Matrices: Definition: An m n matrix, A m n is a rectangular array of numbers with m rows and n columns: a, a, a,n a, a, a,n A m,n =...... a m, a m, a m,n Each a i,j is the entry at the i th row, j th column. E.g 4 0 A m,n = 0 7 5 is a 4 matrix. It has rows and 4 columns. The first row is the horizontal row [ 4 0 ] 4 The second column is the vertical column: The entry at the second row, third column is the number. A matrix with the same number of rows and columns is called a square matrix. 5 E.g. 0 is a square matrix. 8 7 5
In solving systems of linear equations, say: x + z = x + y z = z = We can do the following operation: We can multiply the third equation by (to solve for z): x + z = x + y z = z = 6 We can switch the first equation with the second equation: x + y z = x + z = z = 6 We can multiply to the third equation and add to the second equation to obtain a new equation: x + z = z = x = 5 If we replace the second equation of the original system with this new equation we get: x + y z = x = 5 z = 6 Multiply to the second equation gives: x + y z = x = 5 z = 6 If we switch the first and second equation we have: x = 5 x + y z = z = 6 We can add the second equation with the third equation:
x + y z = x + y z = 6 = 5 Replace the second equation with this new equation gives: x = 5 x + y = 5 z = 6 We now multiply to the first equation and add this to the second equation: x = 0 x + y = 5 y = 5 Replace the second equation with this new equation: x = 5 y = 5 z = 6 Multiply the second equation by gives: x = 5 y = 5 z = 6 At this stage we can readily read out the solution to the system, namely, x = 5, y = 5, z = 6
Going back to the original system of equations: x + z = x + y z = z = If we form a matrix with only the coefficients and the constants of the above system of equations, we get an augmented matrix: 0 0 0 The first column is the coefficient of the x variable, the second column represent the coefficient of the y variable, the third column is the coefficient of the z variable, and the fourth column is the value of the constant on the right of the equation. Notice that if a variable is missing in any of the equation, it is understood that its coefficient is 0. The process that we went through in trying to obtain the solution involved multiplying a (non-zero) constant to any of the equation, switching any two equation, or add a multiple of one equation to another. We know that these operations will not change the solution set of the system. In terms of operations with matrices, these operations correspond to the following elementary row operations one can do to a matrix:. Multiply a non-zero constant to any row of a matrix.. Switch any two rows.. Add a multiple of one row to another. If we look at the equivalent system of equations we get at the end of the previous problem that gave us the easy solution, we see: x = 5 y = 5 z = 6 The corresponding augmented matrix looks like this: 0 0 5 5 0 0 0 0 6
The above matrix is said to be in reduced row-echelon form. The process of trying to use matrix to solve systems of equations is to reduce an augmented matrix to a reduced row-echelon form. We have the following definition for clarification: A matrix is in row-echelon form if it satisfies the following:. The first non-zero number in each row (reading from left to right) is. This is the leading entry.. The leading entry in each row is to the right of the leading entry in the row immediately above it.. All rows rows consisting entirely of zeros are at the bottom of the matrix. If, in addition, every number above and below each leading entry is 0, then the matrix is said to be in reduced row-echelon form. E.g 0 4 5 A = 0 9 0 0 Matrix A is not in row-echelon form, as the leading entries are not all of them, and not in the correct order. 8 4 7 B = 0 0 0 0 Matrix B is in row-echelon form, but not in reduced row-echelon form. All the leading entries are, and they are in the correct oder, but not all the entries above and below the leading entry are 0. 0 0 4 C = 0 0 0 0 5 Matrix C is in reduced row-echelon form. Each leading entry is, and they are in the correct order, and each entry above and below the leading entry is 0. A matrix in reduced row-echelon form allows us to find the solution to a system of equation. We want to use the elementary row operation to change a matrix to row-echelon form. We can do this systematically by the following process: Start by getting a leading in the first row, first column. This can be achieved by multiplying the first row by the reciprocal of the leading entry. If the leading entry is 0, switch the first row with another row that has a non-zero leading entry.
