Answer on Question #62622 – Math – Linear Algebra
D. Solving a System of Linear Equation.
Direction: Use an inverse matrix to solve each system of linear equation.
Question
1. (a)
{ x + 2 y = − 1 ; x − 2 y = 3. \left\{ \begin{array}{l} x + 2y = -1; \\ x - 2y = 3. \end{array} \right. { x + 2 y = − 1 ; x − 2 y = 3.
Solution
A = [ 1 2 1 − 2 ] ; X = [ x y ] ; B = [ − 1 3 ] . A = \begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix}; X = \begin{bmatrix} x \\ y \end{bmatrix}; B = \begin{bmatrix} -1 \\ 3 \end{bmatrix}. A = [ 1 1 2 − 2 ] ; X = [ x y ] ; B = [ − 1 3 ] .
The system
A X = B ( 1 ) AX = B \quad (1) A X = B ( 1 )
is given.
Using the inverse A − 1 A^{-1} A − 1 A A A
X = A − 1 B . X = A^{-1}B. X = A − 1 B .
Next,
det A = ∣ 1 2 1 − 2 ∣ = 1 ⋅ ( − 2 ) − 1 ⋅ 2 = − 4 ≠ 0 ⇒ A − 1 exists . \det A = \begin{vmatrix} 1 & 2 \\ 1 & -2 \end{vmatrix} = 1 \cdot (-2) - 1 \cdot 2 = -4 \neq 0 \Rightarrow A^{-1} \text{ exists}. det A = ∣ ∣ 1 1 2 − 2 ∣ ∣ = 1 ⋅ ( − 2 ) − 1 ⋅ 2 = − 4 = 0 ⇒ A − 1 exists .
Find the minors M i j M_{ij} M ij C i j C_{ij} C ij A A A
M 11 = − 2 , M 12 = 1 , M 21 = 2 , M 22 = 1 , C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = − 2 , M_{11} = -2, M_{12} = 1, M_{21} = 2, M_{22} = 1, C_{11} = (-1)^{1+1} M_{11} = M_{11} = -2, M 11 = − 2 , M 12 = 1 , M 21 = 2 , M 22 = 1 , C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = − 2 , C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = − 1 , C 21 = ( − 1 ) 2 + 1 M 21 = − M 21 = − 2 , C_{12} = (-1)^{1+2} M_{12} = -M_{12} = -1, C_{21} = (-1)^{2+1} M_{21} = -M_{21} = -2, C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = − 1 , C 21 = ( − 1 ) 2 + 1 M 21 = − M 21 = − 2 , C 22 = ( − 1 ) 2 + 2 M 22 = M 22 = 1. C_{22} = (-1)^{2+2} M_{22} = M_{22} = 1. C 22 = ( − 1 ) 2 + 2 M 22 = M 22 = 1.
Write the matrix of cofactors:
C = [ C 11 C 12 C 21 C 22 ] = [ − 2 − 1 − 2 1 ] . C = \begin{bmatrix} C_{11} & C_{12} \\ C_{21} & C_{22} \end{bmatrix} = \begin{bmatrix} -2 & -1 \\ -2 & 1 \end{bmatrix}. C = [ C 11 C 21 C 12 C 22 ] = [ − 2 − 2 − 1 1 ] .
The transposed matrix of cofactors:
C T = [ C 11 C 21 C 12 C 22 ] = [ − 2 − 2 − 1 1 ] . C^T = \begin{bmatrix} C_{11} & C_{21} \\ C_{12} & C_{22} \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ -1 & 1 \end{bmatrix}. C T = [ C 11 C 12 C 21 C 22 ] = [ − 2 − 1 − 2 1 ] .
Find the inverse matrix:
A − 1 = 1 det A C T = − 1 4 ⋅ [ − 2 − 2 − 1 1 ] = [ − 2 − 4 − 2 − 4 − 1 − 4 − 4 − 4 ] = [ 1 2 1 2 1 4 − 1 4 ] . A^{-1} = \frac{1}{\det A} C^T = -\frac{1}{4} \cdot \begin{bmatrix} -2 & -2 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{-2}{-4} & \frac{-2}{-4} \\ \frac{-1}{-4} & \frac{-4}{-4} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & -\frac{1}{4} \end{bmatrix}. A − 1 = det A 1 C T = − 4 1 ⋅ [ − 2 − 1 − 2 1 ] = [ − 4 − 2 − 4 − 1 − 4 − 2 − 4 − 4 ] = [ 2 1 4 1 2 1 − 4 1 ] .
Then the solution of the system (1) is
X = [ x y ] = A − 1 B = [ 1 2 1 2 1 4 − 1 4 ] ⋅ [ − 1 3 ] = [ 1 2 ⋅ ( − 1 ) + 1 2 ⋅ 3 1 4 ⋅ ( − 1 ) − 1 4 ⋅ 3 ] = [ − 1 2 + 3 2 − 1 4 − 3 4 ] = [ 1 − 1 ] . X = \begin{bmatrix} x \\ y \end{bmatrix} = A^{-1}B = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & -\frac{1}{4} \end{bmatrix} \cdot \begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \cdot (-1) + \frac{1}{2} \cdot 3 \\ \frac{1}{4} \cdot (-1) - \frac{1}{4} \cdot 3 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} + \frac{3}{2} \\ -\frac{1}{4} - \frac{3}{4} \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}. X = [ x y ] = A − 1 B = [ 2 1 4 1 2 1 − 4 1 ] ⋅ [ − 1 3 ] = [ 2 1 ⋅ ( − 1 ) + 2 1 ⋅ 3 4 1 ⋅ ( − 1 ) − 4 1 ⋅ 3 ] = [ − 2 1 + 2 3 − 4 1 − 4 3 ] = [ 1 − 1 ] .
Answer: x = 1 , y = − 1 x = 1, y = -1 x = 1 , y = − 1
Question
1. (b)
{ x + 2 y = 10 ; x − 2 y = − 6. \left\{ \begin{array}{l} x + 2 y = 1 0; \\ x - 2 y = - 6. \end{array} \right. { x + 2 y = 10 ; x − 2 y = − 6.
Solution
A = [ 1 2 1 − 2 ] ; X = [ x y ] ; B = [ 10 − 6 ] . A = \left[ \begin{array}{cc} 1 & 2 \\ 1 & -2 \end{array} \right]; X = \left[ \begin{array}{c} x \\ y \end{array} \right]; B = \left[ \begin{array}{c} 1 0 \\ -6 \end{array} \right]. A = [ 1 1 2 − 2 ] ; X = [ x y ] ; B = [ 10 − 6 ] .
The system
A X = B ( 2 ) A X = B \quad (2) A X = B ( 2 )
is given.
Using the inverse A − 1 A^{-1} A − 1 A A A
X = A − 1 B . X = A^{-1} B. X = A − 1 B .
Next,
det A = ∣ 1 2 1 − 2 ∣ = 1 ⋅ ( − 2 ) − 1 ⋅ 2 = − 4 ≠ 0 ⇒ A − 1 exists . \det A = \left| \begin{array}{cc} 1 & 2 \\ 1 & -2 \end{array} \right| = 1 \cdot (-2) - 1 \cdot 2 = -4 \neq 0 \Rightarrow A^{-1} \text{ exists}. det A = ∣ ∣ 1 1 2 − 2 ∣ ∣ = 1 ⋅ ( − 2 ) − 1 ⋅ 2 = − 4 = 0 ⇒ A − 1 exists .
Find the minors M i j M_{ij} M ij C i j C_{ij} C ij A A A
M 11 = − 2 , M 12 = 1 , M 21 = 2 , M 22 = 1 , C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = − 2 , M_{11} = -2, M_{12} = 1, M_{21} = 2, M_{22} = 1, C_{11} = (-1)^{1+1} M_{11} = M_{11} = -2, M 11 = − 2 , M 12 = 1 , M 21 = 2 , M 22 = 1 , C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = − 2 , C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = − 1 , C 21 = ( − 1 ) 2 + 1 M 21 = − M 21 = − 2 , C_{12} = (-1)^{1+2} M_{12} = -M_{12} = -1, C_{21} = (-1)^{2+1} M_{21} = -M_{21} = -2, C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = − 1 , C 21 = ( − 1 ) 2 + 1 M 21 = − M 21 = − 2 , C 22 = ( − 1 ) 2 + 2 M 22 = M 22 = 1. C_{22} = (-1)^{2+2} M_{22} = M_{22} = 1. C 22 = ( − 1 ) 2 + 2 M 22 = M 22 = 1.
