Question

Question: (a) Suppose A and Care non-zero matrices in$M\left(\mathrm{ℝ}\right)$such that AC=0. If is any real number, show that $\mathit{A}\left(kC\right)\mathbf{=}\mathbf{0}$, where is the matrix C with every entry multiplied by k. Hence, the equation AK = 0 has infinitely many solutions.

(b) If$\mathit{A}\mathbf{=}\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]$ , find four solutions of the equation AX =0.

Short answer

Answer:

1. It is proved that has infinitely many solutions.
2. The possible solutions are$\left[\begin{array}{cc}-2& -2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& -4\\ -\frac{1}{2}& 2\end{array}\right]\left[\begin{array}{cc}-6& 1\\ 3& -1\end{array}\right]and\left[\begin{array}{cc}-4& 1\\ 2& -1\end{array}\right]$

Step-by-step solution

Solution to Part (a)

Assume that$A=\left[a_{ij}\right]=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$ and$C=\left[c_{ij}\right]=\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]$ are the matrices from $M\left(\mathrm{ℝ}\right)$, such that $AC=\left[a_{i1}{b}_{1j}+a_{i2}{b}_{2j}\right]=\left[\mathbf{0}\right]=\mathbf{0}$.

$$\begin{gathered} A\left(kC\right)=\left[a_{ij}\right]\left[k{c}_{ij}\right] \\ =\left[a_{i1}k{c}_{1j}+a_{i2}k{c}_{2j}\right] \\ =\left[k\left(a_{i1}{c}_{1j}+a_{i2}{c}_{2j}\right)\right] \\ =\left[k\mathbf{0}\right] \\ =\mathbf{0} \end{gathered}$$

Hence, the statement is proved.

Solution to Part (b)

The given matrix is $A=\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]$. Now, solve AX =0.

$$\begin{gathered} \left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]\left[\begin{array}{cc}{x}_{11}& x_{12}\\ x_{21}& x_{22}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}\mathbf{1}{x}_{11}+2x_{21}& 1x_{12}+2x_{22}\\ 3x_{11}+6x_{12}& 3x_{12}+6x_{22}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Then we get, and , hence, the following possible solutions can be:

$\left[\begin{array}{cc}-2& -2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& -4\\ -\frac{1}{2}& 2\end{array}\right]\left[\begin{array}{cc}-6& 1\\ 3& -1\end{array}\right]and\left[\begin{array}{cc}-4& 1\\ 2& -1\end{array}\right]$