Ch6_Det_&_Inv_Matrix_Ex6.3.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "bfe5227f",
   "metadata": {},
   "source": [
    "<b> Question 1</b>\n",
    "\n",
    "Compute the determinants of the following matrices:\n",
    "\n",
    "(a) $A = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&-10&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}\n",
    "\\qquad$ (b) $B = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&0&1\\\\\n",
    "0&1&0\\\\\n",
    "\\end{pmatrix}\n",
    "\\qquad$(c) $C = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&1&1\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}\\qquad$(d) $D = \\begin{pmatrix}\n",
    "0&0&0&1\\\\\n",
    "0&1&0&0\\\\\n",
    "0&0&1&0\\\\\n",
    "1&0&0&0\\\\\n",
    "\\end{pmatrix}\\qquad$$\\qquad(e)E = \\begin{pmatrix}\n",
    "1&0&0&0\\\\\n",
    "0&1&0&0\\\\\n",
    "0&0&-0.6&0\\\\\n",
    "0&0&0&1\\\\\n",
    "\\end{pmatrix}\\qquad$(f) $F = \\begin{pmatrix}\n",
    "1&0&0&0\\\\\n",
    "0&1&-7&0\\\\\n",
    "0&0&1&0\\\\\n",
    "0&0&0&1\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "cf69b7bc",
   "metadata": {},
   "source": [
    "<b>Solution</b> \n",
    "\n",
    "We know that, if $E$ is an elementary matrix.\n",
    "\n",
    "(i) is obtained from the identity matrix $I$ by multipliying a row by a non-zero scalar $k$ then $det(E) = k$\n",
    "\n",
    "(ii) is obtained from the identity matrix $I$ by adding a multiple of one row to another then $det(E)= 1$\n",
    "\n",
    "(iii) is obtained from the identity matrix $I$ by interchanging two rows then $det(E) = -1$\n",
    "\n",
    "<b>(a)</b> middle row of $I$ is multiplied by -10, therefore, $det(A) = -10$<br>\n",
    "<b>(b)</b> middle and bottom row $I$ are interchanged, therefore, $det(B) = -1$<br>\n",
    "<b>(c)</b> multiple of bottow row by 1 is added to middle row of $I$, therefore $det(C) = 1$<br>\n",
    "<b>(d)</b> row two and row three of $I$ are interchanged, therefore, $det(D) = -1$<br>\n",
    "<b>(e)</b> third rwo is multiplied by -0.6, therefore, $det(E) = -0.6$<br>\n",
    "<b>(f)</b> multiple of third row by -7 is added to second row, therefore $det(F) = 1$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "17eacbfa",
   "metadata": {},
   "source": [
    "------------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "dd3596b9",
   "metadata": {},
   "source": [
    "<b>Question 2</b>\n",
    "\n",
    "Find the determinants of the following matrices:"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3e8c2ad2",
   "metadata": {},
   "source": [
    "(a) $A = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&2&0\\\\\n",
    "0&0&3\\\\\n",
    "\\end{pmatrix}\\qquad$(b)$B = \\begin{pmatrix}\n",
    "1&-1&7\\\\\n",
    "0&2&9\\\\\n",
    "0&0&3\\\\\n",
    "\\end{pmatrix}\\qquad$(c)$C = \\begin{pmatrix}\n",
    "-2&0&0\\\\\n",
    "-98&-3&0\\\\\n",
    "67&17&1\\\\\n",
    "\\end{pmatrix}\\qquad$(d)$D = \\begin{pmatrix}\n",
    "1&-3&5&9\\\\\n",
    "0&-3&67&9\\\\\n",
    "0&0&8&90\\\\\n",
    "0&0&0&3\\\\\n",
    "\\end{pmatrix}\\qquad$\n",
    "\n",
    "(e)$E = \\begin{pmatrix}\n",
    "10&0&0&0\\\\\n",
    "0&20&0&0\\\\\n",
    "0&0&30&0\\\\\n",
    "0&0&0&40\\\\\n",
    "\\end{pmatrix}\\qquad$(f)$F = \\begin{pmatrix}\n",
