tutorials/tutorial02_solution.ipynb

{
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  {
   "cell_type": "markdown",
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   "source": [
    "## CE9010: Introduction to Data Science\n",
    "## Semester 2 2017/18\n",
    "## Xavier Bresson\n",
    "<hr>\n",
    "\n",
    "## Tutorial 2: Linear algebra in Python\n",
    "## Objective\n",
    "### $\\bullet$ Review of linear algebra basics\n",
    "<hr>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "## 1. Matrix, vector and scalar representation\n",
    "<hr>\n",
    "### 1.1 Matrix \n",
    "Example:\n",
    "$$\n",
    "X = \\left[ \n",
    "\\begin{array}{cccc}\n",
    "4.1 & 5.3\\\\ \n",
    "-3.9 & 8.4 \\\\ \n",
    "6.4 & -1.8\n",
    "\\end{array} \n",
    "\\right]\n",
    "$$\n",
    "\n",
    "$X_{ij}$ is the element at the $i^{th}$ row and $j^{th}$ column. Here: $X_{11}=4.1,X_{32}=-1.8$.\n",
    "\n",
    "Dimension of matrix $X$ is the number of rows times the number of columns. <br>\n",
    "Here $dim(X)=3\\times 2$. $X$ is said to be a $3\\times 2$ matrix. \n",
    "\n",
    "The set of all $3\\times 2$ matrices is $\\mathbb{R}^{3\\times 2}$.\n",
    "\n",
    "\n",
    "### 1.2 Vector \n",
    "Example: \n",
    "$$\n",
    "x = \\left[ \n",
    "\\begin{array}{cccc}\n",
    "4.1 \\\\ \n",
    "-3.9 \\\\ \n",
    "6.4 \n",
    "\\end{array} \n",
    "\\right]\n",
    "$$\n",
    "\n",
    "$x_i=i^{th}$ element of $x$. Here: $x_1=4.1,x_3=6.4$.\n",
    "\n",
    "Dimension of vector $x$ is the number of rows. <br>\n",
    "Here $dim(x)=3\\times 1$ or $dim(x)=3$. $x$ is said to be a 3-dim vector.\n",
    "\n",
    "The set of all 3-dim vectors is $\\mathbb{R}^3$.\n",
    "\n",
    "### 1.3 Scalar \n",
    "Example: \n",
    "$$\n",
    "x=5.6\n",
    "$$\n",
    "\n",
    "A scalar has no dimension.\n",
    "\n",
    "The set of all scalars is $\\mathbb{R}$.\n",
    "\n",
    "Note: $x=[5.6]$ is a 1-dim vector, not a scalar.\n",
    "\n",
    "<br>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Question 1: Represent the previous matrix, vector and scalar in Python\n",
    "Hint: You may use numpy library, shape(), type(), dtype."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[ 4.1  5.3]\n",
      " [-3.9  8.4]\n",
      " [ 6.4 -1.8]]\n",
      "(3, 2)\n",
      "<class 'numpy.ndarray'>\n",
      "float64\n",
      "[ 4.1 -3.9  6.4]\n",
      "(3,)\n",
      "5.6\n",
      "()\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "\n",
    "X = np.array([[4.1,5.3],[-3.9,8.4],[6.4,-1.8]]) #YOUR CODE HERE\n",
    "print(X)\n",
    "print(X.shape)\n",
    "print(type(X))\n",
    "print(X.dtype)\n",
    "\n",
    "x = np.array([4.1,-3.9,6.4])\n",
    "print(x)\n",
    "print(x.shape)\n",
    "\n",
    "x = np.array(5.6)\n",
    "print(x)\n",
    "print(x.shape)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "## 2. Matrix addition and scalar-matrix multiplication\n",
    "<hr>\n",
    "\n",
    "### 2.1 Matrix addition\n",
    "Example:\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "4.1 & 5.3\\\\ \n",
    "-3.9 & 8.4 \\\\ \n",
    "6.4 & -1.8\n",
    "\\end{array} \n",
    "\\right] & + & \n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "2.7 & 7.3\\\\ \n",
    "3.5 & 2.4 \\\\ \n",
    "6.0 & -1.1\n",
    "\\end{array} \n",
    "\\right]\n",
    "& = &\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "4.1+2.7 & 5.3+7.3\\\\ \n",
