:
Find the determinant:
Finding a 3 x 3 determinant: SARRUS DIAGRAM METHOD Step 1: Rewrite first two rows of the matrix.
-224 +10 +162 = -52 Step 2: multiply diagonals going up! Step 2: multiply diagonals going down! -126 +12 +240 =126 - (-52) 126 + 52 = 178 Step 3: Bottom minus top!
-18 +50 +6 = 38 Step 2: multiply diagonals going up! Step 2: multiply diagonals going down! 45 - 15 + 8 = 38 38 - 38 =0 Step 3: Bottom minus top!
Let A be the co-efficient matrix of the linear system: ax+by= e & cx+dy= f. IF det A ≠ 0, then the system has exactly one solution. The solution is: The numerators for x and y are the determinant of the matrices formed by using the column of constants as replacements for the coefficients of x and y, respectively.
Use cramer’s rule to solve this system: 8 x+5 y = 2 2 x-4 y = -10
Solution: Evaluate the determinant of the coefficient matrix Apply cramer’s rule since the determinant is not zero. The solution is (-1, 2)
1 2 If a matrix has 24 elements , what are the possible order it can have? Construct a 3 x 3 matrix whose elements are aij=i+j
3. The bookshop of a school has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen books of maths. Their selling price are Rs. 80 , Rs 60 and Rs 40 resp. find the total amount the bookshop will receive from selling all the books? 4. Expand the determinent 2 -4 3 3 1 2 7 6 1 5. find x: x 2 -1 2 5 x -1 2 x
6. evaluate by sarrus diagram: 1 a b+c 1 b c+a 1 c a+b 7. Prove that 1 1 1 a b c = (b-c)(c-a)(a-b)(a+b+c) a 3 b 3 c 3
Solve the equations by crammer rule: 8. 3 x-4 y=1 -2 x+5 y=-3 9. x+y=1 x+z=-6 x-y-2 z=3 10. 2 x-y+z=11 X+2 y+3 z=2 3 x+y-z=6
NOTE: DO ANY THREE
1: find X and Y , if 2 X+3 Y=1 And 5 X+7 Y=2 2: Construct a 2 x 2 Matrix A =[a] where a= I 2 i-3 j. I/2 3 : If 2 4 = 2 x 4 ; Find the value of x 5 1 6 x 4: Evaluate 1 3 5 2 6 10
5. Let f(x)=x-5 x+6. find f(A) A= 2 0 1 2 1 3 1 -1 0 6. Solve the equation : 6 x+y-3 z=5 x+3 y-2 z=5 2 x+y+4 z=8
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