    "-9&0&0&0\\\\\n",
    "67&3&0&0\\\\\n",
    "67&57&7&0\\\\\n",
    "23&78&6&5\\\\\n",
    "\\end{pmatrix}\\qquad$(g)$G = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "2&2&0\\\\\n",
    "-5&3&4\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "cc98f7a2",
   "metadata": {},
   "source": [
    "<b>Solution</b>\n",
    "\n",
    "We know that, <br>\n",
    "The determinant of a triangular or diagonal matrix is the <i>product of the entries along the leading diagonal</i>\n",
    "\n",
    "(a) Diagonal matrix, therefore $det(A) = 1 \\times 2 \\times 3 = 6$<br>\n",
    "(b) Upper triangular matrix, therefore, $det(B) = 1 \\times 2 \\times 3 = 6$<br>\n",
    "(c) Lower triangular matrix, therefore, $det(C) = -2 \\times -3 \\times 1 = 6$<br>\n",
    "(d) Upper triangular matrix, therefore $det(D) = 1 \\times -3 \\times 8 \\times 3 = -72$<br>\n",
    "(e) Diagonal matrix, therefore, $det(E) = 10 \\times 20 \\times 30 \\times 40 = 240000$<br>\n",
    "(f) Lower triangular matrix, therefore, $det(F) = -9 \\times 3 \\times 7 \\times 5 = -945$<br>\n",
    "(g) Lower triangular matrix, therefore, $det(G) = 1 \\times 2 \\times 4 = 8$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f50022b4",
   "metadata": {},
   "source": [
    "----------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f6552b30",
   "metadata": {},
   "source": [
    "<b>Question 3</b>\n",
    "\n",
    "Compute the determinants of the following matrices:\n",
    "\n",
    "(a)$A = \\begin{pmatrix}\n",
    "\\alpha &-\\beta & \\delta\\\\\n",
    "0&\\beta & -\\alpha\\\\\n",
    "0&0&\\gamma\\\\\n",
    "\\end{pmatrix}\\qquad$(b)$B = \\begin{pmatrix}\n",
    "sin(\\theta)&0&0\\\\\n",
    "1&cos(\\theta)&0\\\\\n",
    "-2&sin(\\theta)&2\\\\\n",
    "\\end{pmatrix}\\qquad$(c)$C =\\begin{pmatrix}\n",
    "x&y&z\\\\\n",
    "0&y&0\\\\\n",
    "0&0&z\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b3084539",
   "metadata": {},
   "source": [
    "<b> Solution</b> \n",
    "\n",
    "(a) Upper triangular matrix, therefore $det(A) =\\alpha\\beta\\gamma$<br>\n",
    "(b) Lower triangular matrix, therefore, $det(B) = sin(\\theta)cos(\\theta)2 = 2sin(\\theta)cos(\\theta) = 2sin(\\theta)$<br>\n",
    "(c) Upper triangular matrix, therefore, $det(C) = xyz$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c1504c19",
   "metadata": {},
   "source": [
    "--------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d077b0f2",
   "metadata": {},
   "source": [
    "<b>Question 4</b>\n",
    "\n",
    "Find the values of x which make the following matrix non-invertible(singular):\n",
    "\n",
    "$A = \\begin{pmatrix}\n",
    "1&2&x\\\\\n",
    "3&x&4\\\\\n",
    "5&5&4\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "662e5cb8",
   "metadata": {},
   "source": [
    "<b>Solution</b>\n",
    "\n",
    "We know that, <br>\n",
    "A square matrix $A$ is invertible(has an inverse) $\\iff det(A) \\ne 0$.\n",
    "\n",
    "Therefore, to make a matrix non-invertible $det(A) = 0$\n",
    "\n",
    "So, $det\\begin{pmatrix}\n",
    "1&2&x\\\\\n",
    "3&x&4\\\\\n",
    "5&5&4\\\\\n",
    "\\end{pmatrix} = det\\begin{pmatrix}\n",
    "x&4\\\\\n",
    "5&4\\\\\n",
    "\\end{pmatrix} - 2det\\begin{pmatrix}\n",
    "3&4\\\\\n",