Write the matrix of cofactors:
C = [ C 11 C 12 C 21 C 22 ] = [ − 2 − 1 − 2 1 ] . C = \left[ \begin{array}{cc} C_{11} & C_{12} \\ C_{21} & C_{22} \end{array} \right] = \left[ \begin{array}{cc} -2 & -1 \\ -2 & 1 \end{array} \right]. C = [ C 11 C 21 C 12 C 22 ] = [ − 2 − 2 − 1 1 ] .
The transposed matrix of cofactors:
C T = [ C 11 C 21 C 12 C 22 ] = [ − 2 − 2 − 1 1 ] . C^{T} = \left[ \begin{array}{cc} C_{11} & C_{21} \\ C_{12} & C_{22} \end{array} \right] = \left[ \begin{array}{cc} -2 & -2 \\ -1 & 1 \end{array} \right]. C T = [ C 11 C 12 C 21 C 22 ] = [ − 2 − 1 − 2 1 ] .
Find the inverse matrix:
A − 1 = 1 det A ⋅ C T = − 1 4 [ − 2 − 2 − 1 1 ] = [ − 2 − 4 − 2 − 4 − 1 − 4 1 − 4 ] = [ 1 2 1 2 1 4 − 1 4 ] . A^{-1} = \frac{1}{\det A} \cdot C^{T} = -\frac{1}{4} \left[ \begin{array}{cc} -2 & -2 \\ -1 & 1 \end{array} \right] = \left[ \begin{array}{cc} \frac{-2}{-4} & \frac{-2}{-4} \\ \frac{-1}{-4} & \frac{1}{-4} \end{array} \right] = \left[ \begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & -\frac{1}{4} \end{array} \right]. A − 1 = det A 1 ⋅ C T = − 4 1 [ − 2 − 1 − 2 1 ] = [ − 4 − 2 − 4 − 1 − 4 − 2 − 4 1 ] = [ 2 1 4 1 2 1 − 4 1 ] .
Then the solution of the system (2) is
X = [ x y ] = A − 1 ⋅ B = [ 1 2 1 2 1 4 − 1 4 ] ⋅ [ 10 − 6 ] = [ 1 2 ⋅ 10 + 1 2 ⋅ ( − 6 ) 1 4 ⋅ 10 − 1 4 ⋅ ( − 6 ) ] = [ 10 2 − 6 2 10 4 + 6 4 ] = [ 2 4 ] . X = \left[ \begin{array}{c} x \\ y \end{array} \right] = A^{-1} \cdot B = \left[ \begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & -\frac{1}{4} \end{array} \right] \cdot \left[ \begin{array}{c} 10 \\ -6 \end{array} \right] = \left[ \begin{array}{c} \frac{1}{2} \cdot 10 + \frac{1}{2} \cdot (-6) \\ \frac{1}{4} \cdot 10 - \frac{1}{4} \cdot (-6) \end{array} \right] = \left[ \begin{array}{c} \frac{10}{2} - \frac{6}{2} \\ \frac{10}{4} + \frac{6}{4} \end{array} \right] = \left[ \begin{array}{c} 2 \\ 4 \end{array} \right]. X = [ x y ] = A − 1 ⋅ B = [ 2 1 4 1 2 1 − 4 1 ] ⋅ [ 10 − 6 ] = [ 2 1 ⋅ 10 + 2 1 ⋅ ( − 6 ) 4 1 ⋅ 10 − 4 1 ⋅ ( − 6 ) ] = [ 2 10 − 2 6 4 10 + 4 6 ] = [ 2 4 ] .
Answer: x = 2 , y = 4 x = 2, y = 4 x = 2 , y = 4
Question
2. (a)
{ 2 x − y = − 3 ; 2 x + y = 7. \left\{ \begin{array}{l} 2x - y = -3; \\ 2x + y = 7. \end{array} \right. { 2 x − y = − 3 ; 2 x + y = 7. Solution
A = [ 2 − 1 2 1 ] ; X = [ x y ] ; B = [ − 3 7 ] . A = \left[ \begin{array}{cc} 2 & -1 \\ 2 & 1 \end{array} \right]; X = \left[ \begin{array}{c} x \\ y \end{array} \right]; B = \left[ \begin{array}{c} -3 \\ 7 \end{array} \right]. A = [ 2 2 − 1 1 ] ; X = [ x y ] ; B = [ − 3 7 ] .
The system
A X = B ( 3 ) AX = B \quad (3) A X = B ( 3 )
is given.
Using the inverse A − 1 A^{-1} A − 1 A A A
X = A − 1 B . X = A^{-1}B. X = A − 1 B .
Next,
det A = ∣ 2 − 1 2 1 ∣ = 2 ⋅ 1 − ( − 1 ) ⋅ 2 = 4 ≠ 0 ⇒ A − 1 exists . \det A = \begin{vmatrix} 2 & -1 \\ 2 & 1 \end{vmatrix} = 2 \cdot 1 - (-1) \cdot 2 = 4 \neq 0 \Rightarrow A^{-1} \text{ exists}. det A = ∣ ∣ 2 2 − 1 1 ∣ ∣ = 2 ⋅ 1 − ( − 1 ) ⋅ 2 = 4 = 0 ⇒ A − 1 exists .
Find the minors M i j M_{ij} M ij C i j C_{ij} C ij A A A
M 11 = 1 , M 12 = 2 , M 21 = − 1 , M 22 = 2 , C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = 1 , M_{11} = 1, M_{12} = 2, M_{21} = -1, M_{22} = 2, C_{11} = (-1)^{1+1} M_{11} = M_{11} = 1, M 11 = 1 , M 12 = 2 , M 21 = − 1 , M 22 = 2 , C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = 1 , C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = − 2 , C 21 = ( − 1 ) 2 + 1 M 21 = − M 21 = 1 , C_{12} = (-1)^{1+2} M_{12} = -M_{12} = -2, C_{21} = (-1)^{2+1} M_{21} = -M_{21} = 1, C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = − 2 , C 21 = ( − 1 ) 2 + 1 M 21 = − M 21 = 1 , C 22 = ( − 1 ) 2 + 2 M 22 = M 22 = 2. C_{22} = (-1)^{2+2} M_{22} = M_{22} = 2. C 22 = ( − 1 ) 2 + 2 M 22 = M 22 = 2.
Write the matrix of cofactors:
C = [ C 11 C 12 C 21 C 22 ] = [ 1 − 2 1 2 ] . C = \begin{bmatrix} C_{11} & C_{12} \\ C_{21} & C_{22} \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 1 & 2 \end{bmatrix}. C = [ C 11 C 21 C 12 C 22 ] = [ 1 1 − 2 2 ] .
The transposed matrix of cofactors:
C T = [ C 11 C 21 C 12 C 22 ] = [ 1 1 − 2 2 ] . C^T = \begin{bmatrix} C_{11} & C_{21} \\ C_{12} & C_{22} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ -2 & 2 \end{bmatrix}. C T = [ C 11 C 12 C 21 C 22 ] = [ 1 − 2 1 2 ] .
Find the inverse matrix:
A − 1 = 1 det A C T = 1 4 [ 1 1 − 2 2 ] = [ 1 4 1 4 − 2 4 2 4 ] = [ 1 4 1 4 − 1 2 1 2 ] . A^{-1} = \frac{1}{\det A} C^T = \frac{1}{4} \begin{bmatrix} 1 & 1 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & \frac{1}{4} \\ \frac{-2}{4} & \frac{2}{4} \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & \frac{1}{4} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix}. A − 1 = det A 1 C T = 4 1 [ 1 − 2 1 2 ] = [ 4 1 4 − 2 4 1 4 2 ] = [ 4 1 − 2 1 4 1 2 1 ] .