    "-3.9+3.5 & 8.4+2.4 \\\\ \n",
    "6.4+6.0 & -1.8-1.1\n",
    "\\end{array} \n",
    "\\right]\\\\\n",
    "3 \\times 2 &+& 3 \\times 2 &=& 3 \\times 2\n",
    "\\end{array} \n",
    "$$\n",
    "\n",
    "All matrix and vector operations must satisfy dimensionality properties. For example, it is not allowed to add two matrices of different dimentionalities, such as\n",
    "\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "4.1 & 5.3\\\\ \n",
    "-3.9 & 8.4 \\\\ \n",
    "6.4 & -1.8\n",
    "\\end{array} \n",
    "\\right] & + & \n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "2.7 & 7.3 & 5.0\\\\ \n",
    "3.5 & 2.4 & 2.8 \\\\ \n",
    "\\end{array} \n",
    "\\right]\n",
    "& = &\n",
    "\\textrm{Not allowed}\\\\ \n",
    "3 \\times 2 &+& 2 \\times 3 &=& \\textrm{Not allowed}\n",
    "\\end{array} \n",
    "$$\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "### 2.1 Scalar-matrix multiplication\n",
    "Example:\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "3 &\\times& \\left[ \n",
    "\\begin{array}{cccc}\n",
    "4.1 & 5.3\\\\ \n",
    "-3.9 & 8.4 \\\\ \n",
    "6.4 & -1.8\n",
    "\\end{array} \n",
    "\\right]\n",
    "&=&\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "3 \\times 4.1 & 3 \\times 5.3\\\\ \n",
    "3 \\times -3.9 & 3 \\times 8.4 \\\\ \n",
    "3 \\times 6.4 & 3 \\times  -1.8\n",
    "\\end{array} \n",
    "\\right]\\\\\n",
    "\\textrm{No dim} &+& 3 \\times 2 &=& 3 \\times 2\n",
    "\\end{array} \n",
    "$$\n",
    "\n",
    "\n",
    "<br>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Question 2: Add the two matrices, and perform the multiplication scalar-matrix as above in Python"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[ 4.1  5.3]\n",
      " [-3.9  8.4]\n",
      " [ 6.4 -1.8]]\n",
      "[[ 2.7  3.5]\n",
      " [ 7.3  2.4]\n",
      " [ 5.   2.8]]\n",
      "[[  6.8   8.8]\n",
      " [  3.4  10.8]\n",
      " [ 11.4   1. ]]\n",
      "[[  6.8   8.8]\n",
      " [  3.4  10.8]\n",
      " [ 11.4   1. ]]\n",
      "[[ 27.2  35.2]\n",
      " [ 13.6  43.2]\n",
      " [ 45.6   4. ]]\n",
      "[[ 2.26666667  2.93333333]\n",
      " [ 1.13333333  3.6       ]\n",
      " [ 3.8         0.33333333]]\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "\n",
    "X1 = np.array([[4.1,5.3],[-3.9,8.4],[6.4,-1.8]]) #YOUR CODE HERE\n",
    "X2 = np.array([[2.7,3.5],[7.3,2.4],[5.0,2.8]])\n",
    "X = X1 + X2\n",
    "print(X1)\n",
    "print(X2)\n",
    "print(X)\n",
    "\n",
    "Y1 = 4* X\n",
    "Y2 = X/ 3\n",
    "print(X)\n",
    "print(Y1)\n",
    "print(Y2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "## 3. Matric-vector multiplication\n",
    "<hr>\n",
    "\n",
    "### 3.1 Example\n",
    "Example:\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "4.1 & 5.3\\\\ \n",
    "-3.9 & 8.4 \\\\ \n",
    "6.4 & -1.8\n",
    "\\end{array} \n",
    "\\right]  \n",
    "&\\times &\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "2.7 \\\\ \n",
    "3.5 \\\\ \n",
    "\\end{array} \n",
    "\\right]\n",
    "&=&\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "4.1\\times 2.7 + 5.3\\times 3.5\\\\ \n",
    "-3.9\\times 2.7 + 8.4\\times 3.5 \\\\ \n",
    "6.4\\times  2.7  -1.8\\times 3.5\n",
    "\\end{array} \n",
    "\\right]\\\\\n",
    " 3 \\times 2 && 2 \\times 1 &=&  3 \\times 1\n",
    "\\end{array} \n",
    "$$\n",
    "\n",
    "Dimension of the matric-vector multiplication operation is given by contraction of $3 \\times 2$ with $2 \\times 1$ = $3 \\times 1$.\n",