    "5&4\\\\\n",
    "\\end{pmatrix} +xdet\\begin{pmatrix}\n",
    "3&x\\\\\n",
    "5&5\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (x^2 - 20) -2(3x - 20) + x(15 -5x) = 0$<br>\n",
    "$\\Rightarrow x^2 -20 -6x +40 + 15x -5x^2 = 0$<br>\n",
    "$\\Rightarrow -4x^2 + 9x + 20 = 0$\n",
    "\n",
    "Now, using quadratic equations formula, we get x as: \n",
    "\n",
    "$x = \\frac{-b\\pm \\sqrt{b^2 - 4ac}}{2a}$<br>\n",
    "$x = \\frac{-9\\pm \\sqrt{9^2 - 4(-4)20}}{2(-4)}$<br>\n",
    "\n",
    "$x = \\frac{-9\\pm \\sqrt{81 + 320}}{-8}$<br>\n",
    "$x = \\frac{-9 + \\sqrt{401}}{-8}$ or $x = \\frac{-9 - \\sqrt{401}}{-8}$<br>\n",
    "$x = \\frac{11.025}{-8}$ or $x = \\frac{-29.025}{-8}$\n",
    "\n",
    "$x = -1.38$ or $x = -3.63$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c9da9066",
   "metadata": {},
   "source": [
    "-------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "19f4b681",
   "metadata": {},
   "source": [
    "<b>Question 5</b>\n",
    "\n",
    "Find the determinants of the following matrices:\n",
    "\n",
    "(a)$A = \\begin{pmatrix}\n",
    "0&0&0&9\\\\\n",
    "0&1&4&5\\\\\n",
    "0&0&3&5\\\\\n",
    "1&2&3&4\\\\\n",
    "\\end{pmatrix}\\qquad$(b)$A = \\begin{pmatrix}\n",
    "1&2&3&4\\\\\n",
    "1&3&5&6\\\\\n",
    "1&4&3&7\\\\\n",
    "1&6&1&9\\\\\n",
    "\\end{pmatrix}\\qquad$(c)$C = \\begin{pmatrix}\n",
    "1&2&5&7\\\\\n",
    "3&6&2&8\\\\\n",
    "-1&-2&8&7\\\\\n",
    "-4&-5&1&2\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c5926c1b",
   "metadata": {},
   "source": [
    "<b> Solution </b> \n",
    "\n",
    "<b>(a)</b> We can interchange bottom row with top to make it an upper triangular matrix. Then,\n",
    "\n",
    "$A' = \\begin{pmatrix}\n",
    "1&2&3&4\\\\\n",
    "0&1&4&5\\\\\n",
    "0&0&3&5\\\\\n",
    "0&0&0&9\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now,<br> $det(A') = 1 \\times 1 \\times 3 \\times 9 = 27$ \n",
    "\n",
    "As $det(A') = k det(A)$. Therefore, $det(A) = \\frac{det(A')}{k}$, where $k$ is for interchange operation, i.e, -1\n",
    "\n",
    "Thus, $det(A) = -27$\n",
    "\n",
    "<b>Solution(b)</b>\n",
    "\n",
    "First, we need to make it triangular matrix least or diagonal.\n",
    "\n",
    "Step 1: Substracting $R_{1}$ from every row below:\n",
    "\n",
    "$A' = \\begin{pmatrix}\n",
    "1&2&3&4\\\\\n",
    "0&1&2&2\\\\\n",
    "0&2&0&3\\\\\n",
    "0&4&-2&5\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Step2: Adding $R_{2}$ to $R_{4}$\n",
    "\n",
    "\n",
    "$A' = \\begin{pmatrix}\n",
    "1&2&3&4\\\\\n",
    "0&1&2&2\\\\\n",
    "0&2&0&3\\\\\n",
    "0&5&0&7\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Step3: interchanging col2 and col3\n",
    "\n",
    "\n",
    "$A' = \\begin{pmatrix}\n",
    "1&3&2&4\\\\\n",
    "0&2&1&2\\\\\n",
    "0&0&2&3\\\\\n",
    "0&0&5&7\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Step 4: Multiple $R_{3}$ by 5 and $R_{4}$ by 2 and substract $R_{3}$ from $R_{4}$\n",
    "\n",
    "\n",
    "$A' = \\begin{pmatrix}\n",
    "1&3&2&4\\\\\n",
    "0&2&1&2\\\\\n",
    "0&0&10&15\\\\\n",
    "0&4&0&-1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now $det(A') = 1 \\times 2 \\times 10 \\times -1 = -20$\n",
    "\n",
    "\n",
