Ch7_EigenValuesVector_Ex7.3.ipynb

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    "<b>Question 1</b>\n",
    "\n",
    "For the following matrices find: \n",
    "\n",
    "(i) The eigenvalues and corresponding eigenvectors\n",
    "\n",
    "(ii) Eigenvector matrix $P$ and eigenvalue matrix $D$\n",
    "\n",
    "(a)$A = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}\\quad$(b)$A = \\begin{pmatrix}\n",
    "1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}\\quad$(c)$A = \\begin{pmatrix}\n",
    "3&0\\\\\n",
    "4&4\\\\\n",
    "\\end{pmatrix}\\quad$(d)$A  = \\begin{pmatrix}\n",
    "2&2\\\\\n",
    "1&3\\\\\n",
    "\\end{pmatrix}$"
   ]
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    "<b>Solution (a)</b> \n",
    "\n",
    "(i) Since $A$ is a diagonal matrix, its eigenvalues are $\\lambda_{1} = 1$ and $\\lambda_{2} = 2$\n",
    "\n",
    "For eigenvalue $\\lambda_{1} =1$<br>\n",
    "we get, $\\begin{pmatrix}\n",
    "1-1&0\\\\\n",
    "0&2-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix}\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow y = 0$ and $x$ can take any non-zero value $s$.<br>\n",
    "Thus, eigenvector $u = s\\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "For eigenvalue $\\lambda_{1} =2$<br>\n",
    "we get, $\\begin{pmatrix}\n",
    "1-2&0\\\\\n",
    "0&2-2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "-1&0\\\\\n",
    "0&0\\\\\n",
    "\\end{pmatrix}\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow x = 0$ and $y$ can take any non-zero value $s$.<br>\n",
    "Thus, eigenvector $u = s\\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "(ii) we know that eigenvector matrix is given by $P(u\\quad v)$\n",
    "$\\Rightarrow P(u \\quad v) = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "The diagonal matrix $D$ is given by $D = P^{-1}AP$\n",
    "\n",
    "Here, eigenvector matrix $P$ is identity matrix $I$\n",
    "\n",
    "$\\Rightarrow D = P^{-1}AI =A = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "<b> Solution(b)</b>\n",
    "\n",
    "We can get eigenvalues with $det(A - \\lambda I) = 0$<br>\n",
    "\n",
    "This implies $\\begin{pmatrix}\n",
    "1-\\lambda&1\\\\\n",
    "1&1-\\lambda\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (1-\\lambda)^2 - 1 = 0$<br>\n",
    "$\\Rightarrow 1 - 2\\lambda + \\lambda^2 -1 = 0$<br>\n",
    "$\\Rightarrow \\lambda(\\lambda -2) = 0$\n",
    "\n",
    "So, Eigenvalues are $\\lambda_{1} = 0$ and, $\\lambda_{2} =2$\n",
    "\n",
    "Now, for eigenvector $\\lambda_{1} = 0$<br>\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "1-0&1\\\\\n",
    "1&1-0\\\\\n",
    "\\end{pmatrix} \\cdot\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix} \\cdot\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow x + y = 0$<br>\n",
    "$\\Rightarrow x = -y$\n",
    "\n",
    "$\\Rightarrow u = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "For eigenvalue $\\lambda_{2} = 2$\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "1-2&1\\\\\n",
    "1&1-2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -x + y = 0$<br>\n",
    "\n",
    "$\\Rightarrow x = y = 1$\n",
    "\n",
    "$\\Rightarrow u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "(ii) we know that eigenvector matrix is given by $P(u\\quad v)$\n",
    "$\\Rightarrow P(u \\quad v) = \\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "The diagonal matrix $D$ is given by $D = P^{-1}AP$\n",
    "\n",
    "First, finding $P^{-1} = \\frac{1}{ad - bc}\\begin{pmatrix}\n",
    "d&-b\\\\\n",
    "-c&a\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow P^{-1} = \\frac{1}{-1-1}\\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "-1&-1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\quad = \\frac{1}{-2}\\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "-1&-1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "or $P^{-1} = \\frac{1}{2}\\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now,<br>\n",
    "$D = P^{-1}AP$\n",
    "\n",
    "$\\Rightarrow D = \\frac{1}{2}\\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{2}\\begin{pmatrix}\n",
    "0&0\\\\\n",
    "2&2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{2}\\begin{pmatrix}\n",
    "0&0\\\\\n",
    "-2+2&2+2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{2}\\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&4\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$D = \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "<b>Solution (c)</b>\n",
    "\n",
    "Since it is a lower triangular matrix, eigenvalues are $\\lambda_{1} =3$ and $\\lambda_{2} = 4$\n",
    "\n",
    "For eigenvector, $\\lambda_{1} = 3$, with $(A - \\lambda I)u =0$, <br>\n",
    "we get, $\\begin{pmatrix}\n",
    "3-3&0\\\\\n",
    "4&4-3\\\\\n",
    "\\end{pmatrix} \\cdot\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "4&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow x =1, y = -4$<br>\n",