    "\n",
    "\n",
    "### 3.2 Formalization\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "\\huge[ \n",
    "A\n",
    "]  \n",
    "&\\times &\n",
    "\\huge[ \n",
    "x\n",
    "]\n",
    "&=& \n",
    "\\huge[ \n",
    "y\n",
    "]\n",
    "\\\\\n",
    "m \\times n && n \\times 1 &=&  m \\times 1\n",
    "\\end{array}\n",
    "$$\n",
    "\n",
    "Element $y_i$ is given by multiplying the $i^{th}$ row of $A$ with vector $x$:\n",
    "\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "y_i&=&A_{i}&&x\\\\\n",
    "1 \\times 1 &=&  1 \\times n  &\\times&  n \\times 1\n",
    "\\end{array}\n",
    "$$\n",
    "\n",
    "It is not allowed to multiply a matrix $A$ and a vector $x$ with different $n$ dimensions such as \n",
    "\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "4.1 & 5.3\\\\ \n",
    "-3.9 & 8.4 \\\\ \n",
    "6.4 & -1.8\n",
    "\\end{array} \n",
    "\\right]  \n",
    "&\\times &\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "2.7 \\\\ \n",
    "3.5 \\\\ \n",
    "-7.2 \\\\ \n",
    "\\end{array} \n",
    "\\right]\n",
    "&=&\n",
    "\\quad ?\\\\\n",
    "3 \\times 2 &\\times &  3 \\times 1  &=& \\textrm{not allowed}\n",
    "\\end{array} \n",
    "$$\n",
    "\n",
    "\n",
    "<br>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Question 3: Multiply the matrix and vector above in Python "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[ 4.1  5.3]\n",
      " [-3.9  8.4]\n",
      " [ 6.4 -1.8]]\n",
      "(3, 2)\n",
      "[[ 2.7]\n",
      " [ 3.5]]\n",
      "(2, 1)\n",
      "[[ 29.62]\n",
      " [ 18.87]\n",
      " [ 10.98]]\n",
      "(3, 1)\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "\n",
    "A = np.array([[4.1,5.3],[-3.9,8.4],[6.4,-1.8]]) #YOUR CODE HERE\n",
    "x = np.array([2.7,3.5])\n",
    "x = x[:,None]\n",
    "x = np.array([[2.7],[3.5]])\n",
    "y = A.dot(x)\n",
    "print(A)\n",
    "print(A.shape)\n",
    "print(x)\n",
    "print(x.shape)\n",
    "print(y)\n",
    "print(y.shape)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "## 4. Matrix-matrix multiplication\n",
    "<hr>\n",
    "\n",
    "### 4.1 Example\n",
    "\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "4.1 & 5.3\\\\ \n",
    "-3.9 & 8.4 \\\\ \n",
    "6.4 & -1.8\n",
    "\\end{array} \n",
    "\\right]  \n",
    "&\\times &\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "2.7 & 3.2 \\\\ \n",
    "3.5 & -8.2 \\\\ \n",
    "\\end{array} \n",
    "\\right]\n",
    "&=&\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "4.1\\times 2.7 + 5.3\\times 3.5 & 4.1\\times 3.2 + 5.3\\times -8.2\\\\ \n",
    "-3.9\\times 2.7 + 8.4\\times 3.5 & -3.9\\times 3.2 + 8.4\\times -8.2 \\\\ \n",
    "6.4\\times  2.7  -1.8\\times 3.5 & 6.4\\times  3.2  -1.8\\times -8.2\n",
    "\\end{array} \n",
    "\\right]\\\\\n",
    "3 \\times 2 &\\times &  2 \\times 2 &=&  3 \\times 2\n",
    "\\end{array}\n",
    "$$\n",
    "\n",
    "Dimension of the matrix-matrix multiplication operation is given by contraction of $3 \\times 2$ with $2 \\times 2$ = $3 \\times 2$.\n",
    "\n",
    "\n",
    "### 4.2 Formalization\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "\\huge[ \n",
    "A\n",
    "]  \n",
    "&\\times &\n",
    "\\huge[ \n",
    "X\n",
    "]\n",
    "&=& \n",
    "\\huge[ \n",
    "Y\n",
    "]\n",
    "\\\\\n",
    "m \\times n && n \\times p &=&  m \\times p\n",