    "But, $det(A') = step1_k \\times step2_k \\times step_3 \\times step4_k \\times det(A)$<br>\n",
    "$\\Rightarrow det(A') = 1 \\times 1 \\times -1 \\times (5 \\times 2)(1) \\times det(A)$<br>\n",
    "$\\Rightarrow det(A') -10 \\times det(A)$\n",
    "\n",
    "Therefore, $det(A) = \\frac{-20}{-10} = 2$\n",
    "\n",
    "<b>Solution(c)</b>\n",
    "\n",
    "Step 1: $R_{2} = R_{2} - 3R_{1}, R_{3} = R_{3} + R_{1}$ and $R_{4} = R_{4} + 4R_{1}$\n",
    "\n",
    "Then,\n",
    "$A' = \\begin{pmatrix}\n",
    "1&2&5&7\\\\\n",
    "0&0&-13&-13\\\\\n",
    "0&0&13&14\\\\\n",
    "0&3&21&30\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Step 2: $R_{3} = R_{3} +  R_{2}$\n",
    "Then,\n",
    "$A' = \\begin{pmatrix}\n",
    "1&2&5&7\\\\\n",
    "0&0&-13&-13\\\\\n",
    "0&0&0&1\\\\\n",
    "0&3&21&30\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Step 3: Suppose rows two times to make it upper triangular matrix\n",
    "\n",
    "Then, \n",
    "$A' = \\begin{pmatrix}\n",
    "1&2&5&7\\\\\n",
    "0&3&21&30\\\\\n",
    "0&0&-13&-13\\\\\n",
    "0&0&0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now, <br>$det(A') = 1 \\times 3 \\times (-13) \\times 1 = -39$\n",
    "\n",
    "Here, $det(A') = step1_k \\times step2_k \\times step3_k \\times det(A)$<br>\n",
    "$\\Rightarrow -39 = 1 \\times 1 \\times (-1 \\times -1) \\times det(A)$\n",
    "\n",
    "Hence, $det(A) = -39$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "08fe386b",
   "metadata": {},
   "source": [
    "---------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a6b50a43",
   "metadata": {},
   "source": [
    "<b>Question 6</b> Evaluate the determinant of the following matrices:\n",
    "\n",
    "(a)$A = \\begin{pmatrix}\n",
    "1&1&0&0&1\\\\\n",
    "-1&2&0&0&0\\\\\n",
    "0&0&1&4&16\\\\\n",
    "0&0&1&5&25\\\\\n",
    "0&0&1&2&4\\\\\n",
    "\\end{pmatrix}\\qquad$(b)$A = \\begin{pmatrix}\n",
    "1&2&3&4&5\\\\\n",
    "2&5&8&11&1\\\\\n",
    "7&6&1&9&8\\\\\n",
    "4&10&16&22&2\\\\\n",
    "2&3&7&9&5\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "998bae12",
   "metadata": {},
   "source": [
    "<b>Solution</b>\n",
    "\n",
    "<b>Solution(a)</b>\n",
    "\n",
    "Step 1: $R_{2} = R_{2} + 1 ,R_{4} = R_{4} - R_{3},$ and $R_{5} = R_{5} - R_{3}$\n",
    "\n",
    "$A' = \\begin{pmatrix}\n",
    "1&1&0&0&1\\\\\n",
    "0&3&0&0&1\\\\\n",
    "0&0&1&4&16\\\\\n",
    "0&0&0&1&9\\\\\n",
    "0&0&0&-2&-12\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Step 2: $R_{5} = R_{5} + 2R_{4}$\n",
    "\n",
    "$A' = \\begin{pmatrix}\n",
    "1&1&0&0&1\\\\\n",
    "0&3&0&0&1\\\\\n",
    "0&0&1&4&16\\\\\n",
    "0&0&0&1&9\\\\\n",
    "0&0&0&0&6\\\\\n",
    "\\end{pmatrix}\n",
    "$\n",
    "\n",
    "Therefore, $det(A') = 1 \\times 3 \\times 1 \\times 1 \\times6 = 18$\n",
    "\n",
    "$\\Rightarrow det(A') = step1_k \\times step2_k \\times det(A)$\n",
    "\n",
    "$\\Rightarrow 18 = 1 \\times 1 \\times det(A)$\n",
    "\n",
    "Hence, $det(A) = 18$\n",
    "\n",
    "<b>Solution(b)</b>\n",
    "\n",
    "We know that,<br>\n",
    "If a square matrix $A$ consists of two identical rows, then $det(A) = 0$\n",
    "\n",
    "Here, <br>\n",
    "in step 1: $R_{2} = 2R_{2}$, we get $R_{2} = R_{4}$ as :\n",
