Linear algebra · individual assignment
Given \( A = \begin{bmatrix} 1 & 3 & 0 \\ 4 & -2 & -1 \end{bmatrix},\; B = \begin{bmatrix} 2 & 1 \\ 5 & 3 \\ -2 & 6 \end{bmatrix} \)
▶ \(AB\) (2×3)·(3×2) → 2×2
▶ \(A^T\) (2×3) → (3×2)
▶ \(B^T\) (3×2) → (2×3)
\[ \begin{cases} x_1 + 2x_2 - x_3 = 2 \\ -x_1 - x_2 + 5x_3 = -1 \\ -2x_1 - 4x_2 + x_3 = 5 \end{cases} \]
✏️ Step 1 – Matrix form \(AX = B\)
\[ A = \begin{bmatrix} 1 & 2 & -1 \\ -1 & -1 & 5 \\ -2 & -4 & 1 \end{bmatrix},\quad B = \begin{bmatrix} 2 \\ -1 \\ 5 \end{bmatrix} \]✏️ Step 2 – Find \(A^{-1}\) by Gauss–Jordan (column by column)
Start \([A \mid I]\) & reduce one column completely before next:
\[ \left[\begin{array}{ccc|ccc} 1 & 2 & -1 & 1 & 0 & 0 \\ -1 & -1 & 5 & 0 & 1 & 0 \\ -2 & -4 & 1 & 0 & 0 & 1 \end{array}\right] \]Column 1: \(R_2 \gets R_2+R_1,\; R_3 \gets R_3+2R_1\) →
\[ \left[\begin{array}{ccc|ccc} 1 & 2 & -1 & 1 & 0 & 0 \\ 0 & 1 & 4 & 1 & 1 & 0 \\ 0 & 0 & -1 & 2 & 0 & 1 \end{array}\right] \]Column 2 (pivot row2): \(R_1 \gets R_1-2R_2\) →
\[ \left[\begin{array}{ccc|ccc} 1 & 0 & -9 & -1 & -2 & 0 \\ 0 & 1 & 4 & 1 & 1 & 0 \\ 0 & 0 & -1 & 2 & 0 & 1 \end{array}\right] \]Column 3: normalize \(R_3\) (\( \times -1\)), then \(R_2 \gets R_2-4R_3\), \(R_1 \gets R_1+9R_3\):
\[ \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -19 & -2 & -9 \\ 0 & 1 & 0 & 9 & 1 & 4 \\ 0 & 0 & 1 & -2 & 0 & -1 \end{array}\right] \]Thus \(A^{-1} = \begin{bmatrix} -19 & -2 & -9 \\ 9 & 1 & 4 \\ -2 & 0 & -1 \end{bmatrix}\).
✏️ Step 3 – \(X = A^{-1}B\)
\[ X = \begin{bmatrix} -19 & -2 & -9 \\ 9 & 1 & 4 \\ -2 & 0 & -1 \end{bmatrix} \begin{bmatrix} 2 \\ -1 \\ 5 \end{bmatrix} = \begin{bmatrix} (-19)(2)+(-2)(-1)+(-9)(5) \\ 9(2)+1(-1)+4(5) \\ (-2)(2)+0(-1)+(-1)(5) \end{bmatrix} = \begin{bmatrix} -81 \\ 37 \\ -9 \end{bmatrix} \]\[ A = \begin{bmatrix} 3 & 5 & -2 \\ 0 & 1 & 3 \\ 0 & 0 & 2 \end{bmatrix},\quad B = \begin{bmatrix} 1 & -2 & 5 \\ 0 & 0 & -1 \\ 0 & 0 & 3 \end{bmatrix} \]
📐 (a) \(A^{-1}\) via augmented [A|I] – column-wise reduction
\[ \left[\begin{array}{ccc|ccc} 3 & 5 & -2 & 1 & 0 & 0 \\ 0 & 1 & 3 & 0 & 1 & 0 \\ 0 & 0 & 2 & 0 & 0 & 1 \end{array}\right] \]Column 1: \(R_1 \gets \frac13 R_1\)
\[ \left[\begin{array}{ccc|ccc} 1 & \frac53 & -\frac23 & \frac13 & 0 & 0 \\ 0 & 1 & 3 & 0 & 1 & 0 \\ 0 & 0 & 2 & 0 & 0 & 1 \end{array}\right] \]Column 2 (pivot row2 =1): \(R_1 \gets R_1 - \frac53 R_2\)
\[ \left[\begin{array}{ccc|ccc} 1 & 0 & -\frac{17}{3} & \frac13 & -\frac53 & 0 \\ 0 & 1 & 3 & 0 & 1 & 0 \\ 0 & 0 & 2 & 0 & 0 & 1 \end{array}\right] \]Column 3: normalize \(R_3\) (\(\frac12 R_3\)), then eliminate above:
\[ \begin{aligned} R_3 &\gets \tfrac12 R_3 :\; (0,0,1 \mid 0,0,0.5)\\ R_2 &\gets R_2 - 3R_3,\quad R_1 \gets R_1 + \tfrac{17}{3}R_3 \end{aligned} \] \[ \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac13 & -\frac53 & \frac{17}{6} \\ 0 & 1 & 0 & 0 & 1 & -\frac32 \\ 0 & 0 & 1 & 0 & 0 & \frac12 \end{array}\right] \]\[ A^{-1} = \begin{bmatrix} \frac13 & -\frac53 & \frac{17}{6} \\[4pt] 0 & 1 & -\frac32 \\[4pt] 0 & 0 & \frac12 \end{bmatrix} \]
📐 (b) \(B^{-1}\) existence
\( \det(B) = 1 \cdot (0·3 - (-1)·0) - (-2)(0·3 - (-1)·0) + 5(0·0 - 0·0) = 0 \) (or simply: second diagonal entry 0 ⇒ singular).
Hence \(B^{-1}\) does not exist.
📐 (c) \(AB\)
\[ AB = \begin{bmatrix} 3 & 5 & -2 \\ 0 & 1 & 3 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & -2 & 5 \\ 0 & 0 & -1 \\ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 3·1+5·0-2·0 & 3·(-2)+5·0-2·0 & 3·5+5·(-1)-2·3 \\ 0 & 0·(-2)+1·0+3·0 & 0·5+1·(-1)+3·3 \\ 0 & 0 & 0·5+0·(-1)+2·3 \end{bmatrix} \] \[ = \begin{bmatrix} 3 & -6 & 4 \\ 0 & 0 & 8 \\ 0 & 0 & 6 \end{bmatrix} \]