    "\\end{array}\n",
    "$$\n",
    "\n",
    "Like for matrix-vector multiplication, matrix-matrix multiplication can be carried out only if $A$ and $X$ have the same $n$ dimension.\n",
    "\n",
    "\n",
    "### 4.3 Linear algebra operations can be parallelized/distributed\n",
    "Column $Y_i$ is given by multiplying matrix $A$ with the $i^{th}$ column of $X$:\n",
    "\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "Y_i&=&A &\\times & X_i\\\\\n",
    "1 \\times 1 &=&  1 \\times n  &\\times&  n \\times 1\n",
    "\\end{array}\n",
    "$$\n",
    "\n",
    "Observe that all columns $X_i$ are independent. Consequently, all columns $Y_i$ are also independent. This allows to vectorize/parallelize linear algebra operations on (multi-core) CPUs, GPUs, clouds, and consequently to solve all linear problems (including linear regression) very efficiently, basically with one single line of code ($Y=AX$ for millions/billions of data). With Moore's law (computers speed increases by 100x every decade), it has introduced a computational revolution in data analysis. \n",
    "\n",
    "\n",
    "<br>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Question 4: Multiply the two matrices above in Python"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[ 4.1  5.3]\n",
      " [-3.9  8.4]\n",
      " [ 6.4 -1.8]]\n",
      "(3, 2)\n",
      "[[ 2.7  3.2]\n",
      " [ 3.5 -8.2]]\n",
      "(2, 2)\n",
      "[[ 29.62 -30.34]\n",
      " [ 18.87 -81.36]\n",
      " [ 10.98  35.24]]\n",
      "(3, 2)\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "\n",
    "A = np.array([[4.1,5.3],[-3.9,8.4],[6.4,-1.8]]) #YOUR CODE HERE\n",
    "X = np.array([[2.7,3.2],[3.5,-8.2]])\n",
    "Y = A.dot(X)\n",
    "print(A)\n",
    "print(A.shape)\n",
    "print(X)\n",
    "print(X.shape)\n",
    "print(Y)\n",
    "print(Y.shape)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "## 5. Some linear algebra properties\n",
    "<hr>\n",
    "\n",
    "### 5.1 Matrix multiplication is *not* commutative\n",
    "\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "A &\\times& B &\\not=& B &\\times& A\\\\\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "4.1 & 5.3\\\\ \n",
    "-3.9 & 8.4 \\\\ \n",
    "6.4 & -1.8\n",
    "\\end{array} \n",
    "\\right]  \n",
    "&\\times &\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "2.7 & 3.2 \\\\ \n",
    "3.5 & -8.2 \\\\ \n",
    "\\end{array} \n",
    "\\right]\n",
    "&\\not=&\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "2.7 & 3.2 \\\\ \n",
    "3.5 & -8.2 \\\\ \n",
    "\\end{array} \n",
    "\\right]\n",
    "&\\times &\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "4.1 & 5.3\\\\ \n",
    "-3.9 & 8.4 \\\\ \n",
    "6.4 & -1.8\n",
    "\\end{array} \n",
    "\\right]\n",
    "\\end{array} \n",
    "$$\n",
    "<br>\n",
    "\n",
    "\n",
    "### 5.2 Scalar multiplication is associative\n",
    "\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "\\alpha &\\times& B &=& B &\\times& \\alpha\\\\\n",
    "4.1  &\\times&\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "2.7 & 3.2 \\\\ \n",
    "3.5 & -8.2 \\\\ \n",
    "\\end{array} \n",
    "\\right]\n",
    "&=&\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "2.7 & 3.2 \\\\ \n",
    "3.5 & -8.2 \\\\ \n",
    "\\end{array} \n",
    "\\right]\n",
    "&\\times&\n",
    "4.1\n",
    "\\end{array}\n",
    "$$\n",
    "<br>\n",
    "\n",
    "\n",