    "\n",
    "$A' = \\begin{pmatrix}\n",
    "1&2&3&4&5\\\\\n",
    "4&10&16&22&2\\\\\n",
    "7&6&1&9&8\\\\\n",
    "4&10&16&22&2\\\\\n",
    "2&3&7&9&5\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, $det(A') = 0$\n",
    "\n",
    "$\\Rightarrow det(A') = 2 \\times det(A)$<br>\n",
    "$\\Rightarrow 0 = 2 \\times det(A)$\n",
    "\n",
    "Hence, $det(A) = 0$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a8dc0a37",
   "metadata": {},
   "source": [
    "-------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "12a68664",
   "metadata": {},
   "source": [
    "<b>Question 7</b>\n",
    "\n",
    "(a)Show that $det(A+B) \\ne det(A) + det(B)$\n",
    "\n",
    "(b) Show that the $det(u\\quad v\\quad w) \\ne det(v\\quad u\\quad w)$\n",
    "\n",
    "(c) Show that $det(kA) \\ne kdet(A)$ where $A$ is an n by n matrix and $n \\ge 2$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "01c197ab",
   "metadata": {},
   "source": [
    "<b>Solution</b>\n",
    "\n",
    "(a) Let $A = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix}$ and $B = \\begin{pmatrix}\n",
    "-1&0\\\\\n",
    "0&-1\\\\\n",
    "\\end{pmatrix}$,\n",
    "\n",
    "Then, <br>\n",
    "$A + B = \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$det(A + B) = det\\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&0\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "Separately,<br>\n",
    "$det(A) = 1$ and $det(B) = 1$ as $det(I) = 1$\n",
    "\n",
    "Therefore, <br>\n",
    "$det(A) + det(B) = 1 + 1 = 2$\n",
    "\n",
    "<b>Hence, proved $det(A+B) \\ne det(A) + det(B)$</b>\n",
    "\n",
    "<b>Solution (b)</b><br>\n",
    "If $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$, and $w = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Then, <br>\n",
    "\n",
    "$det(u\\quad v\\quad w) = det\\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&1&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix} = det(I) = 1$\n",
    "\n",
    "\n",
    "$det(v\\quad u\\quad w) = det\\begin{pmatrix}\n",
    "0&1&0\\\\\n",
    "1&0&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix} = det(I) = -1 \\qquad$As interchanging rows make $det(A) = -det(A)$\n",
    "\n",
    "Hence, proved $det(u\\quad v\\quad w) \\ne det(v\\quad u\\quad w)$\n",
    "\n",
    "\n",
    "<b>Solution(c)</b>\n",
    "\n",
    "We know that from proposition 6.16 that<br>\n",
    "If we have square matrix $A$ of $n \\times n$ and $k$ be a scalar, then\n",
    "\n",
    "$det(kA) = k^{n}det(A)$\n",
    "\n",
    "If we have<br>\n",
    "$A = \\begin{pmatrix}\n",
    "1&2&1\\\\\n",
    "3&1&2\\\\\n",
    "1&4&2\\\\\n",
    "\\end{pmatrix}$ and $k$ as 3\n",
    "\n",
    "Then, <br>\n",
    "$det(kA) = 3^{3} det\\begin{pmatrix}\n",
    "1&2&1\\\\\n",
    "3&1&2\\\\\n",
    "1&4&2\\\\\n",
    "\\end{pmatrix}\\qquad$ as it is a $3 \\times 3$ matrix\n",
    "\n",
    "$\\Rightarrow det(kA) = 27\\left[(1det\\begin{pmatrix}\n",
    "1&2\\\\\n",
    "4&2\\\\\n",
    "\\end{pmatrix} -2 det\\begin{pmatrix}\n",
    "3&2\\\\\n",
    "1&2\\\\\n",
    "\\end{pmatrix} + 3 det\\begin{pmatrix}\n",
    "3&1\\\\\n",
    "1&4\\\\\n",
    "\\end{pmatrix}\\right]$\n",
    "\n",
    "$\\Rightarrow 27\\left[(2-8) -2(6-2) (12-1)\\right]$\n",