Obtain zeros below the leading entry by multiplying the first row (the row with the leading entry) by the negative of the entries in the row below and add to those rows. Obtain a leading in the second row, second column by the same method. Repeat this process until the matrix is in row-echelon form. This procedure just described is called the Gaussian elimination process. Once you have a matrix in row-echelon form, you may solve the equation by back-substituion, or you may further reduce the matrix to reduced row-echelon form by adding appropriate multiples (the negative of the entry above the leading entry) with the leading entry and add that to the row above it. If you do that, this is the Gauss-Jordan elimination method.
E.g. Solve the system of equation using matrices: x + y + z = 7 x y = 7 Ans: We write the system of equation and its augmented matrix side-by-side and show the corresponding changes to the system as we perform the matrix operation: 4 x + y + z = 7 7 x y = 7 0 7 Add first equation to second (add first row to second row): 4 R + y + 4z = +R R 0 4 x y = 7 0 7 Multiply first equation (row) by and add to third equation (row): 4 R + y + 4z = +R R 0 4 y z = 5 0 5 Multiply second equation (row) by : + y + 4 z = 7 y z = 5 R 4 4 0 7 0 5 Multiply second equation (row) by and add to third equation: 4 + y + 4 z = 7 R +R R 0 4 7 z = 6 0 0 6
Multiply third equation by + y + 4 z = 7 z = R 4 0 4 7 0 0 Our matrix is now at row-echelon form. We may use back substitution to find the values of x, y, and z if we want, or, we can continue with the Gauss-Jordan elimination process to reduce the matrix to reduced row-echelon form: Multiply 4 to third equation (row), add to second equation (row). 4 + y = 4 R +R R 0 0 z = 0 0 Multiply to third equation/row, add to first equation/row. x + y = 0 R + y = +R R 0 0 z = 0 0 Multiply to second equation/row, add to first equation/row. x + = 0 0 R + y = +R R 0 0 z = 0 0 We can now read out the solution, x =, y =, z =.
Once you are familiar with the Guassian elimination process, you may be able to spot the row operation that is most convenient at a particular stage. For example, in the previous problem, when we get to this stage: + y + 4z = y z = 5 4 0 4 0 5 If we had added the second and third equation instead of multiplied the second equation by, we would have been able to avoid fractions. + y + 4z = z = 6 R +R R 4 0 4 0 0 6 Multiply third equation/row by + y + 4z = z = R 4 0 4 0 0 Multiply 4 to third row, add to second row. 4 4R + y = 9 +R R 0 0 9 z = 0 0 Multiply second row by y = z = R 4 0 0 0 0 We are able to get the matrix to row-echelon form without having to involve fraction in the entries.
In solving system of equations, we will occationally encounter an inconsistent or dependent solution where the system has no solution or infinitely many solutions. You want to be able to recognize these cases with a matrix in row-echelon form. If a matrix in row-echelon form has a row where the leading entry is the only non-zero entry in the row, the system has no solution. For example, 0 0 0 7 0 0 0 is a matrix that represents a system with no solution. The bottom row corresponds to an equation like 0 =, which is impossible. If each variable in the row-echelon form is a leading variable, the system has exactly one solution, we have an independent system. For example, 0 6 0 9 0 0 4 This matrix represents a system with exactly one solution, x =, y =, z = 4. If the variables in the matrix are not all leading variables, and the system is not inconsistent (the matrix has a row of all zeros), then the system has infinitely many solutions. We have a dependent system. E.g. The matrix 0 5 0 4 0 0 0 0 has infinitely many solutions. We may use any value for z, then y = 4 + z, and x = 5 z will be dependent on z.
If A is an n n square matrix, we can define the determinant of A, denoted by det(a), or A, as follows: If A = [ a ] is a matrix, then det(a) = a [ ] a b If A = c d is a matrix, then det(a) = ad bc If A is an n n matrix, where n, then det(a) can be found by expanding along any one of the rows or columns of A, by taking the a ij entries on that row (or column), then multiply to the determinant of the resulting matrix obtained by removing the i th row, j th column, and using a plus or minus sign according to the following pattern: + + + + + +....... 0 0 E.g. Find the determinant of A = 0 0 4 0 Ans: We can expand along the first row: 0 0 det(a) = 0 0 4 0 0 0 0 = 0 ( ) 0 0 + () 0 (0) 0 0 4 0 4 0 0 4 Notice that the first matrix (the one multiplied to ) is obtained by deleting the first row and first column of the 4 4 matrix, the second matrix (the one multiplied to ) is obtained by deleting the first row, second column. Also note the alternating signs that are in front of the numbers. To finish the problem, we need to continue to evaluate the determinant of each of the matrix (we do not need to evalute the last one, since 0 times any number is 0.