Then the solution of the system (3) is
X = [ x y ] = A − 1 B = [ 1 4 1 4 − 1 2 1 2 ] ⋅ [ − 3 7 ] = [ 1 4 ⋅ ( − 3 ) + 1 4 ⋅ 7 − 1 2 ⋅ ( − 3 ) + 1 2 ⋅ 7 ] = [ − 3 4 + 7 4 3 2 + 7 2 ] = [ 1 5 ] . X = \begin{bmatrix} x \\ y \end{bmatrix} = A^{-1}B = \begin{bmatrix} \frac{1}{4} & \frac{1}{4} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix} \cdot \begin{bmatrix} -3 \\ 7 \end{bmatrix} = \begin{bmatrix} \frac{1}{4} \cdot (-3) + \frac{1}{4} \cdot 7 \\ -\frac{1}{2} \cdot (-3) + \frac{1}{2} \cdot 7 \end{bmatrix} = \begin{bmatrix} -\frac{3}{4} + \frac{7}{4} \\ \frac{3}{2} + \frac{7}{2} \end{bmatrix} = \begin{bmatrix} 1 \\ 5 \end{bmatrix}. X = [ x y ] = A − 1 B = [ 4 1 − 2 1 4 1 2 1 ] ⋅ [ − 3 7 ] = [ 4 1 ⋅ ( − 3 ) + 4 1 ⋅ 7 − 2 1 ⋅ ( − 3 ) + 2 1 ⋅ 7 ] = [ − 4 3 + 4 7 2 3 + 2 7 ] = [ 1 5 ] .
Answer: x = 1 , y = 5 x = 1, y = 5 x = 1 , y = 5
Question
2. (b)
{ 2 x − y = − 1 ; 2 x + y = − 3. \begin{cases}
2x - y = -1; \\
2x + y = -3.
\end{cases} { 2 x − y = − 1 ; 2 x + y = − 3. Solution
A = [ 2 − 1 2 1 ] ; X = [ x y ] ; B = [ − 1 − 3 ] . A = \begin{bmatrix} 2 & -1 \\ 2 & 1 \end{bmatrix}; \quad X = \begin{bmatrix} x \\ y \end{bmatrix}; \quad B = \begin{bmatrix} -1 \\ -3 \end{bmatrix}. A = [ 2 2 − 1 1 ] ; X = [ x y ] ; B = [ − 1 − 3 ] .
The system
A X = B ( 4 ) AX = B \quad (4) A X = B ( 4 )
is given.
Using the inverse A − 1 A^{-1} A − 1 A A A
X = A − 1 B . X = A^{-1}B. X = A − 1 B .
Next,
det A = ∣ 2 − 1 2 1 ∣ = 2 ⋅ 1 − ( − 1 ) ⋅ 2 = 4 ≠ 0 ⇒ A − 1 exists . \det A = \begin{vmatrix} 2 & -1 \\ 2 & 1 \end{vmatrix} = 2 \cdot 1 - (-1) \cdot 2 = 4 \neq 0 \Rightarrow A^{-1} \text{ exists}. det A = ∣ ∣ 2 2 − 1 1 ∣ ∣ = 2 ⋅ 1 − ( − 1 ) ⋅ 2 = 4 = 0 ⇒ A − 1 exists .
Find the minors M i j M_{ij} M ij C i j C_{ij} C ij A A A
M 11 = 1 , M 12 = 2 , M 21 = − 1 , M 22 = 2 , C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = 1 , M_{11} = 1, M_{12} = 2, M_{21} = -1, M_{22} = 2, C_{11} = (-1)^{1+1} M_{11} = M_{11} = 1, M 11 = 1 , M 12 = 2 , M 21 = − 1 , M 22 = 2 , C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = 1 , C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = − 2 , C 21 = ( − 1 ) 2 + 1 M 21 = − M 21 = 1 , C_{12} = (-1)^{1+2} M_{12} = -M_{12} = -2, C_{21} = (-1)^{2+1} M_{21} = -M_{21} = 1, C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = − 2 , C 21 = ( − 1 ) 2 + 1 M 21 = − M 21 = 1 , C 22 = ( − 1 ) 2 + 2 M 22 = M 22 = 2. C_{22} = (-1)^{2+2} M_{22} = M_{22} = 2. C 22 = ( − 1 ) 2 + 2 M 22 = M 22 = 2.
Write the matrix of cofactors:
C = [ C 11 C 12 C 21 C 22 ] = [ 1 − 2 1 2 ] . C = \begin{bmatrix} C_{11} & C_{12} \\ C_{21} & C_{22} \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 1 & 2 \end{bmatrix}. C = [ C 11 C 21 C 12 C 22 ] = [ 1 1 − 2 2 ] .
The transposed matrix of cofactors:
C T = [ C 11 C 21 C 12 C 22 ] = [ 1 1 − 2 2 ] . C^T = \begin{bmatrix} C_{11} & C_{21} \\ C_{12} & C_{22} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ -2 & 2 \end{bmatrix}. C T = [ C 11 C 12 C 21 C 22 ] = [ 1 − 2 1 2 ] .
Find the inverse matrix:
A − 1 = 1 det A C T = 1 4 [ 1 1 − 2 2 ] = [ 1 4 1 4 − 2 4 2 4 ] = [ 1 4 1 4 − 1 2 1 2 ] . A^{-1} = \frac{1}{\det A} C^T = \frac{1}{4} \begin{bmatrix} 1 & 1 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & \frac{1}{4} \\ \frac{-2}{4} & \frac{2}{4} \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & \frac{1}{4} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix}. A − 1 = det A 1 C T = 4 1 [ 1 − 2 1 2 ] = [ 4 1 4 − 2 4 1 4 2 ] = [ 4 1 − 2 1 4 1 2 1 ] .
Then the solution of the system (4) is
X = [ x y ] = A − 1 B = [ 1 4 1 4 − 1 2 1 2 ] ⋅ [ − 1 − 3 ] = [ 1 4 ⋅ ( − 1 ) + 1 4 ⋅ ( − 3 ) − 1 2 ⋅ ( − 1 ) + 1 2 ⋅ ( − 3 ) ] = [ − 1 4 − 3 4 1 2 − 3 2 ] = [ − 1 − 1 ] . X = \begin{bmatrix} x \\ y \end{bmatrix} = A^{-1} B = \begin{bmatrix} \frac{1}{4} & \frac{1}{4} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix} \cdot \begin{bmatrix} -1 \\ -3 \end{bmatrix} = \begin{bmatrix} \frac{1}{4} \cdot (-1) + \frac{1}{4} \cdot (-3) \\ -\frac{1}{2} \cdot (-1) + \frac{1}{2} \cdot (-3) \end{bmatrix} = \begin{bmatrix} -\frac{1}{4} - \frac{3}{4} \\ \frac{1}{2} - \frac{3}{2} \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix}. X = [ x y ] = A − 1 B = [ 4 1 − 2 1 4 1 2 1 ] ⋅ [ − 1 − 3 ] = [ 4 1 ⋅ ( − 1 ) + 4 1 ⋅ ( − 3 ) − 2 1 ⋅ ( − 1 ) + 2 1 ⋅ ( − 3 ) ] = [ − 4 1 − 4 3 2 1 − 2 3 ] = [ − 1 − 1 ] .
Answer: x = − 1 , y = − 1 x = -1, y = -1 x = − 1 , y = − 1
Question
3. (a)
{ x + 2 y + z = 2 ; x + 2 y − z = 4 ; x − 2 y + z = − 2. \left\{ \begin{array}{l} x + 2y + z = 2; \\ x + 2y - z = 4; \\ x - 2y + z = -2. \end{array} \right. ⎩ ⎨ ⎧ x + 2 y + z = 2 ; x + 2 y − z = 4 ; x − 2 y + z = − 2. Solution
A = [ 1 2 1 1 2 − 1 1 − 2 1 ] ; X = [ x y z ] ; B = [ 2 4 − 2 ] . A = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 2 & -1 \\ 1 & -2 & 1 \end{bmatrix}; \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}; \quad B = \begin{bmatrix} 2 \\ 4 \\ -2 \end{bmatrix}. A = ⎣ ⎡ 1 1 1 2 2 − 2 1 − 1 1 ⎦ ⎤ ; X = ⎣ ⎡ x y z ⎦ ⎤ ; B = ⎣ ⎡ 2 4 − 2 ⎦ ⎤ .
The system
A X = B ( 5 ) AX = B \quad (5) A X = B ( 5 )
is given.
Using the inverse A − 1 A^{-1} A − 1 A A A
X = A − 1 B . X = A^{-1} B. X = A − 1 B .
