Add maths assignment 2
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\documentclass{article}
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\usepackage{amsmath}
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\usepackage{amssymb}
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\begin{document}
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\title{Mathematics Assignment 2 --- Matrices}
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\author{Ahmad Saalim Lone, 2019BCSE017}
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\date{05 May, 2020}
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\maketitle
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\section*{Question 1}
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If
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\(
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A =
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\begin{bmatrix}
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3 & 3 & 4 \\
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2 & -3 & 4 \\
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0 & -1 & 1
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\end{bmatrix}
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\)
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find $A^{-1}$ and verify that $A^{-1}A = I = AA^{-1}$.
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\subsection*{Solution}
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\begin{align*}
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A &=
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\begin{bmatrix}
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3 & 3 & 4 \\
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2 & -3 & 4 \\
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0 & -1 & 1
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\end{bmatrix} \\
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A^{-1} &= \frac{adj(A)}{|A|} \\
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|A| &= 3 \times (-3 + 4 ) - 3 \times 2 + 4 \times -2 \\
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|A| &= -11 \\
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\text{Cofactor matrix of A } &=
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\begin{bmatrix}
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1 & 2 & -2 \\
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7 & 3 & -3 \\
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24 & 4 & -15
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\end{bmatrix} \\
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adj(A) &=
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\begin{bmatrix}
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1 & 7 & 24 \\
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2 & 3 & 4 \\
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-2 & -3 & -15
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\end{bmatrix} \\
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A^{-1} &=
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\begin{bmatrix}
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\frac{1}{11} & \frac{2}{11} & -\frac{2}{11} \\[6pt]
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\frac{7}{11} & \frac{3}{11} & -\frac{3}{11} \\[6pt]
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\frac{24}{11} & \frac{4}{11} & -\frac{15}{11}
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\end{bmatrix}
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\end{align*}
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\begin{align*}
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A \times A^{-1} &=
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\begin{bmatrix}
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\frac{1}{11} & \frac{2}{11} & -\frac{2}{11} \\[6pt]
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\frac{7}{11} & \frac{3}{11} & -\frac{3}{11} \\[6pt]
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\frac{24}{11} & \frac{4}{11} & -\frac{15}{11}
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\end{bmatrix}
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\begin{bmatrix}
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3 & 3 & 4 \\
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2 & -3 & 4 \\
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0 & -1 & 1
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\end{bmatrix} \\
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&=
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\begin{bmatrix}
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1
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\end{bmatrix} \\
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&= I
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\end{align*}
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Similarly, $A^{-1}\times A = I$
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\section*{Question 2}
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Find the inverse of
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\(
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A =\begin{bmatrix}
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1 & 2 & 1 \\
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3 & 2 & 3 \\
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1 & 1 & 2
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\end{bmatrix}
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\) by applying E-transformation.
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\subsection*{Solution}
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\[
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\begin{bmatrix}
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1 & 2 & 1 \\
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3 & 2 & 3 \\
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1 & 1 & 2
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\end{bmatrix}
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=
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\begin{bmatrix}
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1
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\end{bmatrix}
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A
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\]
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After applying the following transformations
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\begin{align*}
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C_3 &\to C_3 - C_1 \\
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C_2 &\to \frac{C_1}{2} \\
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R_1 &\to R_1 - R_2 \\
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R_1 &\to \frac{R_1}{-2} \\
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C_1 &\to C_1 - C_3 \\
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R_2 &\to R_2 - 3R1 \\
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R_2 &\to R_3 - \frac{R_2}{2}
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\end{align*}
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we get
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\begin{align*}
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\begin{bmatrix}
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1
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\end{bmatrix}
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&=
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\begin{bmatrix}
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-1 & \frac{1}{4} & \frac{1}{2} \\[6pt]
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3 & -\frac{1}{4} & -\frac{1}{2} \\[6pt]
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-\frac{5}{2} & \frac{1}{8} & \frac{3}{4}
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\end{bmatrix}
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\times A \\
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\therefore A^{-1} &=
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\begin{bmatrix}
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-1 & \frac{1}{4} & \frac{1}{2} \\[6pt]
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3 & -\frac{1}{4} & -\frac{1}{2} \\[6pt]
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-\frac{5}{2} & \frac{1}{8} & \frac{3}{4}
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\end{bmatrix}
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\end{align*}
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\section*{Question 3}
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Reduce the matrix
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\(
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A =
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\begin{bmatrix}
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1 & -1 & 2 & -3 \\
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4 & 1 & 0 & 2 \\
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0 & 3 & 0 & 4 \\
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0 & 1 & 0 & 2
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\end{bmatrix}
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\) to the normal form and hence determine its rank.
