17Calculus Precalculus - Matrices
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Basics of Matrices
Matrices are just tables of numbers (or variables) listed in rows and columns, exactly like a table. For example,
\(\displaystyle{ A = \left[ \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ a & b & c \end{array} \right] }\)
This matrix is a 4x3, with 4 rows and 3 columns. The matrix is called A.
To delineate a matrix, it will usually be written with brackets, like we've done here, or with parentheses. However, we never use straight vertical lines unless we are talking about a determinant.
The main idea with matrices is that not only is the number/variable important, but also the position it occupies. For example, matrices C and D below are not equal, even though they are the same size (2x2) and have the same elements, because some of the numbers are in different positions.
\(\displaystyle{ C = \left[ \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} \right] }\) |
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\(\displaystyle{ D = \left[ \begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array} \right] }\) |
So for two matrices to be equal, the dimensions must be the same and the numbers/variables must be in the same position.
If you are just starting to learn about matrices, these first two videos are for you.
Khan Academy - Introduction to matrices [11min-50secs]
video by Khan Academy |
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This second video explains matrices and some basic operations in more depth with plenty of examples.
MIP4U - Matrix Addition, Subtraction, and Scalar Multiplication [9min-36secs]
video by MIP4U |
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Overview of Basic Matrix Operations
Addition - - As described in the last video above, matrix addition requires both matrices to have the same number of rows and columns. We just add the corresponding elements.
Scalar Multiplication - - Multiplication of a matrix by a scalar just involves multiplying each element by that scalar.
Subtraction - - Finally, matrix subtraction is just a combination of addition and multiplication by the scalar \(-1\).
If any of these operations is difficult for you, it will help to go back and watch the last video above and then work some practice problems.
Matrix Multiplication
Unlike with addition, multiplication of two matrices is quite involved and the technique is somewhat counter-intuitive. Here is a good introduction video and example. He explains quite thoroughly how to do matrix multiplication. [More discussion and practice problems on multiplication of matrices can be found on the matrix inverses page.]
Khan Academy - Matrix multiplication introduction [6min-25secs]
video by Khan Academy |
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Matrix Division
The next natural question is, what about matrix division? Well, matrix division is even more complicated that matrix multiplication. Let's look at an analogy with numbers.
An Analogy - - If we want to divide 2 into 3, i.e. 3/2, we can think about it this way. We find the (multiplicative) inverse of 2, written either \(1/2\) or \(2^{-1}\) and then multiply this inverse to 3, i.e. \(3/2=3(2^{-1})\).
That is how we do matrix division. We need to find the multiplicative inverse (usually just called the inverse when talking about matrices) of a matrix and then multiply. Finding the inverse is quite involved and we will discuss this on a separate page.
Okay, now you are ready for some practice problems.
Practice
Instructions - Unless otherwise instructed, perform the indicated operations on these matrices.
Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rr} 3 & -1 \\ 2 & 0 \end{array} \right] + \left[\begin{array}{rr} -7 & 2 \\ 3 & 5 \end{array} \right] }\)
Problem Statement |
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rr} 3 & -1 \\ 2 & 0 \end{array} \right] + \left[\begin{array}{rr} -7 & 2 \\ 3 & 5 \end{array} \right] }\)
Solution |
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1793 video
video by Khan Academy |
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rr} 3 & -1 \\ 2 & 0 \end{array} \right] - \left[\begin{array}{rr} -7 & 2 \\ 3 & 5 \end{array} \right] }\)
Problem Statement |
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rr} 3 & -1 \\ 2 & 0 \end{array} \right] - \left[\begin{array}{rr} -7 & 2 \\ 3 & 5 \end{array} \right] }\)
Solution |
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1794 video
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rrr} 1 & -7 & 5 \\ 0 & 3 & -10 \end{array} \right] + \left[\begin{array}{rrr} 5 & 0 & 3 \\ 11 & -1 & 7 \end{array} \right] }\)
Problem Statement |
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rrr} 1 & -7 & 5 \\ 0 & 3 & -10 \end{array} \right] + \left[\begin{array}{rrr} 5 & 0 & 3 \\ 11 & -1 & 7 \end{array} \right] }\)
Solution |
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1795 video
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rr} 0 & 1 \\ 3 & 2 \end{array} \right] - \left[\begin{array}{rr} -1 & 3 \\ 0 & 5 \end{array} \right] }\)
Problem Statement |
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rr} 0 & 1 \\ 3 & 2 \end{array} \right] - \left[\begin{array}{rr} -1 & 3 \\ 0 & 5 \end{array} \right] }\)
Solution |
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1796 video
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rr} 1 & 2 \\ 3 & 8 \end{array} \right] + \left[\begin{array}{rr} 10 & 0.6 \\ 0.4 & 14 \end{array} \right] }\)
Problem Statement |