Next,
det A = ∣ 1 2 1 1 2 − 1 1 − 2 1 ∣ = ∣ expand along the first row ∣ = 1 ⋅ ∣ 2 − 1 − 2 1 ∣ − 2 ⋅ ∣ 1 − 1 1 1 ∣ + 1 ⋅ ∣ 1 2 1 − 2 ∣ = = ( 2 ⋅ 1 − ( − 2 ) ⋅ ( − 1 ) ) − 2 ⋅ ( 1 ⋅ 1 − 1 ⋅ ( − 1 ) ) + ( 1 ⋅ ( − 2 ) − 1 ⋅ 2 ) = 0 − 4 − 4 = = − 8 ≠ 0 ⇒ A − 1 exists . \begin{aligned}
\det A &= \begin{vmatrix} 1 & 2 & 1 \\ 1 & 2 & -1 \\ 1 & -2 & 1 \end{vmatrix} = | \text{expand along the first row} | \\
&= 1 \cdot \begin{vmatrix} 2 & -1 \\ -2 & 1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 1 & -2 \end{vmatrix} = \\
&= \left(2 \cdot 1 - (-2) \cdot (-1)\right) - 2 \cdot \left(1 \cdot 1 - 1 \cdot (-1)\right) + (1 \cdot (-2) - 1 \cdot 2) = 0 - 4 - 4 = \\
&= -8 \neq 0 \Rightarrow A^{-1} \text{ exists}.
\end{aligned} det A = ∣ ∣ 1 1 1 2 2 − 2 1 − 1 1 ∣ ∣ = ∣ expand along the first row ∣ = 1 ⋅ ∣ ∣ 2 − 2 − 1 1 ∣ ∣ − 2 ⋅ ∣ ∣ 1 1 − 1 1 ∣ ∣ + 1 ⋅ ∣ ∣ 1 1 2 − 2 ∣ ∣ = = ( 2 ⋅ 1 − ( − 2 ) ⋅ ( − 1 ) ) − 2 ⋅ ( 1 ⋅ 1 − 1 ⋅ ( − 1 ) ) + ( 1 ⋅ ( − 2 ) − 1 ⋅ 2 ) = 0 − 4 − 4 = = − 8 = 0 ⇒ A − 1 exists .
Find the minors M i j M_{ij} M ij C i j C_{ij} C ij A A A
M 11 = ∣ 2 − 1 − 2 1 ∣ = 2 ⋅ 1 − ( − 2 ) ⋅ ( − 1 ) = 0 , M_{11} = \left| \begin{array}{cc} 2 & -1 \\ -2 & 1 \end{array} \right| = 2 \cdot 1 - (-2) \cdot (-1) = 0, M 11 = ∣ ∣ 2 − 2 − 1 1 ∣ ∣ = 2 ⋅ 1 − ( − 2 ) ⋅ ( − 1 ) = 0 , M 12 = ∣ 1 − 1 1 1 ∣ = 1 ⋅ 1 − 1 ⋅ ( − 1 ) = 2 , M_{12} = \left| \begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array} \right| = 1 \cdot 1 - 1 \cdot (-1) = 2, M 12 = ∣ ∣ 1 1 − 1 1 ∣ ∣ = 1 ⋅ 1 − 1 ⋅ ( − 1 ) = 2 , M 13 = ∣ 1 2 1 − 2 ∣ = 1 ⋅ ( − 2 ) − 1 ⋅ 2 = − 4 , M_{13} = \left| \begin{array}{cc} 1 & 2 \\ 1 & -2 \end{array} \right| = 1 \cdot (-2) - 1 \cdot 2 = -4, M 13 = ∣ ∣ 1 1 2 − 2 ∣ ∣ = 1 ⋅ ( − 2 ) − 1 ⋅ 2 = − 4 , M 21 = ∣ 2 1 − 2 1 ∣ = 2 ⋅ 1 − ( − 2 ) ⋅ 1 = 4 , M_{21} = \left| \begin{array}{cc} 2 & 1 \\ -2 & 1 \end{array} \right| = 2 \cdot 1 - (-2) \cdot 1 = 4, M 21 = ∣ ∣ 2 − 2 1 1 ∣ ∣ = 2 ⋅ 1 − ( − 2 ) ⋅ 1 = 4 , M 22 = ∣ 1 1 1 1 ∣ = 1 ⋅ 1 − 1 ⋅ 1 = 0 , M_{22} = \left| \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right| = 1 \cdot 1 - 1 \cdot 1 = 0, M 22 = ∣ ∣ 1 1 1 1 ∣ ∣ = 1 ⋅ 1 − 1 ⋅ 1 = 0 , M 23 = ∣ 1 2 1 − 2 ∣ = 1 ⋅ ( − 2 ) − 1 ⋅ 2 = − 4 , M_{23} = \left| \begin{array}{cc} 1 & 2 \\ 1 & -2 \end{array} \right| = 1 \cdot (-2) - 1 \cdot 2 = -4, M 23 = ∣ ∣ 1 1 2 − 2 ∣ ∣ = 1 ⋅ ( − 2 ) − 1 ⋅ 2 = − 4 , M 31 = ∣ 2 1 2 − 1 ∣ = 2 ⋅ ( − 1 ) − 2 ⋅ 1 = − 4 , M_{31} = \left| \begin{array}{cc} 2 & 1 \\ 2 & -1 \end{array} \right| = 2 \cdot (-1) - 2 \cdot 1 = -4, M 31 = ∣ ∣ 2 2 1 − 1 ∣ ∣ = 2 ⋅ ( − 1 ) − 2 ⋅ 1 = − 4 , M 32 = ∣ 1 1 1 − 1 ∣ = 1 ⋅ ( − 1 ) − 1 ⋅ 1 = − 2 , M_{32} = \left| \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right| = 1 \cdot (-1) - 1 \cdot 1 = -2, M 32 = ∣ ∣ 1 1 1 − 1 ∣ ∣ = 1 ⋅ ( − 1 ) − 1 ⋅ 1 = − 2 , M 33 = ∣ 1 2 1 2 ∣ = 1 ⋅ 2 − 1 ⋅ 2 = 0 , M_{33} = \left| \begin{array}{cc} 1 & 2 \\ 1 & 2 \end{array} \right| = 1 \cdot 2 - 1 \cdot 2 = 0, M 33 = ∣ ∣ 1 1 2 2 ∣ ∣ = 1 ⋅ 2 − 1 ⋅ 2 = 0 , C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = 0 , C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = − 2 , C_{11} = (-1)^{1+1} M_{11} = M_{11} = 0, \quad C_{12} = (-1)^{1+2} M_{12} = -M_{12} = -2, C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = 0 , C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = − 2 , C 13 = ( − 1 ) 1 + 3 M 13 = M 13 = − 4 , C 21 = ( − 1 ) 2 + 1 M 21 = − M 21 = − 4 , C_{13} = (-1)^{1+3} M_{13} = M_{13} = -4, \quad C_{21} = (-1)^{2+1} M_{21} = -M_{21} = -4, C 13 = ( − 1 ) 1 + 3 M 13 = M 13 = − 4 , C 21 = ( − 1 ) 2 + 1 M 21 = − M 21 = − 4 , C 22 = ( − 1 ) 2 + 2 M 22 = M 22 = 0 , C 23 = ( − 1 ) 2 + 3 M 23 = − M 23 = 4 , C_{22} = (-1)^{2+2} M_{22} = M_{22} = 0, \quad C_{23} = (-1)^{2+3} M_{23} = -M_{23} = 4, C 22 = ( − 1 ) 2 + 2 M 22 = M 22 = 0 , C 23 = ( − 1 ) 2 + 3 M 23 = − M 23 = 4 , C 31 = ( − 1 ) 3 + 1 M 31 = M 31 = − 4 , C 32 = ( − 1 ) 3 + 2 M 32 = − M 32 = 2 , C_{31} = (-1)^{3+1} M_{31} = M_{31} = -4, \quad C_{32} = (-1)^{3+2} M_{32} = -M_{32} = 2, C 31 = ( − 1 ) 3 + 1 M 31 = M 31 = − 4 , C 32 = ( − 1 ) 3 + 2 M 32 = − M 32 = 2 , C 33 = ( − 1 ) 3 + 3 M 33 = M 33 = 0. C_{33} = (-1)^{3+3} M_{33} = M_{33} = 0. C 33 = ( − 1 ) 3 + 3 M 33 = M 33 = 0.