0 0 0 = 4 0 4 0 0 0 + 0 4 = [0 ( 8)] 0[0 ( )] + [ 0] = 8 0 + = 0 We find the determinant of the other two matrix: 0 0 0 0 = 4 0 4 0 0 0 0 + 0 0 4 = [0 ( 8)] 0[0 ( )] + [0 0] = 8 0 = 0 0 () 0 0 + 0 = [0 ( )] ()[0 ( )] + [0 ] = = We can now finish the problem: 0 0 det(a) = 0 0 4 0 0 0 0 = 0 ( ) 0 0 + () 0 (0) 0 0 4 0 4 0 0 4 = (0) ( )(8) + ()( ) 0 = 40 + 8 9 = 9
It is also valid to expand along another row or column, as long as we follow the ordering of the signs. If we expand along the second column, we will have: 0 0 det(a) = 0 0 = 4 0 0 0 0 0 ( ) 0 0 + () 0 0 () 0 + () 0 4 0 4 0 4 0 0 0 In practice, it is best to expand along the row or column with the most number of zeros to minimize the amount of work one needs to do. In this example, it is best to expand along the third row (with zeros).
Equality of Matrices: If A and B are two matrices, we say that A = B if A and B has the same number of rows and columns (i.e. they have the same dimension) and each of their corresponding entries is equal. i.e. a ij = b ij for all entries of a ij of A and b ij of B. If two matrices have different dimension, they are never equal. E.g 4 5 7 [ ] 4 5 7 A = 0 B = 0 0 0 0 A B because they have different dimensions. One is a matrix while the other is a matrix. 9 A = 5 [ ] B = 9 5 A B because they have different dimensions. One is a 5 matrix while the other is a 5 matrix.
Matrix Addition and Scalar Multiplication If A and B are matrices of the same dimension, then we can define A + B as the matrix of the same dimension whose entries are obtained by adding the corresponding entries of A and B. E.g. 0 5 8 A = 7 4 B = 0 9 5 7 6 0 5 8 + 5 0 + 8 7 8 A + B = 7 4 + 0 9 = 7 + 0 4 + 9 = 5 7 6 + 7 5 + 6 6 If A is a matrix of any dimension and s is a constant (a scalar), we define sa to be the matrix of the same dimension whose entry is obtained by mutliplying every entry of A by s. E.g 6 9 0 4 4 A = 5 s = 4, 0 6 9 0 4( 6) 4(9) 4(0) 4 6 40 4 4 sa = 4 5 = 4() 4( 4) 4(4) 4() 4( ) 4(5) = 8 6 6 44 0 0 4(0) 4() 4() 0 8 4
Matrix Multiplication: If A is a n matrix (a row-matrix ) and B is a n matrix (a column-matrix ), we define A times B, or AB to be the matrix whose only entry is obtained by forming the sume of the product of each a i with b i. Example: [ ] A = 4 6 0 7 B = 9 AB = [ 4() + 6( 7) + ( ) + 0() + (9) ] = [ 8 ] In general, if A is a m n matrix and B is a n k matrix, then AB is the m k matrix whose ij th entry is obtained by multiplying the i th row of A to the j th column of B according to the above rule. Example: A = [ ] 4 7 4 0 5 B = 8 0 5 Since A is a matrix and B is a 4 matrix, AB is defined and is a 4 matrix; its entry in the first row, first column is obtained by multiplying the first row of A with the first column of B: (ab) = [ ] 7 8 = (7) + (8) + () = + 6 = 4 The entry in the first row, second column of AB is obtained by multiplying the first row of A with the second column of B: (ab) = [ ] 4 = ( 4) + ( ) + (0) = = 4 0 Continue in this manner gives us the product of A and B, the matrix AB: [ ] 4 0 5 [ ] 4 4 7 AB = 7 8 = 4 6 7 0 5
Notice that while AB is well-defined, BA is undefined. In general, in order for the matrix product AB to be defined, the number of columns of A must equal to the number of rows of B. Example: [ ] [ ] 5 6 Let A =, let B = 4 7 8 Find AB and BA Ans: Using the definition of the product of two matrices, we have: [ ] [ ] [ ] 5 6 AB = = 4 7 8 4 [ ] [ ] [ ] 5 6 4 BA = = 7 8 4 7 46 Notice that even when AB and BA are both defined, as in the case of the product of two square matrices of the same dimension, they are usually not equal to each other. Multiplication of matrices is not commutative.