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rr} 1 & 2 \\ 3 & 8 \end{array} \right] + \left[\begin{array}{rr} 10 & 0.6 \\ 0.4 & 14 \end{array} \right] }\)
Solution |
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1797 video
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rrr} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array} \right] + \left[\begin{array}{rrr} 10 & 20 & 30 \\ 40 & 50 & 60 \\ 70 & 80 & 90 \end{array} \right] }\)
Problem Statement |
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rrr} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array} \right] + \left[\begin{array}{rrr} 10 & 20 & 30 \\ 40 & 50 & 60 \\ 70 & 80 & 90 \end{array} \right] }\)
Solution |
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1798 video
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rr} 2 & 3 \\ 4 & 5 \\ 6 & 7 \end{array} \right] + \left[\begin{array}{rr} 1 & 5 \\ 8 & 9 \\ 3 & 4 \end{array} \right] }\)
Problem Statement |
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{\left[\begin{array}{rr} 2 & 3 \\ 4 & 5 \\ 6 & 7 \end{array} \right] + \left[\begin{array}{rr} 1 & 5 \\ 8 & 9 \\ 3 & 4 \end{array} \right] }\)
Solution |
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1799 video
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{ \left[\begin{array}{rr} 1 & 2 \\ 3 & 8 \end{array} \right] - \left[\begin{array}{rr} 10 & 0.6 \\ 0.4 & 14 \end{array} \right] }\)
Problem Statement |
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{ \left[\begin{array}{rr} 1 & 2 \\ 3 & 8 \end{array} \right] - \left[\begin{array}{rr} 10 & 0.6 \\ 0.4 & 14 \end{array} \right] }\)
Solution |
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1800 video
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{10 \left[\begin{array}{rr} 2 & 3 \\ 4 & 5 \end{array} \right] }\)
Problem Statement |
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{10 \left[\begin{array}{rr} 2 & 3 \\ 4 & 5 \end{array} \right] }\)
Solution |
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1801 video
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{5 \left[\begin{array}{rrr} -1 & 2 & 8 \\ -3 & -7 & 1 \end{array} \right] }\)
Problem Statement |
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{5 \left[\begin{array}{rrr} -1 & 2 & 8 \\ -3 & -7 & 1 \end{array} \right] }\)
Solution |
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1804 video
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{4 \left[\begin{array}{r} 4 \\ 7 \end{array} \right] + 2 \left[\begin{array}{r} 0 \\ -3 \end{array} \right] }\)
Problem Statement |
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Unless otherwise instructed, perform the indicated operations on these matrices.
\(\displaystyle{4 \left[\begin{array}{r} 4 \\ 7 \end{array} \right] + 2 \left[\begin{array}{r} 0 \\ -3 \end{array} \right] }\)
Solution |
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1805 video
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Instructions - Unless otherwise instructed, use the matrices \(S\) and \(T\) in these problems.
\(\displaystyle{ S = \left[ \begin{array}{rr} 4 & 7 \\ 2 & -1 \end{array} \right] }\)
\(\displaystyle{ T = \left[ \begin{array}{rr} -3 & -7 \\ 5 & 6 \end{array} \right] }\)
Unless otherwise instructed, calculate \( S+T \), where
\(\displaystyle{S = \left[\begin{array}{rr} 4 & 7 \\ 2 & -1 \end{array} \right] }\) | \(\displaystyle{T = \left[\begin{array}{rr} -3 & -7 \\ 5 & 6 \end{array} \right] }\) |
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Problem Statement |
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Unless otherwise instructed, calculate \( S+T \), where
\(\displaystyle{S = \left[\begin{array}{rr} 4 & 7 \\ 2 & -1 \end{array} \right] }\) | \(\displaystyle{T = \left[\begin{array}{rr} -3 & -7 \\ 5 & 6 \end{array} \right] }\) |
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Solution |
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1802 video
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Unless otherwise instructed, calculate \( T-S \), where
\(\displaystyle{S = \left[\begin{array}{rr} 4 & 7 \\ 2 & -1 \end{array} \right] }\) | \(\displaystyle{T = \left[\begin{array}{rr} -3 & -7 \\ 5 & 6 \end{array} \right] }\) |
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Problem Statement |
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Unless otherwise instructed, calculate \( T-S \), where
\(\displaystyle{S = \left[\begin{array}{rr} 4 & 7 \\ 2 & -1 \end{array} \right] }\) | \(\displaystyle{T = \left[\begin{array}{rr} -3 & -7 \\ 5 & 6 \end{array} \right] }\) |
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Solution |
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Unless otherwise instructed, calculate \( T-3S \), where
\(\displaystyle{S = \left[\begin{array}{rr} 4 & 7 \\ 2 & -1 \end{array} \right] }\) | \(\displaystyle{T = \left[\begin{array}{rr} -3 & -7 \\ 5 & 6 \end{array} \right] }\) |
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Problem Statement |
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Unless otherwise instructed, calculate \( T-3S \), where
\(\displaystyle{S = \left[\begin{array}{rr} 4 & 7 \\ 2 & -1 \end{array} \right] }\) | \(\displaystyle{T = \left[\begin{array}{rr} -3 & -7 \\ 5 & 6 \end{array} \right] }\) |
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Solution |
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Really UNDERSTAND Precalculus
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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