    "Thus, eigenvector $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "-4\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "For eigenvector, $\\lambda_{2} = 4$, with $(A - \\lambda I)u =0$, <br>\n",
    "we get, $\\begin{pmatrix}\n",
    "3-4&0\\\\\n",
    "4&4-4\\\\\n",
    "\\end{pmatrix} \\cdot\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-1&0\\\\\n",
    "4&0\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow x = 0$ and $y$ can be any non-zero $s$.\n",
    "\n",
    "Therefore, $u = s\\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "(ii) Eigenvector $P$ can be given as: $P = (u \\quad v)$\n",
    "\n",
    "$\\Rightarrow P = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "-4&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow P^{-1} = \\frac{1}{1(1) -0(4)}\\begin{pmatrix}\n",
    "1&0\\\\\n",
    "4&1\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "4&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Next, Diagonal matrix, $D = P^{-1}AP$\n",
    "\n",
    "$\\Rightarrow D = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "4&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "3&0\\\\\n",
    "4&4\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "-4&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\begin{pmatrix}\n",
    "3&0\\\\\n",
    "16&4\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "-4&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\begin{pmatrix}\n",
    "3&0\\\\\n",
    "0&4\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "<b>Solution (d)</b>\n",
    "\n",
    "Eigenvalue can be computer from $det(A - \\lambda I) = 0$\n",
    "\n",
    "Here, <br>\n",
    "\n",
    "$det \\begin{pmatrix}\n",
    "2 - \\lambda &2 \\\\\n",
    "1 & 3 - \\lambda \\\\\n",
    "\\end{pmatrix} = 0$ \n",
    "\n",
    "$\\Rightarrow (2 - \\lambda)(3 - \\lambda) -2 = 0$<br>\n",
    "$\\Rightarrow 6 - 3\\lambda -2 \\lambda + \\lambda^2 -2 =$<br>\n",
    "$\\Rightarrow \\lambda^2 - 5\\lambda + 4 = 0$<br>\n",
    "$\\Rightarrow (\\lambda -1)(\\lambda -4) = 0$<br>\n",
    "\n",
    "$\\Rightarrow \\lambda_{1} =1$, and $\\lambda_{2} = 4$\n",
    "\n",
    "Now, for eigenvalue $\\lambda_{1} =1$ with, $(A - \\lambda I)u = 0$\n",
    "\n",
    "we get, <br>\n",
    "$\\begin{pmatrix}\n",
    "2-1&2\\\\\n",
    "1&3-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "1&2\\\\\n",
    "1&2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow x + 2y = 0$<br>\n",
    "If $y = 1$ we have $x = -2$\n",
    "\n",
    "Thus, eigenvector $u = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "Now, for eigenvalue $\\lambda_{2} =4$ with, $(A - \\lambda I)u = 0$\n",
    "\n",
    "we get, <br>\n",
    "$\\begin{pmatrix}\n",
    "2-4&2\\\\\n",
    "1&3-4\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-2&2\\\\\n",
    "1&-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -2x + 2y = 0$ or $x - y = 0$<br>\n",
    "\n",
    "$\\Rightarrow x = y= 1$\n",
    "\n",
    "Thus, eigenvector $v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "(ii) Eigenvector matrix, P then can be given by $P(u\\quad v)$\n",
    "\n",
    "$\\Rightarrow P = \\begin{pmatrix}\n",
    "-2&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Diagonal matrix, $D$, can be given by $D = P^{-1}AP$\n",
    "\n",
    "Therefore,<br>\n",
    "first $P^{-1} = \\frac{1}{-2(1) -1(1)}\\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "-1&-2\\\\\n",
    "\\end{pmatrix} = \\frac{1}{-3}\\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "-1&-2\\\\\n",
    "\\end{pmatrix} = \\frac{1}{3}\\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "Now, $ P^{-1}AP = \\frac{1}{3}\\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "2&2\\\\\n",
    "1&3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-2&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{3}\\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "4&8\\\\\n",
    "\\end{pmatrix}\\cdot \\begin{pmatrix}\n",
    "-2&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\qquad = \\frac{1}{3}\\begin{pmatrix}\n",
    "3&0\\\\\n",
    "0&12\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&4\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Thus, the diagonal matrix, $D = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&4\\\n",
    "\\end{pmatrix}$"
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    "-------------------------------------------"
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    "<b>Question 2</b> \n",
    "\n",
    "(i) For the matrices in question 1 find $A^5$\n",
    "\n",
    "(ii) For the matrix in question 1 part C find $A^{-1/2}$"
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    "<b> Solution (i) </b> \n",
    "\n",
    "We know that, if an $n \\times n$ matrix $A$ is diagonalizable with $P^{-1}AP =D$, then $A^{m} = PD^{m}P^{-1}$\n",
    "\n",
