Lesson 09: Matrix Algebra
Objectives
- Learn the basic operations of the matrix algebra.
- Learn how to solve simultaneous equations by using MATLAB matrix operations.
Background
9.1 Matrix Operations and Functions
Transpose
Transpose
The transpose operator changes the rows of a matrix into columns and the columns into rows. In mathematics texts, you will often see the transpose indicated with superscript T (as in AT ). In MATLAB, the transpose operator is a single quote ('), so that the transpose of matrix A is A'.
| Syntax | Description |
|---|---|
| B = A' | Computes the complex conjugate transpose of A. |
| B = ctranspose(A) | This is an alternate way to execute A', but is rarely used. |
| B = A.' | Returns the non-conjugate transpose of A, that is, interchanges the row and column index for each element. |
| B = transpose(A) | This is an alternate way to execute A.' and enables operator overloading for classes. |
Complex Conjugate Transpose
The complex conjugate transpose of a matrix interchanges the row and column index for each element, reflecting the elements across the main diagonal. The operation also negates the imaginary part of any complex numbers.
For example, if B = A' and A(1,2) is 1+1i, then the element B(2,1) is 1-1i.
Complex Non-Conjugate Transpose
The non-conjugate transpose operator, A.', performs a transpose without conjugation. That is, it does not change the sign of the imaginary parts of the elements.
If A contains complex elements, then A.' does not affect the sign of the imaginary parts. For example, if A(3,2) is 1+2i and B = A.', then the element B(2,3) is also 1+2i.
Dot Product
Dot Product
The dot product (sometime called the scalar product) is the sum of the results you obtain when you multiply two vectors together, element by element.
| Syntax | Description |
|---|---|
| C = sum(A.*B) | To multiply two arrays, and then add the elements up to get the dot product. |
| C = dot(A,B) | Returns the scalar dot product of A and B. |
| C = dot(A,B,dim) | Evaluates the dot product of A and B along dimension, dim. The dim input is a positive integer scalar. |
- If A and B are vectors, then they must have the same length.
- If A and B are matrices or multidimensional arrays, then they must have the same size. In this case, the dot function treats A and B as collections of vectors. The function calculates the dot product of corresponding vectors along the first array dimension whose size does not equal 1.
Dimension to operate along, specified as a positive integer scalar. If no value is specified, the default is the first array dimension whose size does not equal 1.
Consider two 2-D input arrays, A and B:
- dot(A,B,1) treats the columns of A and B as vectors and returns the dot products of corresponding columns.
- dot(A,B,2) treats the rows of A and B as vectors and returns the dot products of corresponding rows.
- dot returns conj(A).*B if dim is greater than ndims(A).
The scalar dot product of two real vectors of length n is equal to a mathematics text:
\[A\centerdot B=\sum\limits_{i=i}^{n}{{{A}_{i}}{{B}_{i}}={{A}_{1}}{{B}_{1}}+{{A}_{2}}{{B}_{2}}+\cdots +{{A}_{n}}{{B}_{n}}}\]
This relation is commutative for real vectors, such that dot(A,B) equals dot(B,A). If the dot product is equal to zero, then A and B are perpendicular.
For complex vectors, the dot product involves a complex conjugate. This ensures that the inner product of any vector with itself is real and positive definite.
\[U\centerdot V=\sum\limits_{i=1}^{n}{\overline{{{U}_{i}}} {{V}_{i}}}\]
Unlike the relation for real vectors, the complex relation is not commutative, so dot(U,V) equals conj(dot(V,U)).
Matrix Multiplication
Matrix Multiplication
| Syntax | Description |
|---|---|
| C = A * B | C is the linear algebraic product of the matrices A and B. The number of columns of A must equal the number of rows of B. |
The matrix multiplication operator calculates the product of two matrices with the formula:
\[{{C}_{i,j}}=\sum\limits_{k=1}^{n}{{{A}_{i,k}}{{B}_{k,j}}}\]
Because matrix multiplication is a series of dot products, the number of columns in matrix A must equal the number of rows in matrix B. If matrix A is an m×n matrix, matrix B must be n×p, and the results will be × matrix.
