To show latex rendering in webpage
#Day 2
1.1 no 71 Symmetric
\[A = A^T \quad a_{i,j} = a_{i,j}^T = a_{j,i} \quad a_{i,j} = a_{j,i} \quad \begin{bmatrix} a & b & c \\\ b & e & f \\\ c & f & i \end{bmatrix}\]1.2 no 37
\( u = \begin{bmatrix} 3 \\ 8 \end{bmatrix} , \quad S = \left\lbrace \begin{bmatrix} 1 \\ 2 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \end{bmatrix} \begin{bmatrix} -2 \\ 5 \end{bmatrix} \right\rbrace \quad \quad c_1 \begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_2 \begin{bmatrix} 2 \\ 3 \end{bmatrix} + c_3 \begin{bmatrix} -2 \\ 5 \end{bmatrix} \quad \quad \begin{bmatrix} c_1 + 2c_2 - 2c_3 \\ 2 c_1 + 3c_2 + 5c_3 \end{bmatrix} \quad \begin{bmatrix} 3 \\ 8 \end{bmatrix} \)
System of linear equations:
\( a_1x_1+a_2x_2+\cdots +a_nx_n = k \
a_1,\, a_2,\ \dots,\ a_n\; and \; k \; are \; real \; numbers \
x_1,\, x_2,\ \dots,\ x_n \; are \; the \; variables \)
An equation is non-lin if any var is raised to a power, multiplied together or involve other functions
System eq
\[a_{11} x_1 + a_{12} x_2 + \cdots a_{1n} x_n = b_1 \\\ a_{21} x_1 + a_{22} x_2 + \cdots a_{2n} x_n = b_2 \\\ \vdots \quad \quad \quad \quad \quad \vdots\\\ a_{m1} x_1 + a_{m2} x_2 + \cdots a_{mn} x_n = b_n \\\ x_1 \begin{bmatrix} a_{11} \\\ \vdots \end{bmatrix} + x_2 \begin{bmatrix} a_21 \\\ \vdots \end{bmatrix} + x_3 \begin{bmatrix} a_31 \\\ \vdots \end{bmatrix} = \begin{bmatrix} b \\\ \vdots \end{bmatrix} \\ A = \begin{bmatrix} a_11 \end{bmatrix} , x = \begin{bmatrix} x_1 \\\ x_2 \\\ \vdots \end{bmatrix} , b = \begin{bmatrix} b \\\ \vdots \end{bmatrix} \\\ Ax = b\]x + y +z = 1
x -4y -z = 7
0 + y + 2z = 0
\( \begin{bmatrix} 1 & 1 & 1 \\ 1 & -4 & -1 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 7 \\ 0 \end{bmatrix} \)
Three operations that do not change a system of equations:
- Interchange two equations \( A \xrightarrow{ r_i \leftrightarrow r_j} B \)
- Multiply an equation by a non-zero scaler \( A \xrightarrow{r_i \rightarrow cr_i} B \)
- Multiply an equation by a scaler and add to another equation \( A \xrightarrow{r_i \rightarrow cr_j + r_i} B \)
Interchange two equations \( \begin{bmatrix} 2&-3&4&1\\ -1&2&-1&-2\\ 2&3&1&-2 \end{bmatrix} \xrightarrow{r_1 \leftrightarrow r_2} \begin{bmatrix} -1&2&-1&-2\\ 2&-3&4&1\\ 2&3&1&-2 \end{bmatrix} \)
Multiply an equation by a non-zero scaler \( \begin{bmatrix} -1&2&-1&-2\\ 2&-3&4&1\\ 2&3&1&-2 \end{bmatrix} \xrightarrow{r_2 \rightarrow 2r_1 + r_2} \begin{bmatrix} -1&2&-1&-2\\ 0&1&2&-3\\ 2&3&1&-2 \end{bmatrix} \)
Multiply an equation by a scaler and add to another equation
\( \begin{bmatrix}
-1&2&-1&-2\
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0&1&2&-3\
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2&3&1&-2
\end{bmatrix}
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\xrightarrow{\begin{subarray}{l} r_3 \rightarrow 2r_1 + r_3, \ r_1 \rightarrow -1r_1 \end{subarray}}
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\begin{bmatrix}
1&-2&1&2\
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0&1&2&-3\
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0&7&-1&-6
\end{bmatrix}
\)
Special form of Matrices, goal is to use row ops to get to one of the special forms
The Row-Echelon Form:
- The zero rows are in the bottom of the matrix
- Each leading entry is in a column to the right of the leading entry in the previous row
- The entries under leading entries are equal to zero * not real condition, just an implication
Non-examples:
\[\begin{bmatrix} 2&-3&4\\\ 0&0&0\\\ 0&0&1 \end{bmatrix} \quad and \quad \begin{bmatrix} 2&-3&4 \\\ 0&0&5 \\\ 0&0&1 \end{bmatrix}\]Reduced Row Echelon form:
- All conditions of Row-Echelon form
- All leading entries are one (multiply with coefficient –> row operations to make this true)
- Entries above AND below leading entries are all zero -identity matrix in reduced row echelon
\( \begin{bmatrix} 1&-3&0&1 \\ 0&0&1&-2\\ 0&0&0&0 \end{bmatrix} \quad \begin{bmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix} \
Non\; examples \quad \begin{bmatrix} 1&0&4\\ 0&1&0\\ 0&0&2 \end{bmatrix} \quad \begin{bmatrix} 1&0&0\\ 0&0&1\\ 0&0&1 \end{bmatrix} \)
augmented matrix
x + y +z = 1
x -4y -z = 7
0 + y + 2z = 0
\( \begin{bmatrix} 1 & 1 & 1 & | 1 \\ 1 & -4 & | -1 \\ 0 & 1 & 2 & \0 \end{bmatrix} \quad \xrightarrow \begin{bmatrix} 1&0&0& \frac{7}{4}\\ 0&1&0 & \frac{-3}{2}\\ 0&0&1 & \frac{3}{4} \end{bmatrix} \\
x_1 = \frac{7}{4}, \quad x_2= -\frac{3}{2}, \quad x_3 = \frac{3}{4} \)
\( x = \begin{bmatrix} 2 \\ 1 - 2x_3 \\ x_3 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ -2x_3 \\ x_3 \end{bmatrix} \
= \begin{bmatrix} 2\\ 1 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 0 \\ -2 \\ 1 \end{bmatrix} \)
example b soln x3= 0, x2 = x2, x1 = 5 - 2x2
\( x = \begin{bmatrix} 5 - 2x_2 \\ x_2 \\ 0 \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} -2x2 x2 0…… \)
example c
no solution because
last row gives self contradictory solution 0 = 4