    "(a) We have, $P = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix}, P^{-1} = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix}$, and $D = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, $A^5 $<br>\n",
    "$\\qquad= \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix}\\cdot \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}^5 \\cdot\n",
    "\\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad= \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix}\\cdot \\begin{pmatrix}\n",
    "1^5&0\\\\\n",
    "0&2^5\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&32\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "(b) We have, $P = \\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}, P^{-1} = \\frac{1}{2}\\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$, and $D = \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, $A^5 = PD^5P^{-1}$\n",
    "\n",
    "$\\Rightarrow A^5 = \\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}\\cdot \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}^5 \\cdot \\frac{1}{2}\\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{2}\\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}\\cdot \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&32\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\qquad = \\frac{1}{2} \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&32\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$ \n",
    "\n",
    "$\\qquad  = \\frac{1}{2}\\begin{pmatrix}\n",
    "32&32\\\\\n",
    "32&32\\\\\n",
    "\\end{pmatrix} $\n",
    "\n",
    "$\\qquad  = \\begin{pmatrix}\n",
    "16&16\\\\\n",
    "16&16\\\\\n",
    "\\end{pmatrix} $\n",
    "\n",
    "\n",
    "(c) We have, $P = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "-4&1\\\\\n",
    "\\end{pmatrix}, P^{-1} = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "4&1\\\\\n",
    "\\end{pmatrix}$, and $D = \\begin{pmatrix}\n",
    "3&0\\\\\n",
    "0&4\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, $A^5 =  \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "-4&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "3&0\\\\\n",
    "0&4\\\\\n",
    "\\end{pmatrix}^5 \\cdot \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "4&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad  = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "-4&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "243&0\\\\\n",
    "0&1024\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "4&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad  =  \\begin{pmatrix}\n",
    "243&0\\\\\n",
    "-972&1024\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "4&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad  =  \\begin{pmatrix}\n",
    "243&0\\\\\n",
    "3124&1024\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "(d) We have, $P = \\begin{pmatrix}\n",
    "-2&1\\\\\n",
    "1&-1\\\\\n",
    "\\end{pmatrix}, P^{-1} = \\frac{1}{3}\\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&2\\\\\n",
    "\\end{pmatrix}$, and $D = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&4\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, $A^5 = \\begin{pmatrix}\n",
    "2&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&4\\\\\n",
    "\\end{pmatrix}^5 \\cdot \\frac{1}{3}\\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "1&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{3} \\begin{pmatrix}\n",
    "2&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&1024\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "1&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{3} \\begin{pmatrix}\n",
    "2&1024\\\\\n",
    "-1&1024\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "1&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{3} \\begin{pmatrix}\n",
    "1026&2046\\\\\n",
    "1023&2049\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\begin{pmatrix}\n",
    "342&682\\\\\n",
    "341&683\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "<b> Solution (ii)</b> \n",
    "\n",
    "$A^{-1/2} = PD^{-1/2}P^{-1}$\n",
    "\n",
    "$\\qquad = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "-4&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "3&0\\\\\n",
    "0&4\\\\\n",
    "\\end{pmatrix}^{-1/2} \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "4&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "-4&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1/\\sqrt{3}&0\\\\\n",
    "0&1/2\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "4&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\qquad = \\begin{pmatrix}\n",
    "1/\\sqrt{3}&0\\\\\n",
    "-4/\\sqrt{3}&1/2\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "4&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\begin{pmatrix}\n",
    "1/\\sqrt{3}&0\\\\\n",
    "2 -4/\\sqrt{3}&1/2\\\\\n",
    "\\end{pmatrix} $"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1c224534",
   "metadata": {},
   "source": [