For example: A is a 2×3 matrix and B is a 3×3 matrix. The result is a 2×3 matrix. In this example, we have
The two inner numbers must match, and the two outer numbers determine the size of the resulting matrix.
Matrix multiplication is not in general commutative, which means that $A*B\ne B*A$. If both matrices are square, we can indeed calculate an answer for $A*B$ and an answer for $B*A$, but the answers are nor the same.
To see this, you can calculate the product of two matrices.
$A=\left[ \begin{matrix}1 & 3 \\2 & 4 \\\end{matrix} \right]$, $B=\left[ \begin{matrix}3 & 0 \\1 & 5 \\\end{matrix} \right]$
>> A * B
ans =
6 15
10 20
>> B * A
ans =
3 9
11 23
The previous matrix product is not equal to the following element-wise product.
>> A .* B
ans =
3 0
2 20
Matrix Power
Matrix Power
| Syntax | Description |
|---|---|
| C = A^B | Computes A to the B power and returns the result in C. |
| C = mpower(A,B) | This is an alternate way to execute A^B, but is rarely used. |
Raising a matrix to a power is equivalent to multiplying the matrix by itself the requisite number of times. For example, ${{A}^{2}}$ is the same as $A*A$; ${{A}^{3}}$ is the same as $A*A*A$.
Recalling that the number of columns in the first matrix of a multiplication must be equal to the number of rows in the second matrix, we see that in order to raise a matrix to a power, the matrix must be square (have the same number of rows and columns). Consider the matrix
\[A=\left[ \begin{matrix}\begin{matrix}1 & 2 & 3 \\\end{matrix} \\\begin{matrix}4 & 5 & 6 \\\end{matrix} \\\end{matrix} \right]\]
If we tried to square this matrix, we would get an error statement because of the rows and columns mismatch:
Matrix Inverse
Matrix Inverse
A matrix X is invertible if there exists a matrix Y of the same size such that $XY=YX={{I}_{n}}$, where ${{I}_{n}}$ is the n×n identity matrix. The matrix Y is called the inverse of X.
A matrix that has no inverse is singular. A square matrix is singular only when its determinant is exactly zero.
| Syntax | Description |
|---|---|
| Y = X^(-1) | X^(-1) is equivalent to inv(X). |
| Y = inv(X) | Computes the inverse of square matrix X. |
Determinants
Determinants
| Syntax | Description |
|---|---|
| d = det(A) | Returns the determinant of square matrix A. |
Determinants are used in linear algebra and are related to the matrix inverse. If the determinant of a matrix is 0, the matrix does not have an inverse, and we say that it is singular. Determinants are calculated by multiplying together the elements along the matrix’s left-to-right diagonals and subtracting the product of the right-to-left diagonals. For example, for a 2×2 matrix:
\[A = \left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\{{A_{21}}}&{{A_{22}}}\end{array}} \right]\]
the determinant is $\left| A \right| = {A_{11}}{A_{22}} - {A_{12}}{A_{21}}$
Thus, for $A = \left[ {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right]$
$\left| A \right| = (1)(4) - (2)(3) = - 2$
For a 3×3 matrix, $A = \left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}&{{A_{13}}}\\{{A_{21}}}&{{A_{22}}}&{{A_{23}}}\\{{A_{31}}}&{{A_{32}}}&{{A_{33}}}\end{array}} \right]$
- Multiply each left-to-right diagonal and add them up:
- Multiply each right-to-left diagonal and add them up.
- Finally, subtract the second calculation from the first. For example, $A = \left[ {\begin{array}{*{20}{c}}
1&2&3\\
4&5&6\\
7&8&9
\end{array}} \right]$ , we might have
\[\begin{array}{l}
\left| A \right| = (1 \times 5 \times 9) + (2 \times 6 \times 7) + (3 \times 4 \times 8) - \left[ {(3 \times 5 \times 7) + (2 \times 4 \times 9) + (1 \times 6 \times 8)} \right]\\
= 225 - 225 = 0
\end{array}\]
Use MTLAB for the same calculation:
>> A = [1 2 3; 4 5 6; 7 8 9];
>> det(A)
ans = 0
The matrix with a determinant of zero do not have inverses. If you try to find the inverse of A, you will get a warning message:
>> inv(A)
Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 1.541976e-18.