Write the matrix of cofactors:
C = [ C 11 C 12 C 13 C 21 C 22 C 23 C 31 C 32 C 33 ] = [ 0 − 2 − 4 − 4 0 4 − 4 2 0 ] . C = \left[ \begin{array}{ccc} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{array} \right] = \left[ \begin{array}{ccc} 0 & -2 & -4 \\ -4 & 0 & 4 \\ -4 & 2 & 0 \end{array} \right]. C = ⎣ ⎡ C 11 C 21 C 31 C 12 C 22 C 32 C 13 C 23 C 33 ⎦ ⎤ = ⎣ ⎡ 0 − 4 − 4 − 2 0 2 − 4 4 0 ⎦ ⎤ .
The transposed matrix of cofactors:
C T = [ C 11 C 21 C 31 C 12 C 22 C 32 C 13 C 23 C 33 ] = [ 0 − 4 − 4 − 2 0 2 − 4 4 0 ] . C^T = \left[ \begin{array}{ccc} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{array} \right] = \left[ \begin{array}{ccc} 0 & -4 & -4 \\ -2 & 0 & 2 \\ -4 & 4 & 0 \end{array} \right]. C T = ⎣ ⎡ C 11 C 12 C 13 C 21 C 22 C 23 C 31 C 32 C 33 ⎦ ⎤ = ⎣ ⎡ 0 − 2 − 4 − 4 0 4 − 4 2 0 ⎦ ⎤ .
Find the inverse matrix:
A − 1 = 1 det A C T = − 1 8 [ 0 − 4 − 4 − 2 0 2 − 4 4 0 ] = [ 0 − 8 − 4 − 8 − 4 − 8 − 2 − 8 0 − 8 2 − 8 − 4 − 8 4 − 8 0 − 8 ] = [ 0 1 2 1 2 1 0 − 1 4 1 2 − 1 2 0 ] . A^{-1} = \frac{1}{\det A} C^T = -\frac{1}{8} \left[ \begin{array}{ccc} 0 & -4 & -4 \\ -2 & 0 & 2 \\ -4 & 4 & 0 \end{array} \right] = \left[ \begin{array}{ccc} \frac{0}{-8} & \frac{-4}{-8} & \frac{-4}{-8} \\ \frac{-2}{-8} & \frac{0}{-8} & \frac{2}{-8} \\ \frac{-4}{-8} & \frac{4}{-8} & \frac{0}{-8} \end{array} \right] = \left[ \begin{array}{ccc} 0 & \frac{1}{2} & \frac{1}{2} \\ 1 & 0 & \frac{-1}{4} \\ \frac{1}{2} & -\frac{1}{2} & 0 \end{array} \right]. A − 1 = det A 1 C T = − 8 1 ⎣ ⎡ 0 − 2 − 4 − 4 0 4 − 4 2 0 ⎦ ⎤ = ⎣ ⎡ − 8 0 − 8 − 2 − 8 − 4 − 8 − 4 − 8 0 − 8 4 − 8 − 4 − 8 2 − 8 0 ⎦ ⎤ = ⎣ ⎡ 0 1 2 1 2 1 0 − 2 1 2 1 4 − 1 0 ⎦ ⎤ .
Then the solution of the system (5) is
X = [ x y z ] = A − 1 B = [ 0 1 2 1 2 1 4 0 − 1 4 1 2 − 1 2 0 ] ⋅ [ 2 4 − 2 ] = [ 0 ⋅ 2 + 1 2 ⋅ 4 + 1 2 ⋅ ( − 2 ) 1 4 ⋅ 2 + 0 ⋅ 4 − 1 4 ⋅ ( − 2 ) 1 2 ⋅ 2 − 1 2 ⋅ 4 + 0 ⋅ ( − 2 ) ] = = [ 0 + 2 − 1 1 2 + 0 + 1 2 1 − 2 − 0 ] = [ 1 1 − 1 ] \begin{array}{l}
X = \left[ \begin{array}{l} x \\ y \\ z \end{array} \right] = A^{-1} B = \left[ \begin{array}{ccc} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & 0 & -\frac{1}{4} \\ \frac{1}{2} & -\frac{1}{2} & 0 \end{array} \right] \cdot \left[ \begin{array}{l} 2 \\ 4 \\ -2 \end{array} \right] = \left[ \begin{array}{c} 0 \cdot 2 + \frac{1}{2} \cdot 4 + \frac{1}{2} \cdot (-2) \\ \frac{1}{4} \cdot 2 + 0 \cdot 4 - \frac{1}{4} \cdot (-2) \\ \frac{1}{2} \cdot 2 - \frac{1}{2} \cdot 4 + 0 \cdot (-2) \end{array} \right] = \\
= \left[ \begin{array}{c} 0 + 2 - 1 \\ \frac{1}{2} + 0 + \frac{1}{2} \\ 1 - 2 - 0 \end{array} \right] = \left[ \begin{array}{c} 1 \\ 1 \\ -1 \end{array} \right]
\end{array} X = ⎣ ⎡ x y z ⎦ ⎤ = A − 1 B = ⎣ ⎡ 0 4 1 2 1 2 1 0 − 2 1 2 1 − 4 1 0 ⎦ ⎤ ⋅ ⎣ ⎡ 2 4 − 2 ⎦ ⎤ = ⎣ ⎡ 0 ⋅ 2 + 2 1 ⋅ 4 + 2 1 ⋅ ( − 2 ) 4 1 ⋅ 2 + 0 ⋅ 4 − 4 1 ⋅ ( − 2 ) 2 1 ⋅ 2 − 2 1 ⋅ 4 + 0 ⋅ ( − 2 ) ⎦ ⎤ = = ⎣ ⎡ 0 + 2 − 1 2 1 + 0 + 2 1 1 − 2 − 0 ⎦ ⎤ = ⎣ ⎡ 1 1 − 1 ⎦ ⎤
Answer: x = 1 , y = 1 , z = − 1 x = 1, y = 1, z = -1 x = 1 , y = 1 , z = − 1
Question
3. (b)
{ x + 2 y + z = 1 ; x + 2 y − z = 3 ; x − 2 y + z = − 3. \left\{ \begin{array}{l} x + 2y + z = 1; \\ x + 2y - z = 3; \\ x - 2y + z = -3. \end{array} \right. ⎩ ⎨ ⎧ x + 2 y + z = 1 ; x + 2 y − z = 3 ; x − 2 y + z = − 3. Solution
A = [ 1 2 1 1 2 − 1 1 − 2 1 ] ; X = [ x y z ] ; B = [ 1 3 − 3 ] . A = \left[ \begin{array}{ccc} 1 & 2 & 1 \\ 1 & 2 & -1 \\ 1 & -2 & 1 \end{array} \right]; X = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]; B = \left[ \begin{array}{c} 1 \\ 3 \\ -3 \end{array} \right]. A = ⎣ ⎡ 1 1 1 2 2 − 2 1 − 1 1 ⎦ ⎤ ; X = ⎣ ⎡ x y z ⎦ ⎤ ; B = ⎣ ⎡ 1 3 − 3 ⎦ ⎤ .
The system
A X = B ( 6 ) AX = B \quad (6) A X = B ( 6 )
is given.
Using the inverse A − 1 A^{-1} A − 1 A A A
X = A − 1 B . X = A^{-1} B. X = A − 1 B .
Next,
det A = ∣ 1 2 1 1 2 − 1 1 − 2 1 ∣ = ∣ expand along the first row ∣ = 1 ⋅ ∣ 2 − 1 − 2 1 ∣ − 2 ⋅ ∣ 1 − 1 1 1 ∣ + 1 ⋅ ∣ 1 2 1 − 2 ∣ = = 0 − 4 − 4 = − 8 ≠ 0 ⇒ A − 1 exists . \begin{array}{l}
\det A = \left| \begin{array}{ccc} 1 & 2 & 1 \\ 1 & 2 & -1 \\ 1 & -2 & 1 \end{array} \right| = | \text{expand along the first row} | \\
= 1 \cdot \left| \begin{array}{cc} 2 & -1 \\ -2 & 1 \end{array} \right| - 2 \cdot \left| \begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array} \right| + 1 \cdot \left| \begin{array}{cc} 1 & 2 \\ 1 & -2 \end{array} \right| = \\
= 0 - 4 - 4 = -8 \neq 0 \Rightarrow A^{-1} \text{ exists}.
\end{array} det A = ∣ ∣ 1 1 1 2 2 − 2 1 − 1 1 ∣ ∣ = ∣ expand along the first row ∣ = 1 ⋅ ∣ ∣ 2 − 2 − 1 1 ∣ ∣ − 2 ⋅ ∣ ∣ 1 1 − 1 1 ∣ ∣ + 1 ⋅ ∣ ∣ 1 1 2 − 2 ∣ ∣ = = 0 − 4 − 4 = − 8 = 0 ⇒ A − 1 exists .