    "--------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "664d6e8d",
   "metadata": {},
   "source": [
    "<b>Question 3</b>\n",
    "\n",
    "For the following matrices find :\n",
    "\n",
    "(i) The eigenvalues and corresponding eigenvectors \n",
    "\n",
    "(ii) Matrices $P$ and $D$ where $P$ is the invertible (non-singular) matrix and $D = P^{-1}AP$ is the diagonal matrix. To find $P^{-1}$ you may use matlab$\n",
    "\n",
    "(iii) Determine $A^4$ in each case by using the results of parts(i) and (ii)\n",
    "\n",
    "(a)$A = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&2&0\\\\\n",
    "0&0&3\\\\\n",
    "\\end{pmatrix}\\quad$(b)$A = \\begin{pmatrix}\n",
    "-1&4&0\\\\\n",
    "0&4&3\\\\\n",
    "0&0&5\\\\\n",
    "\\end{pmatrix}\\quad$(c)$A = \\begin{pmatrix}\n",
    "2&0&0\\\\\n",
    "1&5&0\\\\\n",
    "1&2&6\\\\\n",
    "\\end{pmatrix}$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "340c78e7",
   "metadata": {},
   "source": [
    "<b> Solution (a) </b>\n",
    "\n",
    "<b>(i)</b>\n",
    "\n",
    "Since $A$ is a diagonal matrix, its eigenvalues are $\\lambda_{1} = 1, \\lambda_{2} = 2$ and $\\lambda_{3} =3$\n",
    "\n",
    "Next for eigenvector when $\\lambda_{1} = 1$ with $(A - \\lambda I)u = 0$\n",
    "\n",
    "we get, <br>\n",
    "$\\begin{pmatrix}\n",
    "1-1&0&0\\\\\n",
    "0&2-1&0\\\\\n",
    "0&0&3-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow  \\begin{pmatrix}\n",
    "0&0&0\\\\\n",
    "0&1&0\\\\\n",
    "0&0&2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow y = 0$ and $z = 0$ and $x$ can be any non-zero $s$. \n",
    "\n",
    "Therefore, $u  =\\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "Next for eigenvector when $\\lambda_{2} = 2$ with $(A - \\lambda I)u = 0$\n",
    "\n",
    "we get, <br>\n",
    "$\\begin{pmatrix}\n",
    "1-2&0&0\\\\\n",
    "0&2-2&0\\\\\n",
    "0&0&3-2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow  \\begin{pmatrix}\n",
    "-1&0&0\\\\\n",
    "0&0&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow z = 0$ and $x = 0$ and $y$ can be any non-zero $s$. \n",
    "\n",
    "Therefore, $u  =\\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Next for eigenvector when $\\lambda_{3} = 3$ with $(A - \\lambda I)u = 0$\n",
    "\n",
    "we get, <br>\n",
    "$\\begin{pmatrix}\n",
    "1-3&0&0\\\\\n",
    "0&2-3&0\\\\\n",
    "0&0&3-3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow  \\begin{pmatrix}\n",
    "-2&0&0\\\\\n",
    "0&-1&0\\\\\n",
    "0&0&0\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow x = 0$ and $y = 0$ and $z$ can be any non-zero $s$. \n",
    "\n",
    "Therefore, $u  =\\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "<b>(ii)</b>\n",
    "\n",
    "Eigenvector matrix, $P$ is given by $P = (u \\quad v \\quad w)$\n",
    "\n",
    "Therefore, here $P = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&1&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix} = I$\n",
    "\n",
    "So, $P^{-1} = I$\n",
    "\n",
    "Therfore, $D = I^{-1}AI = A$\n",
    "\n",
    "$\\Rightarrow D =  \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&2&0\\\\\n",
    "0&0&3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "<b>(iii)</b>\n",
    "\n",
    "$A^4 = PD^4 P^{-1}$\n",
    "\n",
    "Here, $P =I$, and $P^{-1} = I$<br>\n",
    "Therefore, $A^4 = D^4$<br>\n",
    "$\\Rightarrow A^4 = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&2&0\\\\\n",
    "0&0&3\\\\\n",
    "\\end{pmatrix}^4$ \n",
    "\n",
    "$\\Rightarrow A^4 = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&16&0\\\\\n",
    "0&0&81\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "<b> Solution (b) </b>\n",
    "\n",
    "<b>(i)</b> Since, $A$ is an upper triangular matrix, its eigenvalues are $\\lambda_{1} = -1, \\lambda_{2} = 4$ and $\\lambda_{3} = 5$\n",
    "\n",
    "For eigenvector ,when $\\lambda_{1} = -1$, with $(A - \\lambda I)u = 0$<br>\n",
    "we get, <br>\n",
    "$\\begin{pmatrix}\n",
    "-1+1&4&0\\\\\n",
    "0&4+1&3\\\\\n",
    "0&0&5+1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "0&4&0\\\\\n",
    "0&5&3\\\\\n",
    "0&0&6\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow z= 0$ and $y = 0$ and $x$ can be any non-zero $s$.\n",
    "\n",
    "Therefore, $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "For eigenvector ,when $\\lambda_{2} = 4$, with $(A - \\lambda I)u = 0$<br>\n",
    "we get, <br>\n",
    "$\\begin{pmatrix}\n",
    "-1-4&4&0\\\\\n",
    "0&4-4&3\\\\\n",
    "0&0&5-4\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-5&4&0\\\\\n",
    "0&0&3\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow z= 0$ and $y = 5$ and $x = 4$\n",
    "\n",
    "Therefore, $u = \\begin{pmatrix}\n",
    "4\\\\\n",
    "5\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "For eigenvector ,when $\\lambda_{3} = 5$, with $(A - \\lambda I)u = 0$<br>\n",
    "we get, <br>\n",
    "$\\begin{pmatrix}\n",