    "\n",
    "$\\Rightarrow det(kA) = 27(-3)$<br>\n",
    "where we have $det(A) = -3$\n",
    "\n",
    "\n",
    "Now, lets solve $kdet(A)$,\n",
    "\n",
    "Here, we have $k = 3$ and $det(A) = -3$\n",
    "\n",
    "Therefore, $kdet(A) = 3(-3) = -9$\n",
    "\n",
    "<b>Hence proved $det(kA) \\ne kdet(A)$ </b>"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f8a6ea43",
   "metadata": {},
   "source": [
    "-------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "2a789265",
   "metadata": {},
   "source": [
    "<b>Question 8</b>\n",
    "\n",
    "Evaluate the determinants of the following matrices\n",
    "\n",
    "(a)$A = \\begin{pmatrix}\n",
    "1/2&1/2&-1/2\\\\\n",
    "2&3&4\\\\\n",
    "1&-1&1\\\\\n",
    "\\end{pmatrix}\\qquad$(b)$B = \\begin{pmatrix}\n",
    "1/2&1/3&1\\\\\n",
    "1/7&2/21&1/21\\\\\n",
    "2/3&1/3&-4/3\\\\\n",
    "\\end{pmatrix}\\qquad$(c)$C = \\begin{pmatrix}\n",
    "10&20&30\\\\\n",
    "-4&5&-6\\\\\n",
    "-70&80&-90\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b559acd3",
   "metadata": {},
   "source": [
    "<b>Solution 8</b>\n",
    "\n",
    "We know that, \n",
    "\n",
    "$det(B) = \\begin{cases}\n",
    "det(A)\\quad \\text{if a multiple of one row is added to another}\\\\\n",
    "-det(A)\\quad \\text{if two rows have been interchanged}\\\\\n",
    "k det(A)\\quad \\text{if a row has been multiplied by non-zero k}\\\\\n",
    "\\end{cases}$\n",
    "\n",
    "<b>(a)</b> Here, $A$ can be written as:<br>\n",
    "\n",
    "$A = \\begin{pmatrix}\n",
    "1/2(1)&1/2(1)&1/2(-1)\\\\\n",
    "2&3&4\\\\\n",
    "1&-1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, \n",
    "\n",
    "$det(A) = \\frac{1}{2}det \\begin{pmatrix}\n",
    "1&1&-1\\\\\n",
    "2&3&4\\\\\n",
    "1&-1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$= \\frac{1}{2}\\left(det\\begin{pmatrix}\n",
    "3&4\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} - det\\begin{pmatrix}\n",
    "2&4\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix} - det\\begin{pmatrix}\n",
    "2&3\\\\\n",
    "1&-1\\\\\n",
    "\\end{pmatrix}\\right)$\n",
    "\n",
    "$= \\frac{1}{2}(3+4 -(2-4)-(-2-3))$\n",
    "\n",
    "$\\Rightarrow det(A) = 7$\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "<b>Solution(b)</b>\n",
    "\n",
    "Step 1: $R_1 = 6R_{1}, R_{2} = 21R_{2}$ and $R_{3} = 3R_{3}$\n",
    "\n",
    "We have, $B' = \\begin{pmatrix}\n",
    "3&2&6\\\\\n",
    "3&2&1\\\\\n",
    "2&1&4\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Then, $det(B') = 6 \\times 21\n",
    "\\times 3 \\times det(B)$\n",
    "\n",
    "Lets calculate $det(B')$\n",
    "\n",
    "$det(B') = 3 det\\begin{pmatrix}\n",
    "2&1\\\\\n",
    "1&-4\\\\\n",
    "\\end{pmatrix} - 2 det\\begin{pmatrix}\n",
    "3&1\\\\\n",
    "2&-4\\\\\n",
    "\\end{pmatrix} + 6 det\\begin{pmatrix}\n",
    "3&2\\\\\n",
    "2&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow det(B') = 3(-8-1) -2(-12-2) +6(3-4)$\n",
    "\n",
    "$\\Rightarrow det(B')= -5$\n",
    "\n",
    "Therefore, <br>\n",
    "$det(B') =  6 \\times 21\n",
    "\\times 3 \\times det(B) $ <br>\n",
    "\n",
    "$ -5 = 378 \\times det(B)$\n",
    "\n",
    "Hence, $det(B) = \\frac{-5}{378}$\n",
    "\n",