Find the minors M i j M_{ij} M ij C i j C_{ij} C ij A A A
M 11 = ∣ 2 − 1 − 2 1 ∣ = 2 ⋅ 1 − ( − 2 ) ⋅ ( − 1 ) = 0 , M_{11} = \left| \begin{array}{cc} 2 & -1 \\ -2 & 1 \end{array} \right| = 2 \cdot 1 - (-2) \cdot (-1) = 0, M 11 = ∣ ∣ 2 − 2 − 1 1 ∣ ∣ = 2 ⋅ 1 − ( − 2 ) ⋅ ( − 1 ) = 0 , M 12 = ∣ 1 − 1 1 1 ∣ = 1 ⋅ 1 − 1 ⋅ ( − 1 ) = 2 , M_{12} = \left| \begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array} \right| = 1 \cdot 1 - 1 \cdot (-1) = 2, M 12 = ∣ ∣ 1 1 − 1 1 ∣ ∣ = 1 ⋅ 1 − 1 ⋅ ( − 1 ) = 2 , M 13 = ∣ 1 2 1 − 2 ∣ = 1 ⋅ ( − 2 ) − 1 ⋅ 2 = − 4 , M_{13} = \left| \begin{array}{cc} 1 & 2 \\ 1 & -2 \end{array} \right| = 1 \cdot (-2) - 1 \cdot 2 = -4, M 13 = ∣ ∣ 1 1 2 − 2 ∣ ∣ = 1 ⋅ ( − 2 ) − 1 ⋅ 2 = − 4 , M 21 = ∣ 2 1 − 2 1 ∣ = 2 ⋅ 1 − ( − 2 ) ⋅ 1 = 4 , M_{21} = \left| \begin{array}{cc} 2 & 1 \\ -2 & 1 \end{array} \right| = 2 \cdot 1 - (-2) \cdot 1 = 4, M 21 = ∣ ∣ 2 − 2 1 1 ∣ ∣ = 2 ⋅ 1 − ( − 2 ) ⋅ 1 = 4 , M 22 = ∣ 1 1 1 1 ∣ = 1 ⋅ 1 − 1 ⋅ 1 = 0 , M_{22} = \left| \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right| = 1 \cdot 1 - 1 \cdot 1 = 0, M 22 = ∣ ∣ 1 1 1 1 ∣ ∣ = 1 ⋅ 1 − 1 ⋅ 1 = 0 , M 23 = ∣ 1 2 1 − 2 ∣ = 1 ⋅ ( − 2 ) − 1 ⋅ 2 = − 4 , M_{23} = \left| \begin{array}{cc} 1 & 2 \\ 1 & -2 \end{array} \right| = 1 \cdot (-2) - 1 \cdot 2 = -4, M 23 = ∣ ∣ 1 1 2 − 2 ∣ ∣ = 1 ⋅ ( − 2 ) − 1 ⋅ 2 = − 4 , M 31 = ∣ 2 1 2 − 1 ∣ = 2 ⋅ ( − 1 ) − 2 ⋅ 1 = − 4 , M_{31} = \left| \begin{array}{cc} 2 & 1 \\ 2 & -1 \end{array} \right| = 2 \cdot (-1) - 2 \cdot 1 = -4, M 31 = ∣ ∣ 2 2 1 − 1 ∣ ∣ = 2 ⋅ ( − 1 ) − 2 ⋅ 1 = − 4 , M 32 = ∣ 1 1 1 − 1 ∣ = 1 ⋅ ( − 1 ) − 1 ⋅ 1 = − 2 , M_{32} = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = 1 \cdot (-1) - 1 \cdot 1 = -2, M 32 = ∣ ∣ 1 1 1 − 1 ∣ ∣ = 1 ⋅ ( − 1 ) − 1 ⋅ 1 = − 2 , M 33 = ∣ 1 2 1 2 ∣ = 1 ⋅ 2 − 1 ⋅ 2 = 0 , M_{33} = \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = 1 \cdot 2 - 1 \cdot 2 = 0, M 33 = ∣ ∣ 1 1 2 2 ∣ ∣ = 1 ⋅ 2 − 1 ⋅ 2 = 0 , C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = 0 , C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = − 2 , C_{11} = (-1)^{1+1} M_{11} = M_{11} = 0, \quad C_{12} = (-1)^{1+2} M_{12} = -M_{12} = -2, C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = 0 , C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = − 2 , C 13 = ( − 1 ) 1 + 3 M 13 = M 13 = − 4 , C 21 = ( − 1 ) 2 + 1 M 21 = − M 21 = − 4 , C_{13} = (-1)^{1+3} M_{13} = M_{13} = -4, \quad C_{21} = (-1)^{2+1} M_{21} = -M_{21} = -4, C 13 = ( − 1 ) 1 + 3 M 13 = M 13 = − 4 , C 21 = ( − 1 ) 2 + 1 M 21 = − M 21 = − 4 , C 22 = ( − 1 ) 2 + 2 M 22 = M 22 = 0 , C 23 = ( − 1 ) 2 + 3 M 23 = − M 23 = 4 , C_{22} = (-1)^{2+2} M_{22} = M_{22} = 0, \quad C_{23} = (-1)^{2+3} M_{23} = -M_{23} = 4, C 22 = ( − 1 ) 2 + 2 M 22 = M 22 = 0 , C 23 = ( − 1 ) 2 + 3 M 23 = − M 23 = 4 , C 31 = ( − 1 ) 3 + 1 M 31 = M 31 = − 4 , C 32 = ( − 1 ) 3 + 2 M 32 = − M 32 = 2 , C_{31} = (-1)^{3+1} M_{31} = M_{31} = -4, \quad C_{32} = (-1)^{3+2} M_{32} = -M_{32} = 2, C 31 = ( − 1 ) 3 + 1 M 31 = M 31 = − 4 , C 32 = ( − 1 ) 3 + 2 M 32 = − M 32 = 2 , C 33 = ( − 1 ) 3 + 3 M 33 = M 33 = 0. C_{33} = (-1)^{3+3} M_{33} = M_{33} = 0. C 33 = ( − 1 ) 3 + 3 M 33 = M 33 = 0.
Write the matrix of cofactors:
C = [ C 11 C 12 C 13 C 21 C 22 C 23 C 31 C 32 C 33 ] = [ 0 − 2 − 4 − 4 0 4 − 4 2 0 ] . C = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix} = \begin{bmatrix} 0 & -2 & -4 \\ -4 & 0 & 4 \\ -4 & 2 & 0 \end{bmatrix}. C = ⎣ ⎡ C 11 C 21 C 31 C 12 C 22 C 32 C 13 C 23 C 33 ⎦ ⎤ = ⎣ ⎡ 0 − 4 − 4 − 2 0 2 − 4 4 0 ⎦ ⎤ .
The transposed matrix of cofactors:
C T = [ C 11 C 21 C 31 C 12 C 22 C 32 C 13 C 23 C 33 ] = [ 0 − 4 − 4 − 2 0 2 − 4 4 0 ] . C^T = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix} = \begin{bmatrix} 0 & -4 & -4 \\ -2 & 0 & 2 \\ -4 & 4 & 0 \end{bmatrix}. C T = ⎣ ⎡ C 11 C 12 C 13 C 21 C 22 C 23 C 31 C 32 C 33 ⎦ ⎤ = ⎣ ⎡ 0 − 2 − 4 − 4 0 4 − 4 2 0 ⎦ ⎤ .