    "-1-5&4&0\\\\\n",
    "0&4-5&3\\\\\n",
    "0&0&5-5\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-6&4&0\\\\\n",
    "0&-1&3\\\\\n",
    "0&0&0\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow x= 2, y = 3$ and $z = 1$\n",
    "\n",
    "Therefore, $u = \\begin{pmatrix}\n",
    "2\\\\\n",
    "3\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "<b>(ii)</b> Eigenvector matrix, $P$ can be given as $P = (u \\quad v \\quad w)$:\n",
    "\n",
    "$P = \\begin{pmatrix}\n",
    "1&4&2\\\\\n",
    "0&5&3\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "1878abaf",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      "Inverse of P:\n",
      "[[ 1.  -0.8  0.4]\n",
      " [ 0.   0.2 -0.6]\n",
      " [ 0.   0.   1. ]]\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "from scipy.linalg import inv\n",
    "\n",
    "P = np.array([[1,4, 2], [0, 5, 3],[0, 0, 1]])\n",
    "\n",
    "P_inv = inv(P)\n",
    "print(\"\\nInverse of P:\")\n",
    "print(P_inv)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "87664fff",
   "metadata": {},
   "source": [
    "<b>(ii) continued</b>\n",
    "\n",
    "Now, we have $P^{-1}$, calculating diagonal matrix, $D$\n",
    "\n",
    "with $D = P^{-1}AP$\n",
    "\n",
    "$\\Rightarrow D = \\begin{pmatrix}\n",
    "1&-0.8&0.4\\\\\n",
    "0&0.2&-0.6\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-1&4&0\\\\\n",
    "0&4&3\\\\\n",
    "0&0&5\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&4&2\\\\\n",
    "0&5&3\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow D = \\begin{pmatrix}\n",
    "-1& 0.8 & -0.4\\\\\n",
    "0 & 0.8 & -2.4\\\\\n",
    "0& 0 & 5\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&4&2\\\\\n",
    "0&5&3\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow D = \\begin{pmatrix}\n",
    "-1+0+0&-4+4+0&-2+2.4-0.4\\\\\n",
    "0&4&0+2.4-2.4\\\\\n",
    "0&0&5\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow D = \\begin{pmatrix}\n",
    "-1&0&0\\\\\n",
    "0&4&0\\\\\n",
    "0&0&5\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "<b>(iii)</b>\n",
    "\n",
    "$A^4 = PD^4P^{-1}$\n",
    "\n",
    "$ \\Rightarrow A^4 = \\begin{pmatrix}\n",
    "1&4&2\\\\\n",
    "0&5&3\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix} \\cdot\\begin{pmatrix}\n",
    "-1&0&0\\\\\n",
    "0&4&0\\\\\n",
    "0&0&5\\\\\n",
    "\\end{pmatrix}^4 \\cdot \\begin{pmatrix}\n",
    "1&-0.8&0.4\\\\\n",
    "0&0.2&-0.6\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow A^4 = \\begin{pmatrix}\n",
    "1&4&2\\\\\n",
    "0&5&3\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix} \\cdot\\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&256&0\\\\\n",
    "0&0&625\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&-0.8&0.4\\\\\n",
    "0&0.2&-0.6\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow A^4 = \\begin{pmatrix}\n",
    "1&1024&1250\\\\\n",
    "0&1280&1875\\\\\n",
    "0&0&625\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&-0.8&0.4\\\\\n",
    "0&0.2&-0.6\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow A^4 = \\begin{pmatrix}\n",
    "1&-0.8 + 204.8 &0.4 - 614.4 + 1250\\\\\n",
    "0&256&-768 + 1875\\\\\n",
    "0&0&625\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow A^4 = \\begin{pmatrix}\n",
    "1&204&636\\\\\n",
    "0&256&1107\\\\\n",
    "0&0&625\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1fabd7a3",
   "metadata": {},
   "source": [
    "----------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "95e1cbfb",
   "metadata": {},
   "source": [
    "<b> Question 4</b> \n",
    "\n",
    "For the following matrices determine whether they are diagonalizable\n",
    "\n",
    "(a)$A = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&1&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}\\quad$(b)$A = \\begin{pmatrix}\n",
    "-1&2&3\\\\\n",
    "0&2&5\\\\\n",
    "0&0&8\\\\\n",
    "\\end{pmatrix}\\quad$(c)$A = \\begin{pmatrix}\n",
    "\\sqrt{2}&0&0&0\\\\\n",
    "1&\\sqrt{3}&0&0\\\\\n",
    "6&7&1/2&0\\\\\n",
    "2&9&7&-5\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "313b873b",
   "metadata": {},
   "source": [
    "<b>Solution</b> \n",
    "\n",
    "<b>(a)</b>\n",
    "\n",
    "Here, the invertible matrix $P =I$. Therefore, $P^{-1}AP = III = D$, the diagonal matrix with eigenvalue 1 along the leading diagonal. Therefore, diagonalizable.\n",
    "\n",
    "<b>(b)</b>\n",
    "\n",
    "Since, A is an upper triangular matrix, its eigenvalues are $\\lambda_{1} = -1, \\lambda_{2} = 2$ and $\\lambda_{3} = 8$. <br>\n",
    "Also, this $3 \\times 3$ matrix $A$ has 3 distinct eigenvalues, it is diagonalizable.\n",
    "\n",
    "<b>(c)</b>\n",
    "\n",
    "Again, $A$ is a lower triangular matrix, its eigenvalues are $\\lambda_{1} = \\sqrt{2}, \\lambda_{2} = \\sqrt{3}, \\lambda_{3} = 1/2$, and $\\lambda_{4} = -5$\n",
    "\n",