ans =
1.0e+16 *
-0.4504 0.9007 -0.4504
0.9007 -1.8014 0.9007
-0.4504 0.9007 -0.4504
Cross Products
Cross Products
| Syntax | Description |
|---|---|
| C = cross(A,B) | Returns the cross product of A and B. |
| C = cross(A,B,dim) | Evaluates the cross product of arrays A and B along dimension, dim. The dim input is a positive integer scalar. |
- If A and B are vectors, then they must have a length of 3.
- If A and B are matrices or multidimensional arrays, then they must have the same size. In this case, the cross function treats A and B as collections of three-element vectors. The function calculates the cross product of corresponding vectors along the first array dimension whose size equals 3.
Dimension to operate along, specified as a positive integer scalar. The size of dimension dim must be 3. If no value is specified, the default is the first array dimension whose size equals 3.
Consider two 2-D input arrays, A and B:
- cross(A,B,1) treats the columns of A and B as vectors and returns the cross products of corresponding columns.
- cross(A,B,2) treats the rows of A and B as vectors and returns the cross products of corresponding rows.
Cross returns an error if dim is greater than ndims(A).
Cross products are sometimes called vector products, because, unlike dot products, which return a scalar, the result of a cross product is a vector. The resulting vector is always at right angles (normal) to the plane defined by the two input vectors — a property that is called orthogonality.
We have two vectors in three-dimensional space that represent both a direction and a magnitude.
$\vec i$, $\vec j$ and $\vec k$ are unit vectors in x, y, z directions.
$\vec A \times \vec B = \left| {\matrix{ i & j & k \cr {{a_1}} & {{a_2}} & {{a_3}} \cr {{b_1}} & {{b_2}} & {{b_3}} \cr } } \right| = ({a_2}{b_3} - {a_3}{b_2})i - ({a_1}{b_3} - {a_3}{b_1})j + ({a_1}{b_2} - {a_2}{b_1})k$
In MATLAB, the cross product is found using the function cross, which requires two inputs, the vectors A and B. Each of these MATLAB vectors must have three elements, since they represent the vector components in three-dimensional space.
9.2 Solutions of System of Linear Equations
Consider the following equations with three unknowns:
$$ 3x + 2y - z = 10$$
$$ - x + 3y + 2z = 5$$
$$ x - y - z = - 1$$
The equations can be rewritten by the matrices:
$$\underbrace {\left[ {\matrix{3 & 2 & -1 \cr { - 1} & 3 & 2 \cr 1 & { - 1} & { - 1} \cr} } \right]}_A\underbrace {\left[ {\matrix{x \cr y \cr z \cr} } \right]}_X = \underbrace {\left[ {\matrix{{10} \cr 5 \cr { - 1} \cr} } \right]}_B\quad \Rightarrow \quad AX = B$$
Solution Using the Matrix Inverse
Solution Using the Matrix Inverse
Probably the most straightforward way of solving this system of equations is to use the matrix inverse.
$AX = B\quad \Rightarrow \quad X = {A^{ - 1}}B$
As in all matrix mathematics, the order of multiplication is important. Since A is a 3×3 matrix, its inverse A-1 is also a 3×3 matrix. The multiplication A-1B :
works because the dimensions match up. The result is the 3×1 matrix X . If we change the order to BA-1 the dimensions would no longer match, and the operation would be impossible.
>> A = [3 2 -1; -1 3 2; 1 -1 -1];
>> B = [10; 5; -1];
>> X = inv(A)*B
X =
-2.0000
5.0000
-6.0000
>> X = A^-1*B
X =
-2.0000
5.0000
-6.0000
Solution Using Matrix Left Division
Solution Using Matrix Left Division
In MATLAB, we can use left division to solve the problem.
$$AX = B\quad \Rightarrow \quad X = {1 \over A}B\quad \Rightarrow X = A\backslash B$$
>> A = [3 2 -1; -1 3 2; 1 -1 -1];
>> B = [10; 5; -1];
>> X = A\B
X =
-2.0000
5.0000
-6.0000
Solution Using the Reverse Row Echelon Function
Solution Using the Reverse Row Echelon Function
| Syntax | Description |
|---|---|
| R = ref(A) | Return the reduced row echelon form of A using Gauss-Jordan elimination with partial pivoting. |
The linear equations can be solved by using the reduced row echelon function, rref. This function uses the Gauss-Jordan elimination technique.
