Question

4 Let A12 and b4 14 (a) Find A-1 and use it solve the four equations Ax-b1, Ax b2 Ax b3, and Ax b4 (b) The four equations in part (a) can be solved by the same set of operations, since the coefficient matrix is the same in each case Solve the four equations in part (a) by row reducing the augmented matrix [A bj b2 b3 b4

Answer 1

Solution -

$A= \begin{bmatrix} 1 &2 \\ 5& 12 \end{bmatrix}, b_{1} = \begin{bmatrix} -3\\ -9 \end{bmatrix}, b_{2} = \begin{bmatrix} 4\\ 14 \end{bmatrix},b_{3} = \begin{bmatrix} 2\\ 6 \end{bmatrix}, b_{4} = \begin{bmatrix} 1\\ -7\end{bmatrix}$

  $A^{-1}= \begin{bmatrix} 6 &-1 \\ -\frac{5}{2}& \frac{1}{2}\end{bmatrix}$

  $Ax= b_{1} \Rightarrow x = A^{-1}b_{1}$

  $x = \begin{bmatrix} 6 &-1 \\ -\frac{5}{2}& \frac{1}{2} \end{bmatrix}\times \begin{bmatrix} -3\\ -9 \end{bmatrix}$

  $x = \begin{bmatrix} -9\\ 3 \end{bmatrix}$

  $Ax= b_{2} \Rightarrow x = A^{-1}b_{2}$

$x = \begin{bmatrix} 6 &-1 \\ -\frac{5}{2}& \frac{1}{2} \end{bmatrix}\times \begin{bmatrix} 4\\ 14 \end{bmatrix}$

  $x = \begin{bmatrix} 10\\ -3 \end{bmatrix}$

  $Ax= b_{3} \Rightarrow x = A^{-1}b_{3}$

  $x = \begin{bmatrix} 6 &-1 \\ -\frac{5}{2}& \frac{1}{2} \end{bmatrix}\times \begin{bmatrix} 2\\ 6 \end{bmatrix}$

$x = \begin{bmatrix} 6\\ -2 \end{bmatrix}$

  $Ax= b_{4} \Rightarrow x = A^{-1}b_{4}$

  $x = \begin{bmatrix} 6 &-1 \\ -\frac{5}{2}& \frac{1}{2} \end{bmatrix}\times \begin{bmatrix} 1\\ -7 \end{bmatrix}$

  

13 -б

b)

$\left [ Ab_{1}b_{1}b_{2}b_{3}b_{4} \right ] =\begin{bmatrix} 1 & 2 &-3&4 & 2& 1 \\ 5 & 12 & -9& 14& 6 & -7 \end{bmatrix}$

Operating  $R_{2}\rightarrow R_{2}-5R_{1}$

  $=\begin{bmatrix} 1 & 2 &-3&4 & 2& 1 \\ 0 & 2 & 6& -6& -4 & -12 \end{bmatrix}$

  $R_{1}\rightarrow R_{1}-R_{2}$

$=\begin{bmatrix} 1 & 0 &-9&10 & 6& 13 \\ 0 & 2 & 6& -6& -4 & -12 \end{bmatrix}$

Operating $R_{2} \rightarrow \frac{1}{2}R_{2}$

  $=\begin{bmatrix} 1 & 0 &-9&10 & 6& 13 \\ 0 & 1 & 3& -3& -2 & -6 \end{bmatrix}$

So we got same solution .