    "As, these 4 eigenvalues of $4 \\times 4$ matrix are distinct, this is a diagonalizable matrix"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7499feaf",
   "metadata": {},
   "source": [
    "----------------------- "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a692307b",
   "metadata": {},
   "source": [
    "<b> Question 5</b>\n",
    "\n",
    "In Example 7.17, take $P = (v\\quad u)$ and determine the diagonal matrix $D$ is given by $P^{-1}AP$ where $A = \\begin{pmatrix}\n",
    "1&4\\\\\n",
    "2&3\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "58621e1c",
   "metadata": {},
   "source": [
    "<b> Solution</b> \n",
    "\n",
    "From example 7.17<br>\n",
    "\n",
    "we get $u = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, eigenvector matrix, $P(v \\quad u) = \\begin{pmatrix}\n",
    "1&-2\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "and, $P^{-1} = \\frac{1}{2+1}\\begin{pmatrix}\n",
    "1&2\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} = \\frac{1}{3}\\begin{pmatrix}\n",
    "1&2\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Then, the diagonal matrix, $D = P^{-1}AP $\n",
    "\n",
    "$\\qquad = \\frac{1}{3}\\begin{pmatrix}\n",
    "1&2\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix}\\cdot \\begin{pmatrix}\n",
    "1&4\\\\\n",
    "2&3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&-2\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad =  \\frac{1}{3}\\begin{pmatrix}\n",
    "5&10\\\\\n",
    "1&-1\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "1&-2\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{3}\\begin{pmatrix}\n",
    "15&0\\\\\n",
    "0&-3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\begin{pmatrix}\n",
    "5&0\\\\\n",
    "0&-1\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "2ede8513",
   "metadata": {},
   "source": [
    "-------------------------------- "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "72e071c9",
   "metadata": {},
   "source": [
    "<b> Question 6 </b>\n",
    "\n",
    "Let $A$ be a $3 \\times 3$ matrix with the following eigenvalues and eigenvectors\n",
    "\n",
    "$\\lambda_{1} = -2, u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}\\quad,\\lambda_{2} = -5, v =  \\begin{pmatrix}\n",
    "5\\\\\n",
    "4\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}\\quad$ and $\\lambda_{3} = -1, w = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Is the matrix $A$ diagonalizable? If it is then find the diagonal eigenvalue matrix $D$ which is similar to the matrix $A$ and also determine the invertible matrix $P$ such that $P^{-1}AP = D$<br>\n",
    "Find $A^3.$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1287b398",
   "metadata": {},
   "source": [
    "<b> Solution </b>\n",
    "\n",
    "Since, its eigenvalues, $\\lambda_{1}=-2, \\lambda_{2} = -5$ and $\\lambda_{3} = -1$ are distinct, matrix $A$ is diagonalizable.\n",
    "\n",
    "Diagonal Eigenvalue matrix have eigenvalues along the leading diagonal of matrix. $D$ can be written as:\n",
    "$$D = \\begin{pmatrix}\n",
    "-2&0&0\\\\\n",
    "0&-5&0\\\\\n",
    "0&0&-1\\\\\n",
    "\\end{pmatrix}$$\n",
    "\n",
    "invertible matrix or eigenvector matrix, $P$ is given by $P(u \\quad v \\quad w)$\n",
    "\n",
    "Therefore, $P = \\begin{pmatrix}\n",
    "1&5&0\\\\\n",
    "2&4&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Next, getting $P^{-1}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "c03912d4",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      "Symbolic Inverse of P:\n",
      "⎡-2/3  5/6   0⎤\n",
      "⎢             ⎥\n",
      "⎢1/3   -1/6  0⎥\n",
      "⎢             ⎥\n",
      "⎣ 0     0    1⎦\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "import sympy as sp\n",
    "\n",
    "P = np.array([[1, 5, 0], [2, 4, 0], [0, 0, 1]])\n",
    "\n",
    "P_sym = sp.Matrix(P)\n",
    "P_inv_sym = P_sym.inv()\n",
    "\n",
    "# Print the symbolic inverse matrix\n",
    "print(\"\\nSymbolic Inverse of P:\")\n",
    "sp.pprint(P_inv_sym)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "dff2a3f6",
   "metadata": {},
   "source": [
    "we can get $A^3$ by $A^m = PD^mP^{-1}$\n",
    "\n",
    "$\\Rightarrow A^3 = PD^3P^{-1}$\n",
    "\n",
    "$\\Rightarrow A^3 = \\begin{pmatrix}\n",
    "1&5&0\\\\\n",
    "2&4&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-2&0&0\\\\\n",
    "0&-5&0\\\\\n",
    "0&0&-1\\\\\n",
    "\\end{pmatrix}^3 \\cdot \\begin{pmatrix}\n",
    "-2/3&5/6&0\\\\\n",
    "1/3&-1/6&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\Rightarrow A^3 = \\begin{pmatrix}\n",
    "1&5&0\\\\\n",
    "2&4&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-8&0&0\\\\\n",
    "0&-125&0\\\\\n",
    "0&0&-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-2/3&5/6&0\\\\\n",
    "1/3&-1/6&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\Rightarrow A^3 =  \\begin{pmatrix}\n",
    "-8&-625&0\\\\\n",
    "-16&-500&0\\\\\n",