9 Consider A = 6 8 0 5 0 0 0 Let I = 0 0 0 0 9 0 0 9 Notice that AI = 6 8 0 0 = 6 8 = A 0 5 0 0 0 0 5 0 0 0 9 9 and IA = 0 0 6 8 = 6 8 = A 0 0 0 5 0 0 5 0 If A is any n n square matrix, let I be the n n square matrix with s in the 0 0 0 0 0 0 diagonal and 0 s everywhere else. i.e I = 0 0 0...... then AI = IA = A. We call I the n n identity matrix. It plays the same role in matrix multiplication as the number in number multiplication. 6 6 5 6 Let A = 0 B = 6 6 5 6 Then AB = 0 0 0 = 0 0 0 0 and BA = 6 6 5 6 0 0 0 = 0 0 0 0
Let A be a n n matrix. If there exists another n n matrix B such that AB = BA = I, then we say that B is the inverse matrix of A and write B = A The inverse of a matrix, A, plays the same role in matrix multiplication as that of the reciprocal of a number in number multiplication. 6 6 5 6 In the above example, A = 0 A = It is important to note that not every square matrix has an inverse. If A is a square matrix and its inverse matrix A exists, then we say that A is invertible. The following theorem tells us under what circumstances does a matrix have an inverse: Theorem: A square matrix A is invertible if and only if det(a) 0.
Given a square matrix A, we can find A by putting A and the identity matrix, I, side-by-side. We then use the row-operations to reduce the matrix A into the identity matrix. The resulting matrix that the original identity matrix I becomes is the inverse of A. Example: 0 Let A = 4. Find A 7 0 5 Ans: We place A and I side-by-side, then perform row operation to turn A into the identity matrix: 0 0 0 4 0 0 R +R R 7 0 5 0 0 0 0 0 0 4 0 7R +R R 7 0 5 0 0 0 0 0 0 4 0 0 0 7 0 0 0 0 6 0 4 0 7 0 0 0 5 0 0 0 6 0 0 76 4 76 7 0 0 0 R 4 R 0 0 0 0 4 0 0 0 7 0 0 0 0 6 0 0 76 4 76 7 0 0 0 ( )R 0 0 5 0 6 0 0 76 4 76 7 0 0 0 R +R R ( )R +R R
Now that A has been changed to the identity matrix, the resulting matrix at the right (from the original identity matrix) is the inverse of A. In other words, A = 5 0 6 76 4 76 7 0 We can multiply to confirm that A and A are indeed inverses of each other: 0 5 0 0 0 AA = 4 6 76 4 76 = 0 0 = I 7 0 5 0 0 A A = 5 0 6 76 4 76 7 0 7 0 0 4 7 0 5 0 0 = 0 0 = I 0 0
Consider the system of equations x y + z = x + y z = y + z = Now that we have matrix multiplication defined, we can write this system as an equation in matrix multiplication form: x y = 0 z x If we let A =, X = y, and K =, 0 z then we can write the equation as a matrix equation: AX = K Since det(a) = 0, A exists. We find A = 7 5 6 4 5 If we multiply both sides of the matrix equation by A we get: A AX = A K X = A K 7 5 7 x 6 4 y = 5 0 z 7 5 0 0 x 0 0 y = 6 4 0 0 z 5 7 5 x y = 6 4 z 5 5 6 4 5
x y = z 9 Comparing the two matrices we can conclude that: x = y = z = 9 Using a matrix equation to solve system of equations can save some work if we have to solve multiple systems where A remains the same, but the constants, K, changes often.