    "<b>Solution(c)</b>\n",
    "\n",
    "Here, $C$ can be written as:\n",
    "\n",
    "$C = \\begin{pmatrix}\n",
    "10(1)&10(2)&10(3)\\\\\n",
    "-4&5&-6\\\\\n",
    "10(-7)&10(8)&10(-9)\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, <br>\n",
    "\n",
    "$det(C) = 10 \\times 10 det\\begin{pmatrix}\n",
    "1&2&3\\\\\n",
    "-4&5&-6\\\\\n",
    "-7&8&-9\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$= 100 \\left(1det\\begin{pmatrix}\n",
    "5&-6\\\\\n",
    "8&-9\\\\\n",
    "\\end{pmatrix} -2 det\\begin{pmatrix}\n",
    "-4&-8\\\\\n",
    "-7&-9\\\\\n",
    "\\end{pmatrix} + 3 det\\begin{pmatrix}\n",
    "-4&5\\\\\n",
    "-7&8\\\\\n",
    "\\end{pmatrix}\\right)$\n",
    "\n",
    "$= 100\\left[1(5(-9) -8(-6) -2(-4(-9) - (-7)(-6))  -3((-4)8 - (-7)5)\\right]$<br>\n",
    "$= 100\\left[(-45 + 48) -2(36 - 42) - 3(-32 + 35)\\right]$<br>\n",
    "$= 100(3 + 12 -9)$\n",
    "\n",
    "\n",
    "$= 600$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d6812530",
   "metadata": {},
   "source": [
    "--------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c4c1d572",
   "metadata": {},
   "source": [
    "<b>Question 9</b>\n",
    "\n",
    "Let $A$, $B$ be 3 by 3 matrices with $det(A) = 3$, $det(B) = -4$. Determine:\n",
    "\n",
    "\n",
    "(a) $det(-2AB)\\qquad$(b)$B = det(A^5B^6)\\qquad$(c)$det(A^{-1}A^T)$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "417e19e7",
   "metadata": {},
   "source": [
    "<b> Solution</b>\n",
    "\n",
    "<b>(a)</b><br>\n",
    "$det(-2AB) = det((-2)AB)$\n",
    "\n",
    "We know that $det(kA) = k^{n}det(A)$. Therefore, <br>\n",
    "$det(-2AB) = (-2)^3 det(AB)$<br>\n",
    "$\\Rightarrow det(-2AB) = -8 det(A)det(B)$\n",
    "\n",
    "$\\Rightarrow det(-2AB) = -8 \\times 3 \\times -4$<br>\n",
    "$\\Rightarrow det(-2AB) = -96$\n",
    "\n",
    "\n",
    "<b>(b)</b>\n",
    "$det(A^5B^6) = det(A^5)det(B^6)$\n",
    "\n",
    "\n",
    "We know that, $det(A^n) = det(A)^n$. Therefore, <br>\n",
    "$det(A^5B^6) = det(A)^5 det(B)^6$<br>\n",
    "$ = 3^5(-4)^6$<br>\n",
    "\n",
    "Hence, $det(A^5B^6) = 995328$\n",
    "\n",
    "\n",
    "<b>(c)</b>\n",
    "\n",
    "We know that, \n",
    "\n",
    "$det(A^T) = det(A)$, and<br>\n",
    "$det(A^{-1} = \\frac{1}{det(A)}$\n",
    "\n",
    "Therefore, <br>\n",
    "$det(A^{-1}A^T) = det(A^{-1})det(A^T)$\n",
    "\n",
    "$= \\frac{1}{det(A)}det(A)$<br>$ = 1$\n",
    "\n",
    "$\\Rightarrow det(A^{-1}A^T) = 1$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0a579d98",
   "metadata": {},
   "source": [
    "---------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "344ca2d7",
   "metadata": {},
   "source": [
    "<b> Question 10</b>\n",
    "\n",
    "Let $A = \\begin{pmatrix}\n",
    "1&2\\\\\n",
    "3&4\\\\\n",
    "\\end{pmatrix}, B = \\begin{pmatrix}\n",
    "5&6\\\\\n",
    "7&8\\\\\n",
    "\\end{pmatrix}$, and $C = \\begin{pmatrix}\n",
    "9&10\\\\\n",
    "11&12\\\\\n",
    "\\end{pmatrix}$. Decide whether the matrix $A \\times B \\times C$ is invertible"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8640c712",
   "metadata": {},
   "source": [
    "<b>Solution</b>\n",
    "\n",
    "We know that, <br>\n",