    "0&0&-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-2/3&5/6&0\\\\\n",
    "1/3&-1/6&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow A^3 =  \\begin{pmatrix}\n",
    "-203&97.5&0\\\\\n",
    "-156&70&0\\\\\n",
    "0&0&-1\\\\\n",
    "\\end{pmatrix} $"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "61369e7a",
   "metadata": {},
   "source": [
    "---------------------------- "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ca8e2c39",
   "metadata": {},
   "source": [
    "<b> Question 7 </b> \n",
    "\n",
    "Show that the following matrices are not diagonalizable:\n",
    "\n",
    "(a)$A = \\begin{pmatrix}\n",
    "2&-1\\\\\n",
    "1&4\\\\\n",
    "\\end{pmatrix}\\quad$(b)$A = \\begin{pmatrix}\n",
    "-2&4\\\\\n",
    "-1&-6\\\\\n",
    "\\end{pmatrix}\\quad$(c)$A = \\begin{pmatrix}\n",
    "1&2&3\\\\\n",
    "0&1&3\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "4bee2389",
   "metadata": {},
   "source": [
    "<b> Solution (a) </b> \n",
    "\n",
    "Eigenvalues of $A$ is given by $det(A - \\lambda I) = 0$\n",
    "\n",
    "$\\Rightarrow det\\begin{pmatrix}\n",
    "2-\\lambda&1\\\\\n",
    "1&4-\\lambda\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (2 - \\lambda)(4 - \\lambda) + 1 =0$<br>\n",
    "$\\Rightarrow 8 - 6\\lambda + \\lambda^2 + 1 = 0$<br>\n",
    "$\\Rightarrow \\lambda^2 - 6\\lambda + 9 = 0$<br>\n",
    "\n",
    "$\\Rightarrow (\\lambda - 3)^2 = 0$<br>\n",
    "\n",
    "$\\Rightarrow \\lambda = 3$\n",
    "\n",
    "Therefore, eigenvector can be given by $(A - \\lambda I)u = 0$\n",
    "\n",
    "$\\begin{pmatrix} \n",
    "2-3&-1\\\\\n",
    "1&4-3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\begin{pmatrix} \n",
    "-1&-1\\\\\n",
    "0&0\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -x - y = 0$<br>\n",
    "$\\Rightarrow x = 1$ and $y = -1$ and eigenvector $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Since, there is only one non-zero equation and 2 unknowns, we have 2-1 = 1 free variable.\n",
    "therefore, $u$ is the only independent eigenvector.\n",
    "\n",
    "We know that, <br>\n",
    "Any $n \\times n$ matrix $A$ is diagonalizable $\\iff$ it has $n$ linearly independent\n",
    "\n",
    "Here, we have only one not 2, therefore <b>not diagonalizable</b>\n",
    "\n",
    "\n",
    "<b>Solution (b)</b>\n",
    "\n",
    "Eigenvalues of $A$ is given by $det(A - \\lambda I) = 0$\n",
    "\n",
    "$\\Rightarrow det\\begin{pmatrix}\n",
    "-2-\\lambda&4\\\\\n",
    "-1&-6-\\lambda\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (-2-\\lambda)(-6 - \\lambda) + 4 = 0$\n",
    "\n",
    "$\\Rightarrow 12 + 8\\lambda + \\lambda^2 + 4 = 0$\n",
    "\n",
    "$\\Rightarrow \\lambda^2 + 8\\lambda + 16 = 0$\n",
    "\n",
    "$\\Rightarrow (\\lambda + 4)^2 = 0$<br>\n",
    "$\\Rightarrow \\lambda = -4$\n",
    "\n",
    "Therefore, eigenvector can be given by $(A - \\lambda I)u = 0$\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "-2+4&4\\\\\n",
    "-1&-6+4\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "2&4\\\\\n",
    "-1&-2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow 2x + 4y = 0$<br>\n",
    "and, $-x - 2y = 0$\n",
    "\n",
    "$\\Rightarrow x = -2$, and $y =1$. Thus, eigenvector $u = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "We know that, <br>\n",
    "Any $n \\times n$ matrix $A$ is diagonalizable $\\iff$ it has $n$ linearly independent\n",
    "\n",
    "Here, we have only one not 2, therefore <b>not diagonalizable</b>\n",
    "\n",
    "\n",
    "<b> Solution (c)</b>\n",
    "\n",
    "Since, $A$ is an upper triangular matrix, its eigenvalues are entries on leading diagonal i.e, $\\lambda_{1,2,3} = 1$\n",
    "\n",
    "Now, to get eigenvector, we use $(A - \\lambda I)u = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "1-1&2&3\\\\\n",
    "0&1-1&3\\\\\n",
    "0&0&1-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "0&2&3\\\\\n",
    "0&0&3\\\\\n",
    "0&0&0\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow z = 0, y = $ and $x$ can be any non-zero $s$. Thus, eigenvector is $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "We know that, <br>\n",
    "Any $n \\times n$ matrix $A$ is diagonalizable $\\iff$ it has $n$ linearly independent\n",
    "\n",
    "Here, we have only one not 2, therefore <b>not diagonalizable</b>"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7014f6fa",
   "metadata": {},
   "source": [
    "----------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "cd81a7af",
   "metadata": {},
   "source": [
    "<b> Question 8 </b>\n",
    "\n",
    "Let $A = \\begin{pmatrix}\n",
    "-4&2\\\\\n",
    "-9&5\\\\\n",
    "\\end{pmatrix}$. Determine (i) $A^{11}\\quad$(ii)$A^{-1}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "142dc75b",
   "metadata": {},
   "source": [
    "<b> Solution (i) </b> \n",
    "\n",
    "Eigenvalue can be obtained by $det(A - \\lambda I) = 0$\n",
    "\n",