Find the inverse matrix:
A − 1 = 1 det A C T = − 1 8 [ 0 − 4 − 4 − 2 0 2 − 4 4 0 ] = [ 0 − 8 − 4 − 8 − 4 − 8 − 2 − 8 0 − 8 2 − 8 − 4 − 8 4 − 8 0 − 8 ] = [ 0 1 2 1 2 1 4 0 − 1 4 1 2 − 1 2 0 ] . A^{-1} = \frac{1}{\det A} C^T = -\frac{1}{8} \begin{bmatrix} 0 & -4 & -4 \\ -2 & 0 & 2 \\ -4 & 4 & 0 \end{bmatrix} = \begin{bmatrix} \frac{0}{-8} & \frac{-4}{-8} & \frac{-4}{-8} \\ \frac{-2}{-8} & \frac{0}{-8} & \frac{2}{-8} \\ \frac{-4}{-8} & \frac{4}{-8} & \frac{0}{-8} \end{bmatrix} = \begin{bmatrix} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & 0 & -\frac{1}{4} \\ \frac{1}{2} & -\frac{1}{2} & 0 \end{bmatrix}. A − 1 = det A 1 C T = − 8 1 ⎣ ⎡ 0 − 2 − 4 − 4 0 4 − 4 2 0 ⎦ ⎤ = ⎣ ⎡ − 8 0 − 8 − 2 − 8 − 4 − 8 − 4 − 8 0 − 8 4 − 8 − 4 − 8 2 − 8 0 ⎦ ⎤ = ⎣ ⎡ 0 4 1 2 1 2 1 0 − 2 1 2 1 − 4 1 0 ⎦ ⎤ .
Then the solution of the system (6) is
X = [ x y ] = A − 1 B = [ 0 1 2 1 2 1 4 0 − 1 4 1 2 − 1 2 0 ] ⋅ [ 1 3 − 3 ] = [ 0 ⋅ 1 + 1 2 ⋅ 3 + 1 2 ⋅ ( − 3 ) 1 4 ⋅ 1 + 0 ⋅ 3 − 1 4 ⋅ ( − 3 ) 1 2 ⋅ 1 − 1 2 ⋅ 3 + 0 ⋅ ( − 3 ) ] = = [ 0 + 3 2 − 3 2 1 4 + 0 + 3 4 1 2 − 3 2 − 0 ] = [ 0 1 − 1 ] X = \begin{bmatrix} x \\ y \end{bmatrix} = A^{-1} B = \begin{bmatrix} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & 0 & -\frac{1}{4} \\ \frac{1}{2} & -\frac{1}{2} & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 3 \\ -3 \end{bmatrix} = \begin{bmatrix} 0 \cdot 1 + \frac{1}{2} \cdot 3 + \frac{1}{2} \cdot (-3) \\ \frac{1}{4} \cdot 1 + 0 \cdot 3 - \frac{1}{4} \cdot (-3) \\ \frac{1}{2} \cdot 1 - \frac{1}{2} \cdot 3 + 0 \cdot (-3) \end{bmatrix} = \\
= \begin{bmatrix} 0 + \frac{3}{2} - \frac{3}{2} \\ \frac{1}{4} + 0 + \frac{3}{4} \\ \frac{1}{2} - \frac{3}{2} - 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} X = [ x y ] = A − 1 B = ⎣ ⎡ 0 4 1 2 1 2 1 0 − 2 1 2 1 − 4 1 0 ⎦ ⎤ ⋅ ⎣ ⎡ 1 3 − 3 ⎦ ⎤ = ⎣ ⎡ 0 ⋅ 1 + 2 1 ⋅ 3 + 2 1 ⋅ ( − 3 ) 4 1 ⋅ 1 + 0 ⋅ 3 − 4 1 ⋅ ( − 3 ) 2 1 ⋅ 1 − 2 1 ⋅ 3 + 0 ⋅ ( − 3 ) ⎦ ⎤ = = ⎣ ⎡ 0 + 2 3 − 2 3 4 1 + 0 + 4 3 2 1 − 2 3 − 0 ⎦ ⎤ = ⎣ ⎡ 0 1 − 1 ⎦ ⎤
Answer: x = 0 , y = 1 , z = − 1 x = 0, y = 1, z = -1 x = 0 , y = 1 , z = − 1
E. Let A, B and C be:
A = [ 1 2 − 3 0 1 2 − 1 2 0 ] ; B = [ − 1 2 0 0 1 2 1 2 − 3 ] ; C = [ 0 4 − 3 0 1 2 − 1 2 0 ] . A = \begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 2 \\ -1 & 2 & 0 \end{bmatrix}; \quad B = \begin{bmatrix} -1 & 2 & 0 \\ 0 & 1 & 2 \\ 1 & 2 & -3 \end{bmatrix}; \quad C = \begin{bmatrix} 0 & 4 & -3 \\ 0 & 1 & 2 \\ -1 & 2 & 0 \end{bmatrix}. A = ⎣ ⎡ 1 0 − 1 2 1 2 − 3 2 0 ⎦ ⎤ ; B = ⎣ ⎡ − 1 0 1 2 1 2 0 2 − 3 ⎦ ⎤ ; C = ⎣ ⎡ 0 0 − 1 4 1 2 − 3 2 0 ⎦ ⎤ .
1. Find an elementary matrix E such that EA = B
2. Find an elementary matrix E such that EC = A
3. Find an elementary matrix E such that EB = A
Remark
Type I
E 1 = [ 0 1 0 1 0 0 0 0 1 ] ; E 1 A = [ 0 1 0 1 0 0 0 0 1 ] × [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ] = [ a 21 a 22 a 23 a 11 a 12 a 13 a 31 a 32 a 33 ] . E _ {1} = \left[ \begin{array}{c c c} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right]; E _ {1} A = \left[ \begin{array}{c c c} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{c c c} a _ {1 1} & a _ {1 2} & a _ {1 3} \\ a _ {2 1} & a _ {2 2} & a _ {2 3} \\ a _ {3 1} & a _ {3 2} & a _ {3 3} \end{array} \right] = \left[ \begin{array}{c c c} a _ {2 1} & a _ {2 2} & a _ {2 3} \\ a _ {1 1} & a _ {1 2} & a _ {1 3} \\ a _ {3 1} & a _ {3 2} & a _ {3 3} \end{array} \right]. E 1 = ⎣ ⎡ 0 1 0 1 0 0 0 0 1 ⎦ ⎤ ; E 1 A = ⎣ ⎡ 0 1 0 1 0 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 ⎦ ⎤ = ⎣ ⎡ a 21 a 11 a 31 a 22 a 12 a 32 a 23 a 13 a 33 ⎦ ⎤ .
Type II
E 1 = [ 1 0 0 0 1 0 0 0 k ] ; E 1 A = [ 1 0 0 0 1 0 0 0 k ] × [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ] = [ a 11 a 12 a 13 a 21 a 22 a 23 k a 31 k a 32 k a 33 ] . E _ {1} = \left[ \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & k \end{array} \right]; E _ {1} A = \left[ \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & k \end{array} \right] \times \left[ \begin{array}{c c c} a _ {1 1} & a _ {1 2} & a _ {1 3} \\ a _ {2 1} & a _ {2 2} & a _ {2 3} \\ a _ {3 1} & a _ {3 2} & a _ {3 3} \end{array} \right] = \left[ \begin{array}{c c c} a _ {1 1} & a _ {1 2} & a _ {1 3} \\ a _ {2 1} & a _ {2 2} & a _ {2 3} \\ k a _ {3 1} & k a _ {3 2} & k a _ {3 3} \end{array} \right]. E 1 = ⎣ ⎡ 1 0 0 0 1 0 0 0 k ⎦ ⎤ ; E 1 A = ⎣ ⎡ 1 0 0 0 1 0 0 0 k ⎦ ⎤ × ⎣ ⎡ a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 ⎦ ⎤ = ⎣ ⎡ a 11 a 21 k a 31 a 12 a 22 k a 32 a 13 a 23 k a 33 ⎦ ⎤ .
