ECON 441: Introduction to Mathematical Economics
Lecture 4
Only possible to multiply two matrices, \(A_{m \times n}\) and \(B_{p \times q}\) to get \(AB\) if \(n = p\) i.e. \[\text{number of columns in } A = \text{ number of rows in } B\]
So how to actually multiply these matrices? \[C = AB\] \[c_{ij} = a_{i1} b_{1j} + a_{i2} b_{2j} + ... + a_{in} b_{nj} = \sum_{k=1}^n a_{ik} b_{kj}\]
The element \(c_{ij}\) is obtained by multiplying term-by-term the entries of the \(i\)th row of \(A\) and \(j\)th column of \(B\).
\[A = \begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
\end{bmatrix}_{2 \times 3}
B = \begin{bmatrix}
b_{11} & b_{12} \\
b_{21} & b_{22} \\
b_{31} & b_{32} \\
\end{bmatrix}_{3 \times 2}\]
Here, \[C = AB = \begin{bmatrix}
a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31} & a_{11}b_{12} + a_{12}b_{22} + a_{13}b_{32} \\
a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31} & a_{21}b_{12} + a_{22}b_{22} + a_{23}b_{32} \\
\end{bmatrix}_{2 \times 2}\]
\[A = \begin{bmatrix} 2 & 3 & 1 \\ 4 & -6 & -2 \end{bmatrix}_{2 \times 3} \quad B = \begin{bmatrix} 1 & 8 \\ -2 & 3 \end{bmatrix}_{2 \times 2}\]
\(AB\) is not defined here (why?), but \(BA\) is. Find \(BA\).
One more. \[A = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}_{3 \times 1} \quad B = \begin{bmatrix} 2 & 0 & 1 \end{bmatrix}_{1 \times 3}\]
Both \(AB\) and \(BA\) are defined. Find both. What are their dimensions?
\[A = \begin{bmatrix}
1 & 2 \\
1 & -3
\end{bmatrix} \quad
x = \begin{bmatrix}
q \\
p
\end{bmatrix} \quad
b = \begin{bmatrix}
100 \\
20
\end{bmatrix}\]
What is \(Ax\)? \[Ax = \begin{bmatrix}
q + 2p\\
q-3p
\end{bmatrix}\]
Setting \(Ax=b\) gives us back our demand and supply equations.
Matrices with only one column: column vectors \[x = \begin{bmatrix} x_1\\ x_2 \\ \vdots \\ x_n \end{bmatrix}\]
Matrices with only one row: row vectors \[x^T = \begin{bmatrix} x_1 & x_2 & \cdots & x_n \end{bmatrix}\]
A set of vectors is said to be linearly dependent if and only if any one of them can be expressed as a linear combination of the remaining vectors.
Example.\[v_1 = \begin{bmatrix}
1\\
2 \\
\end{bmatrix} \quad \quad v_2 = \begin{bmatrix}
2\\
4 \\
\end{bmatrix}\]
A set of vectors is said to be linearly dependent if and only if any one of them can be expressed as a linear combination of the remaining vectors.
Example.\[v_1 = \begin{bmatrix}
3\\
2 \\
\end{bmatrix} \quad \quad v_2 = \begin{bmatrix}
1\\
3 \\
\end{bmatrix} \quad \quad v_3 = \begin{bmatrix}
1\\
-4 \\
\end{bmatrix}\]
A set of \(m\)-vectors \(v_1, v_2, ...,v_n\) is linearly dependent if and only if there exists a set of scaler \(k_1, k_2, ..., k_n\) (not all zero) such that: \[\sum_{i=1}^n k_i v_i = 0 \quad (m \times 1)\]
Square matrix with \(1\)s in its principal diagonal and \(0\)s elsewhere
A \(2 \times 2\) identity matrix: \[I_2 = \begin{bmatrix}
1 & 0\\
0 & 1 \\
\end{bmatrix}\] A \(3 \times 3\) identity matrix: \[I_3 = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}\]
Acts like 1, \[AI = IA = A\]
Example. \[A = \begin{bmatrix} 2 & 3 & 1 \\ 4 & -6 & 2 \end{bmatrix}\]
A matrix is an idempotent matrix if it remains unchanged when multiplied by itself any number of times.
\(A\) is idempotent if and only if \(A = A^k\).
Is an identity matrix idempotent?
A null matrix is a matrix with all elements \(0\).
\[\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}\]
\(A + 0 = A\)
\(A 0 = 0\)
Transpose of A (\(A^T\), also written \(A'\)): interchange rows and columns
\[A = \begin{bmatrix}
2 & 3 & 1 \\
4 & -6 & 2
\end{bmatrix}\]
A matrix \(A\) is said to be symmetric if \[A^T=A\]
A matrix \(A\) is said to be skew-symmetric if \[A^T=-A\]
A matrix \(A\) is said to be orthogonal if \[A^T A=I\]
\[A=\left[\begin{array}{rrr} 1 & 2 & 0 \\ 2 & 3 & -5 \\ 0 & -5 & 4 \end{array}\right]\]
\[A=\left[\begin{array}{ccc}0 & -1 & 3 \\ 1 & 0 & -4 \\ -3 & 4 & 0\end{array}\right]\]
\[A=\left[\begin{array}{cc} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array}\right]\]
\[\begin{array}{l} \left(A^T\right)^T=A \\~\\ (A+B)^T=A^T+B^T \\~\\ (A B)^T=B^T A^T \\~\\ \end{array}\] Example: \(A=\left[\begin{array}{ll}4 & 1 \\ 9 & 0\end{array}\right] \quad B=\left[\begin{array}{ll}2 & 0 \\ 7 & 1\end{array}\right]\)
Exercise 4.4: 5, 7
Exercise 4.5: 1, 4
Exercise 4.6: 2, 6
Exercise 5.1: 3, 4
Textbook reference: 4.2-4.6
ECON 441 · Introduction to Mathematical Economics