    "$\\Rightarrow det\\begin{pmatrix}\n",
    "4 - \\lambda& 2\\\\\n",
    "-9 & 5 - \\lambda\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (-4 - \\lambda)(5 - \\lambda) - (-9)2 = 0$\n",
    "\n",
    "$\\Rightarrow - 20 + 4\\lambda - 5\\lambda + \\lambda^2 + 18 = 0$<br>\n",
    "$\\Rightarrow \\lambda^2 - \\lambda -2 = 0$<br>\n",
    "$\\Rightarrow (\\lambda + 1)(\\lambda -2) = 0$\n",
    "\n",
    "$\\Rightarrow \\lambda = -1$ or $\\lambda = 2$\n",
    "\n",
    "For eigenvector $\\lambda_{1} = -1$, eigenvector can be given by $(A - \\lambda I)u =0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-4 + 1&2\\\\\n",
    "-9 & 5+1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-3&2\\\\\n",
    "-9&6\\\\\n",
    "\\end{pmatrix} \\cdot\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -3x + 2y =0$<br>\n",
    "and, $-9x + 6y = 0$\n",
    "\n",
    "$\\Rightarrow x = 2$ and $y = 3$. Thus eigenvector $u = \\begin{pmatrix}\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "For $\\lambda_{2} = 2$, again eigenvector can be given by $(A - \\lambda )u = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-4 -2& 2\\\\\n",
    "-9 & 5-2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-6&2\\\\\n",
    "-9&3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -6x + 2y = 0$<br>\n",
    "and, $-9x + 3y = 0$\n",
    "\n",
    "$\\Rightarrow x =1$ and $y =3$. Thus, eigenvector $ v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, eigenvector matrix, $P$ is given by $P = (u \\quad v)$\n",
    "\n",
    "$\\Rightarrow P = \\begin{pmatrix}\n",
    "2&1\\\\\n",
    "3&3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now, $P^{-1} = \\frac{1}{2(3) - 3(1)}\\begin{pmatrix}\n",
    "3&-1\\\\\n",
    "-3&2\\\\\n",
    "\\end{pmatrix} = \\frac{1}{3}\\begin{pmatrix}\n",
    "3&-1\\\\\n",
    "-3&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Also, diagonal matrix, $D$, is given by eigenvalues along the leading diagonal.\n",
    "\n",
    "Then, $D = \\begin{pmatrix}\n",
    "-1&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "We also know that, $A^m = PD^m P^{-1}$\n",
    "\n",
    "Therefore, $A^{11} = \\begin{pmatrix}\n",
    "2&1\\\\\n",
    "3&3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-1&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}^{11} \\cdot \\frac{1}{3}\\begin{pmatrix}\n",
    "3&-1\\\\\n",
    "-3&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{3} \\begin{pmatrix}\n",
    "2&1\\\\\n",
    "3&3\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "-1&0\\\\\n",
    "0&2048\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "3&-1\\\\\n",
    "-3&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{3}\\begin{pmatrix}\n",
    "-2&2048\\\\\n",
    "-3&6144\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "3&-1\\\\\n",
    "-3&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{3}\\begin{pmatrix}\n",
    "-6150&4098\\\\\n",
    "-18441&12291\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow A^{11} = \\begin{pmatrix}\n",
    "-2050&1366\\\\\n",
    "-6147&4097\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "<b> Solution (ii)</b>\n",
    "\n",
    "$A^{-1} = PD^{-1}P^{-1}$\n",
    "\n",
    "$\\qquad = \\begin{pmatrix}\n",
    "2&1\\\\\n",
    "3&3\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "-1&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}^{-1} \\cdot \\frac{1}{3}\\begin{pmatrix}\n",
    "3&-1\\\\\n",
    "-3&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{3}\\begin{pmatrix}\n",
    "2&1\\\\\n",
    "3&3\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "-1&0\\\\\n",
    "0&1/2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "3&-1\\\\\n",
    "-3&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{6}\\begin{pmatrix}\n",
    "2&1\\\\\n",
    "3&3\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "-2&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "3&-1\\\\\n",
    "-3&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{6} \\begin{pmatrix}\n",
    "-4&1\\\\\n",
    "-6&3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "3&-1\\\\\n",
    "-3&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{6} \\begin{pmatrix}\n",
    "-15&6\\\\\n",
    "-27&12\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "269c80aa",
   "metadata": {},
   "source": [
    "---------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e5f5c0e7",
   "metadata": {},
   "source": [
    "<b> Question 9 </b> \n",
    "\n",
    "Prove that if $D$ is a diagonal matrix then the matrix $D^m$ is simply a diagonal matrix with its leading diagonal entries raised to the power $m$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "106b5a43",
   "metadata": {},
   "source": [
    "---------------------------------"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "66354385",
   "metadata": {},
   "outputs": [],
   "source": []
  }
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