    "### 5.3 Transpose matrix\n",
    "\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "X^T_{ij} &=& X_{ji}\\\\\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "2.7 & 3.2 & 5.4 \\\\ \n",
    "3.5 & -8.2 & -1.7\\\\ \n",
    "\\end{array} \n",
    "\\right]^T\n",
    "&=&\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "2.7 & 3.5\\\\\n",
    "3.2 & -8.2 \\\\ \n",
    "5.4 & -1.7\\\\ \n",
    "\\end{array} \n",
    "\\right]\n",
    "\\end{array}\n",
    "$$\n",
    "<br>\n",
    "\n",
    "\n",
    "### 5.4 Identity matrix\n",
    "\n",
    "$$\n",
    "I=I_{n}=Diag([1,1,...,1])\n",
    "$$\n",
    "\n",
    "such that\n",
    "\n",
    "$$\n",
    "I\\times A = A \\times I\n",
    "$$\n",
    "\n",
    "Examples:\n",
    "$$\n",
    "I_2=\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "1 & 0\\\\\n",
    "0 & 1  \n",
    "\\end{array} \n",
    "\\right]\n",
    "$$\n",
    "\n",
    "$$\n",
    "I_3=\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "1 & 0 & 0\\\\\n",
    "0 & 1 & 0\\\\\n",
    "0 & 0 & 1  \n",
    "\\end{array} \n",
    "\\right]\n",
    "$$\n",
    "<br>\n",
    "\n",
    "### 5.5 Matrix inverse\n",
    "\n",
    "For any square $n\\times n$ matrix $A$, the matrix inverse $A^{-1}$ is defined as\n",
    "\n",
    "$$\n",
    "AA^{-1} = A^{-1}A = I\n",
    "$$\n",
    "\n",
    "Example:\n",
    "\n",
    "$$\n",
    "\\begin{array}{cccc}\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "2.7 & 3.5\\\\\n",
    "3.2 & -8.2 \n",
    "\\end{array} \n",
    "\\right]\n",
    "&\\times &\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "0.245 & 0.104 \\\\ \n",
    "0.095 & -0.080\n",
    "\\end{array} \n",
    "\\right]\n",
    "&=&\n",
    "\\left[ \n",
    "\\begin{array}{cccc}\n",
    "1 & 0\\\\\n",
    "0 & 1  \n",
    "\\end{array} \n",
    "\\right]\\\\\n",
    "A&\\times &A^{-1} &=& I\n",
    "\\end{array}\n",
    "$$\n",
    "\n",
    "\n",
    "Some matrices do not hold an inverse such as zero matrices. They are called degenerate or singular.\n",
    "<br>\n",
    "\n",
    "\n",
    "\n",
    "<br>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Question 5: Compute the matrix transpose as above in Python. Determine also the matrix inverse in Python.\n",
    "Hint: You may use numpy attribute *.T* and function *np.linalg.inv()*."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[ 2.7 -8.2]\n",
      " [ 3.5  5.4]\n",
      " [ 3.2 -1.7]]\n",
      "(2, 3)\n",
      "(3, 2)\n",
      "[[ 2.7  3.5]\n",
      " [ 3.2 -8.2]]\n",
      "(2, 2)\n",
      "[[ 0.24595081  0.104979  ]\n",
      " [ 0.0959808  -0.0809838 ]]\n",
      "(2, 2)\n",
      "[[  1.00000000e+00   9.02056208e-17]\n",
      " [ -3.96603366e-18   1.00000000e+00]]\n",
      "(2, 2)\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "\n",
    "A = np.array([[2.7,3.5,3.2],[-8.2,5.4,-1.7]]) #YOUR CODE HERE\n",
    "AT = A.T\n",
    "print(AT)\n",
    "print(A.shape)\n",
    "print(AT.shape)\n",
    "\n",
    "A = np.array([[2.7,3.5],[3.2,-8.2]])\n",
    "Ainv = np.linalg.inv(A)\n",
    "AAinv = A.dot(Ainv)\n",
    "print(A)\n",
    "print(A.shape)\n",
    "print(Ainv)\n",
    "print(Ainv.shape)\n",
    "print(AAinv)\n",
    "print(AAinv.shape)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
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    "collapsed": true
   },
   "outputs": [],
   "source": [
    "   "
   ]
  }
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