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\subsection*{Solution}
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To reduce the matrix we apply the following operations
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\begin{align*}
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R_1 &\to R_1 + R_3 \\
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R_4 &\to R_4 - R_2 \\
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R_2 &\to R_2 + R_4 \\
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C_2 &\to C_2 - C_3 \\
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C_4 &\to C_4 - C_2 \\
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R_3 &\to R_3 - R_2 \\
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C_2 &\rightleftharpoons C_3 \\
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C_2 &\to \frac{C_2}{2} \\
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R_3 &\to \frac{R_3}{2} \\
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R_4 &\to \frac{R_4}{2} \\
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C_4 &\to C_4 - C_2 \\
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C_3 &\to C_3 - C_4 \\
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C_4 &\rightleftharpoons C_2 \\
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R_4 &\to R_4 - R_1 \\
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R_4 &\to -R_4 \\
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R_1 &\to R_1 - R_4
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\end{align*}
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At the end we arrive at
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\[
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\begin{bmatrix}
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1 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 \\
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0 & 0 & 1 & 0 \\
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0 & 0 & 0 & 1
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\end{bmatrix}
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\]
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i.e. $[I_4]$
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$rank = 4$
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\section*{Question 4}
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Reduce the matrix \(A =
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\begin{bmatrix}
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-2 & -1 & -3 & -1 \\
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1 & 2 & 3 & -1 \\
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1 & 0 & 1 & 1 \\
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0 & 1 & 1 & -1
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\end{bmatrix}
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\) to Echelon form and find its rank.
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\subsection*{Solution}
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To convert it into Echelon form, we apply the following transformations.
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\begin{align*}
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R_1 &\to R_1 + R_2 + R_3 \\
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R_2 &\to R_2 - R_3 \\
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C_2 &\to C_2 + C_4 \\
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C_4 &\to C_4 + C_3 \\
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C_4 &\to C_4 - 2 C_1 \\
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C_3 &\to C_3 - C_2 \\
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C_2 &\to C_2 - C_1 \\
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R_2 &\to \frac{R_2}{2} \\
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R_1 &\leftrightharpoons R_4 \\
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C_4 &\leftrightharpoons C_1 \\
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R_4 &\to R_4 - R_2
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\end{align*}
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We get
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\[
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\begin{bmatrix}
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0 & 1 & 0 & 0 \\
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0 & 0 & 1 & 0 \\
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0 & 0 & 0 & 1 \\
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0 & 0 & 0 & 0
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\end{bmatrix}
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\]
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\[
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rank = 3
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\]
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\section*{Question 5}
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Solve
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\begin{align*}
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x + y + z &= 9 \\
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2x + 5y + 7z &= 52 \\
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2x + y - z &= 0
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\end{align*}
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\subsection*{Solution}
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\begin{align*}
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\Delta &=
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\begin{vmatrix}
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1 & 1 & 1 \\
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2 & 5 & 7 \\
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2 & 1 & -1
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\end{vmatrix}
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= -4
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\\
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\Delta_1 &=
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\begin{vmatrix}
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9 & 1 & 1 \\
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52 & 5 & 7 \\
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0 & 1 & -1
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\end{vmatrix}
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= -4
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\\
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\Delta_2 &=
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\begin{vmatrix}
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1 & 9 & 1 \\
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2 & 52 & 7 \\
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2 & 0 & -1
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\end{vmatrix}
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= -12
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\\
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\Delta_3 &=
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\begin{vmatrix}
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1 & 1 & 9 \\
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2 & 5 & 52 \\
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2 & 1 & 0
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\end{vmatrix}
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= -20
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\\
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x &= \frac{\Delta_1}{\Delta} = 1 \\
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y &= \frac{\Delta_2}{\Delta} = 3 \\
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z &= \frac{\Delta_3}{\Delta} = 5
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\end{align*}
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\section*{Question 6}
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Test the consistency of:
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\begin{align*}
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3x - y + 2z &= 3 \\
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2x + y + 3z &= 5 \\
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x - 2y - z &= 1
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\end{align*}
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\subsection*{Solution}
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\[
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\text{Augemented matrix} =
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\begin{bmatrix}
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3 & -1 & 2 & : & 3 \\
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2 & 1 & 3 & : & 5 \\
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1 & -2 & -1 & : & 1
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\end{bmatrix}
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\]
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\[
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\Delta =
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\begin{vmatrix}
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3 & -1 & 2 \\
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2 & 1 & 3 \\
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1 & -2 & -1
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\end{vmatrix}
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= 10
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\]
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\[
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\Delta \ne 0 \therefore \text{it is consistent with the unique solution}
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\]
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\section*{Question 7}
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Solve the equations
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\begin{align*}
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x + 3y - 2z &= 0 \\
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2x -y + 4z &= 0 \\
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x - 11y + 14z &= 0
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\end{align*}
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\subsection*{Solution}
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\begin{align*}
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\Delta =\begin{vmatrix}
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1 & 3 & -2 \\
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2 & -1 & 4 \\
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1 & -11 & 14
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\end{vmatrix}
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&= 0 \\
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\Delta_1 = \Delta_2 = \Delta_3 &= 0 \\
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\therefore \text{unique solution is } x = y = z &= 0
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\end{align*}
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\end{document}
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