This ﬁrst part treats vectors in Euclidean space as well as matrices, matrix algebra and systems of linear equations. It turns out that we can use linear transformations to solve linear systems of equations. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Recall the definition of a linear transformation discussed above. Constraints involving two linear equation like toe above (going through the origin) ax1+bx2+cx3+dx4= 0 Ax1+Bx2+Cx3+Dx4= 0, which is eﬀectively a two dimensional plane. We have spent a lot of time finding solutions to systems of equations in general, as well as homogeneous systems. We may also refer to the null space as the kernel of \(T\), and we write \(ker\left(T\right)\). This is also called the null space of \(T\). Equivalently, find the solutions to the system of equations \(A\vec{x}=\vec{0}\). Linear Algebra - Questions with Solutions Linear algebra questions with solutions and detailed explanations on matrices, spaces, subspaces and vectors, determinants, systems of linear equations and online linear algebra calculators are included. Definition \(\PageIndex{2}\): Null Space or Kernel of a Linear Transformation. Notes on Proof by Mathematical Induction. Find the general solution to the linear system, \[\left ( \begin{array}{rrrr} 1 & 2 & 3 & 0 \\ 2 & 1 & 1 & 2 \\ 4 & 5 & 7 & 2 \end{array} \right ) \left ( \begin{array}{r} x \\ y \\ z \\ w \end{array} \right ) =\left ( \begin{array}{r} 9 \\ 7 \\ 25 \end{array} \right )\]. Of course, this is much harder but also more rewarding. However, we know that every linear transformation \(T\) is determined by some matrix \(A\). This book is the ﬁrst part of a three-part series titled Problems, Theory and Solutions in Linear Algebra. The solutions to the system \(T\left(\vec{x}\right)=\vec{b}\) are given by \(\vec{x}_{p}+\ker \left( T\right)\) where \(\vec{x}_{p}\) is a particular solution to \(T\left(\vec{x}\right)=\vec{b}\). Supplementary Problems with Brief Solutions are at the following link: Supplementary Problems. By this, we mean that we replace \(\vec{b}\) by \(\vec{0}\) and look at \(A\vec{x}=\vec{0}\). Hence, \(T\left(\vec{y}\right)-T\left( \vec{x}_{p} \right) =\vec{b} - \vec{b} = \vec{0}\). (You can enter any name and any number like 123 for your student number. Have questions or comments? If you are an academic and would like complete solutions to these problems then email me at k.singh@herts.ac.uk. Since \(\vec{y}\) and \(\vec{x}_{p}\) are both solutions to the system, it follows that \(T\left(\vec{y}\right)= \vec{b}\) and \(T\left(\vec{x}_p\right) = \vec{b}\). Remember that \(T\left(\vec{x}\right) = A\vec{x}\). Suppose we look at a system given by \(A\vec{x}=\vec{b}\), and consider the related homogeneous system. The following video explains why linear algebra is important. Missed the LibreFest? This yields \(x= \frac{1}{3}z- \frac{4}{3}w\) and \(y= \frac{2}{3}w- \frac{5}{3}z.\) Since \(\mathrm{null} \left( A\right)\) consists of the solutions to this system, it consists vectors of the form, \[\left ( \begin{array}{c} \frac{1}{3}z- \frac{4}{3}w \\ \frac{2}{3}w- \frac{5}{3}z \\ z \\ w \end{array} \right ) =z \left ( \begin{array}{r} \frac{1}{3} \\ - \frac{5}{3} \\ 1 \\ 0 \end{array} \right ) +w \left ( \begin{array}{r} - \frac{4}{3} \\ \frac{2}{3} \\ 0 \\ 1 \end{array} \right )\], Example \(\PageIndex{3}\): A General Solution. Therefore, we can also speak about the null space of a matrix. Challenging Problems on Linear Algebra with complete solutions are Here. A good way to revise for examination is to try past examination papers.