Question 3.12 please linearly dependent or linearly independent in the 3.12. For two matrices A and B for which AB can be defined, prove the following state- ments. (1) If both A and B have linearly independent column vectors, then the column Vectors of AB are also linearly independent. 2) If both A and B have linearly independent row vectors, then the row vectors of AB are also linearly independent. (3) If the column vectors of B are linearly dependent, then the column vectors of AB are also linearly dependent

2et A and B be given two matriusLet us suppose that A and B have linearly independent Column vectors.Let us tuke simple examples to check the given strtement.\text { Let } \begin{aligned}A & =\left[\begin{array}{ll}1 & 1 \\0 & 1\end{array}\right] \\B & =\left[\begin{array}{ll}0 & 2 \\1 & 3\end{array}\right]\end{aligned}Cleacly A has linearly independent Colnmn vectors and B also has linearly independent column vectors.NowA B=\left[\begin{array}{ll}1 & 1 \\0 & 1\end{array}\right]\left[\begin{array}{ll}0 & 2 \\1 & 3\end{array}\right]=\left[\begin{array}{ll}1 & 5 \\1 & 3\end{array}\right]Let ws check whether columns of A B are linearly independent or not.Let v_{1}=(1,1)^{\top}, \quad v_{2}=(5,3)^{\top}Let a, b \in \mathbb{R} \quad \Rightarrow \quad a v_{1}+b v_{2}=0Now a(1,1)+b(5,3)=0\Rightarrow a+5 b=0 \text { and } a+3 b=0\begin{array}{l}a+3 b=0 \\a+3 b=0 \\(-1 \quad+-1 \quad \\2 b=0 \Rightarrow b=0\end{array}We see that when b=0, a+5 b=0 \Rightarrow a=0 \therefore both a and b are rero Hance columns of A B are also linearly independent.(2) Lef ws take the same matrices\begin{array}{l}A=\left[\begin{array}{ll}1 & 1 \\0 & 1\end{array}\right] \\B=\left[\begin{array}{ll}0 & 2 \\1 & 3\end{array}\right]\end{array}clemkly A and B has linearly independent row vertors.A B=\left[\begin{array}{ll}1 & 1 \\0 & 1\end{array}\right]\left[\begin{array}{ll}0 & 2 \\1 & 3\end{array}\right]=\left[\begin{array}{ll}1 & 5 \\1 & 3\end{array}\right]Let \alpha_{1} and \alpha_{2} be such that\begin{aligned}& \alpha_{1}(1,1)+\alpha_{2}(5,3)=0 \\\Rightarrow & \alpha_{1}+5 \alpha_{2}=0 \quad \text { and } \alpha_{1}+3 \alpha_{2}=0\end{aligned}Clervily \quad \alpha_{1}=0 and \alpha_{2}=03\therefore Row vectors of A B are also linearly indefendent.(3). Let the Columns of B be linearly dependent Let B=\left[b_{1}, b_{2}, \cdots, b_{k}\right]. Be cause the columns of B are linearly dependent, there exists Constants c_{1}, c_{2}, \cdots, c_{k} not all rero, such thatc_{1} b_{1}+c_{2} b_{2}+\cdots+c_{k} b_{k}=0NowA B=A\left[b_{1}, b_{2}, \ldots, b_{k}\right]=\left[A b_{1}, A b_{2}, \ldots, A b_{k}\right]Hene A b_{1}, A b_{2}, \ldots, A b_{k} are the columns of matrit A B.Now2et us multiply equation (1) by Ac_{1} A b_{1}+c_{2} A b_{2}+\cdots+C_{k} A b_{k}=0thus we see that we have constants c_{1}, \ldots, c_{k} not all rero such thatc_{1} A b_{1}+\cdots+c_{k} A b_{k}=0Therefore, by definition columns of A B=\left[A b_{1}, \ldots, A b_{k}\right] are linearly dependunt. ...