    "A square matrix $A$ is invertible $\\iff det(A) \\ne 0$. So, need to check $A \\times B \\times C \\ne 0$ or not.\n",
    "\n",
    "We have, <br>\n",
    "\n",
    "$det(A) = \\begin{pmatrix}\n",
    "1&2\\\\\n",
    "3&4\\\\\n",
    "\\end{pmatrix} = 4 - 6 = -2$\n",
    "\n",
    "$det(B) = \\begin{pmatrix}\n",
    "5&6\\\\\n",
    "7&8\\\\\n",
    "\\end{pmatrix} = 40 - 42 = -2$\n",
    "\n",
    "$det(C) = \\begin{pmatrix}\n",
    "9&10\\\\\n",
    "11&12\\\\\n",
    "\\end{pmatrix} = 108 -110 = -2$\n",
    "\n",
    "Now, $det(A) \\times det(B) \\times det(C) = -2 \\times -2 \\times -2 = -8 \\ne 0$\n",
    "\n",
    "Hence, it is invertible"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "eeba70d9",
   "metadata": {},
   "source": [
    "-----------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "de493b61",
   "metadata": {},
   "source": [
    "<b>Question 11</b>\n",
    "\n",
    "Without evaluating the determinants of the following matrices, decide whether they are positive, negative or zero:\n",
    "\n",
    "(a)$A = \\begin{pmatrix}\n",
    "0&0&0&61\\\\\n",
    "0&89&0&0\\\\\n",
    "0&0&98&0\\\\\n",
    "21&0&0&0\\\\\n",
    "\\end{pmatrix}\\qquad$(b)$B = \\begin{pmatrix}\n",
    "2&5&7&8\\\\\n",
    "7&8&3&1\\\\\n",
    "12&30&42&48\\\\\n",
    "10&51&41&44\\\\\n",
    "\\end{pmatrix}\\qquad$(c)$C = \\begin{pmatrix}\n",
    "0&0&25&1\\\\\n",
    "0&0&-5&0\\\\\n",
    "0&2&0&0\\\\\n",
    "23&0&0&0\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "19e2300f",
   "metadata": {},
   "source": [
    "<b>Solution</b><br>\n",
    "\n",
    "(a)\n",
    "\n",
    "\n",
    "We need to swipe first row with the last row, and we know that swapping multiplies $det$ with $-1$,\n",
    "\n",
    "we have now, $\\begin{pmatrix}\n",
    "21&0&0&0\\\\\n",
    "0&89&0&0\\\\\n",
    "0&0&98&0\\\\\n",
    "0&0&0&61\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Determinant of this will be position as each diagonal elements are positive. Therefore, <b>net determinant wil be negative</b>\n",
    "\n",
    "\n",
    "(b)\n",
    "\n",
    "If we will multiply the first row by six, we will have:<br>\n",
    "$\\begin{pmatrix}\n",
    "12&30&42&48\\\\\n",
    "7&8&3&1\\\\\n",
    "12&30&42&48\\\\\n",
    "10&51&41&44\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "We know that, if we have identical rows, then determinant is zero. Therefore, <b>$det(B)$ is Zero</b>\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c47ec74b",
   "metadata": {},
   "source": [
    "----------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "97c0e1ca",
   "metadata": {},
   "source": [
    "<b>Question 12</b>\n",
    "\n",
    "Prove that the determinant of a lower triangular matrix is the product of the entries along the leading diagonal"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e4299c27",
   "metadata": {},
   "source": [
    "<b> Question 13</b>\n",
    "\n",
    "\n",
    "Prove that the determinant of a diagonal matrix is the product of the entries along the leading diagonal"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "005bf65a",
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3 (ipykernel)",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.9.9"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 5
}