>> A = [3 2 -1; -1 3 2; 1 -1 -1];
>> B = [10; 5; -1];
>> C = [A,B]
C =
3 2 -1 10
-1 3 2 5
1 -1 -1 -1
>> rref(C)
ans =
1 0 0 -2
0 1 0 5
0 0 1 -6
The solution to our problem is represented by the last column in the output array, and corresponds to the results achieved with the other methods.
Questions
Dot Product
- Compute the dot product of the following pairs of vectors, and then show that $A \cdot B = B \cdot A$
- $A = \left[ {\begin{array}{*{20}{c}}1&3&5\end{array}} \right]$ $B = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 2}&4\end{array}} \right]$
- $A = \left[ {\begin{array}{*{20}{c}}0&{ - 1}&{ - 4}&{ - 8}\end{array}} \right]$ $B = \left[ {\begin{array}{*{20}{c}}4&{ - 2}&{ - 3}&{24}\end{array}} \right]$
- Use a dot product and the shopping list in the following Table to determine your total bill at the grocery store.
Item Number Needed Cost Milk 2 gallons $\$ 3.50$ per gallon Eggs 1 dozen $\$ 1.25$ per dozen Cereal 2 boxes $\$ 4.25$ per box Soup 5 cans $\$ 1.55$ per can Cookies 1 package $\$ 3.15$ per package - Bomb calorimeters are used to determine the energy released during chemical reactions. The total heat capacity of a bomb calorimeter is defined as the sum of the products of the mass of each component and the specific heat capacity of each component, or
$$CP = \sum\limits_{i = 1}^n {{m_i}{C_i}} $$
where
${{m_i}} = $mass of component i, g
${{C_i}} = $ heat capacity of component, i , J/(gK)
$CP = $ total heat capacity, J/K
Find the total heat capacity of a bomb calorimeter, using the thermal data in the following Table.
Component Mass, g Heat Capacity, J?(gK) Steel 250 0.45 Water 100 4.2 Aluminum 10 0.90
Matrix Multiplication
- Compute the matrix product $A*B$ of the following pairs of matrices:
- $A = \left[ {\begin{array}{*{20}{c}}{12}&4\\3&{ - 5}\end{array}} \right]$ $B = \left[ {\begin{array}{*{20}{c}}2&{12}\\0&0\end{array}} \right]$
- $A = \left[ {\begin{array}{*{20}{c}}1&3&5\\2&4&6\end{array}} \right]$ $B = \left[ {\begin{array}{*{20}{c}}{ - 2}&4\\3&8\\{12}&{ - 2}\end{array}} \right]$
Determinants and Inverses
- Recall that not all matrices have an inverse. A matrix is singular (i.e., it doesn't have an inverse) if its determinant equals 0 (i.e., $\left| A \right| = 0$). Use the determinant function to test whether each of the following matrices has an inverse:
$A = \left[ {\begin{array}{*{20}{c}}2&{ - 1}\\2&5\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{c}}4&2\\2&1\end{array}} \right]$, $C = \left[ {\begin{array}{*{20}{c}}2&0&0\\1&2&2\\5&{ - 4}&0\end{array}} \right]$
If an inverse exists, compute it.
Solving Linear Systems of Equations
- Solve the following system of equations, using both matrix left division and the inverse matrix method:
- $ - 2x + y = 3$
$ x+y=10$ - $ 5x+3y-z=10$
$ 3x+2y+z=4$
$ 4x-y+3z=12$ - $ 3x+y+z+w=24$
$ x-3y+7z+w=12$
$ 2x+2y-3z+4w=17$
$ x+y+z+w=0$
- $ - 2x + y = 3$