Type III
E 1 = [ 1 0 m 0 1 0 0 0 1 ] ; E 1 A = [ 1 0 m 0 1 0 0 0 1 ] × [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ] = = [ a 11 + m a 31 a 12 + m a 32 a 13 + m a 33 a 21 a 22 a 23 a 31 a 32 a 33 ] \begin{array}{l} E _ {1} = \left[ \begin{array}{c c c} 1 & 0 & m \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]; E _ {1} A = \left[ \begin{array}{c c c} 1 & 0 & m \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{c c c} a _ {1 1} & a _ {1 2} & a _ {1 3} \\ a _ {2 1} & a _ {2 2} & a _ {2 3} \\ a _ {3 1} & a _ {3 2} & a _ {3 3} \end{array} \right] = \\ = \left[ \begin{array}{c c c} a _ {1 1} + m a _ {3 1} & a _ {1 2} + m a _ {3 2} & a _ {1 3} + m a _ {3 3} \\ a _ {2 1} & a _ {2 2} & a _ {2 3} \\ a _ {3 1} & a _ {3 2} & a _ {3 3} \end{array} \right] \\ \end{array} E 1 = ⎣ ⎡ 1 0 0 0 1 0 m 0 1 ⎦ ⎤ ; E 1 A = ⎣ ⎡ 1 0 0 0 1 0 m 0 1 ⎦ ⎤ × ⎣ ⎡ a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 ⎦ ⎤ = = ⎣ ⎡ a 11 + m a 31 a 21 a 31 a 12 + m a 32 a 22 a 32 a 13 + m a 33 a 23 a 33 ⎦ ⎤ Question
1. Let A A A B B B
A = [ 1 2 − 3 0 1 2 − 1 2 0 ] ; B = [ − 1 2 0 0 1 2 1 2 − 3 ] . A = \left[ \begin{array}{c c c} 1 & 2 & - 3 \\ 0 & 1 & 2 \\ - 1 & 2 & 0 \end{array} \right]; B = \left[ \begin{array}{c c c} - 1 & 2 & 0 \\ 0 & 1 & 2 \\ 1 & 2 & - 3 \end{array} \right]. A = ⎣ ⎡ 1 0 − 1 2 1 2 − 3 2 0 ⎦ ⎤ ; B = ⎣ ⎡ − 1 0 1 2 1 2 0 2 − 3 ⎦ ⎤ .
Find an elementary matrix E E E E A = B EA = B E A = B
Solution
Type I: row1 and row3 of A A A B B B
E = [ 0 0 1 0 1 0 1 0 0 ] E A = [ 0 0 1 0 1 0 1 0 0 ] × [ 1 2 − 3 0 1 2 − 1 2 0 ] = [ − 1 2 0 0 1 2 1 2 − 3 ] = B . \begin{array}{l} E = \left[ \begin{array}{c c c} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array} \right] \\ E A = \left[ \begin{array}{c c c} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array} \right] \times \left[ \begin{array}{c c c} 1 & 2 & - 3 \\ 0 & 1 & 2 \\ - 1 & 2 & 0 \end{array} \right] = \left[ \begin{array}{c c c} - 1 & 2 & 0 \\ 0 & 1 & 2 \\ 1 & 2 & - 3 \end{array} \right] = B. \\ \end{array} E = ⎣ ⎡ 0 0 1 0 1 0 1 0 0 ⎦ ⎤ E A = ⎣ ⎡ 0 0 1 0 1 0 1 0 0 ⎦ ⎤ × ⎣ ⎡ 1 0 − 1 2 1 2 − 3 2 0 ⎦ ⎤ = ⎣ ⎡ − 1 0 1 2 1 2 0 2 − 3 ⎦ ⎤ = B .
Thus, E A = B EA = B E A = B
Answer: E = [ 0 0 1 0 1 0 1 0 0 ] E = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} E = ⎣ ⎡ 0 0 1 0 1 0 1 0 0 ⎦ ⎤
Question
2. Let A A A C C C
A = [ 1 2 − 3 0 1 2 − 1 2 0 ] ; C = [ 0 4 − 3 0 1 2 − 1 2 0 ] . A = \left[ \begin{array}{c c c} 1 & 2 & - 3 \\ 0 & 1 & 2 \\ - 1 & 2 & 0 \end{array} \right]; C = \left[ \begin{array}{c c c} 0 & 4 & - 3 \\ 0 & 1 & 2 \\ - 1 & 2 & 0 \end{array} \right]. A = ⎣ ⎡ 1 0 − 1 2 1 2 − 3 2 0 ⎦ ⎤ ; C = ⎣ ⎡ 0 0 − 1 4 1 2 − 3 2 0 ⎦ ⎤ .
Find an elementary matrix E E E E C = A EC = A EC = A
Solution
Type III: m = − 1 m = -1 m = − 1 C C C C C C A A A
E = [ 1 0 − 1 0 1 0 0 0 1 ] E = \left[ \begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] E = ⎣ ⎡ 1 0 0 0 1 0 − 1 0 1 ⎦ ⎤ E C = [ 1 0 − 1 0 1 0 0 0 1 ] × [ 0 4 − 3 0 1 2 − 1 2 0 ] = [ 1 2 − 3 0 1 2 − 1 2 0 ] = A . EC = \left[ \begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 0 & 4 & -3 \\ 0 & 1 & 2 \\ -1 & 2 & 0 \end{array} \right] = \left[ \begin{array}{ccc} 1 & 2 & -3 \\ 0 & 1 & 2 \\ -1 & 2 & 0 \end{array} \right] = A. EC = ⎣ ⎡ 1 0 0 0 1 0 − 1 0 1 ⎦ ⎤ × ⎣ ⎡ 0 0 − 1 4 1 2 − 3 2 0 ⎦ ⎤ = ⎣ ⎡ 1 0 − 1 2 1 2 − 3 2 0 ⎦ ⎤ = A .
Thus, E C = A EC = A EC = A
Answer: E = [ 1 0 − 1 0 1 0 0 0 1 ] E = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} E = ⎣ ⎡ 1 0 0 0 1 0 − 1 0 1 ⎦ ⎤
Question
3. Let A A A B B B
A = [ 1 2 − 3 0 1 2 − 1 2 0 ] ; B = [ − 1 2 0 0 1 2 1 2 − 3 ] . A = \left[ \begin{array}{ccc} 1 & 2 & -3 \\ 0 & 1 & 2 \\ -1 & 2 & 0 \end{array} \right]; \quad B = \left[ \begin{array}{ccc} -1 & 2 & 0 \\ 0 & 1 & 2 \\ 1 & 2 & -3 \end{array} \right]. A = ⎣ ⎡ 1 0 − 1 2 1 2 − 3 2 0 ⎦ ⎤ ; B = ⎣ ⎡ − 1 0 1 2 1 2 0 2 − 3 ⎦ ⎤ .
Find an elementary matrix E E E E B = A EB = A EB = A
Solution
Type I: row1 and row3 of B B B A A A
E = [ 0 0 1 0 1 0 1 0 0 ] E = \left[ \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array} \right] E = ⎣ ⎡ 0 0 1 0 1 0 1 0 0 ⎦ ⎤ E A = [ 0 0 1 0 1 0 1 0 0 ] × [ − 1 2 0 0 1 2 1 2 − 3 ] = [ 1 2 − 3 0 1 2 − 1 2 0 ] = A . EA = \left[ \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array} \right] \times \left[ \begin{array}{ccc} -1 & 2 & 0 \\ 0 & 1 & 2 \\ 1 & 2 & -3 \end{array} \right] = \left[ \begin{array}{ccc} 1 & 2 & -3 \\ 0 & 1 & 2 \\ -1 & 2 & 0 \end{array} \right] = A. E A = ⎣ ⎡ 0 0 1 0 1 0 1 0 0 ⎦ ⎤ × ⎣ ⎡ − 1 0 1 2 1 2 0 2 − 3 ⎦ ⎤ = ⎣ ⎡ 1 0 − 1 2 1 2 − 3 2 0 ⎦ ⎤ = A .
Thus, E B = A EB = A EB = A
Answer: E = [ 0 0 1 0 1 0 1 0 0 ] E = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} E = ⎣ ⎡ 0 0 1 0 1 0 1 0 0 ⎦ ⎤
F. Find the LU Factorization of the matrix.
Question
4.
A = [ 1 0 − 2 1 ] A = \left[ \begin{array}{cc} 1 & 0 \\ -2 & 1 \end{array} \right] A = [ 1 − 2 0 1 ] Solution
Add twice row1 to row2 in the Gauss elimination method:
[ 1 0 2 1 ] × [ 1 0 − 2 1 ] = [ 1 0 0 1 ] ( ∗ ) \left[ \begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array} \right] \times \left[ \begin{array}{cc} 1 & 0 \\ -2 & 1 \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] (*) [ 1 2 0 1 ] × [ 1 − 2 0 1 ] = [ 1 0 0 1 ] ( ∗ )
According to ( ∗ ) (^{*}) ( ∗ ) [ 1 0 2 1 ] \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} [ 1 2 0 1 ] [ 1 0 − 2 1 ] \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} [ 1 − 2 0 1 ]
Let's multiply both sides of the equation ( ∗ ) (^{*}) ( ∗ ) [ 1 0 − 2 1 ] \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} [ 1 − 2 0 1 ]
[ 1 0 − 2 1 ] × [ 1 0 2 1 ] × [ 1 0 − 2 1 ] = [ 1 0 − 2 1 ] × [ 1 0 0 1 ] \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} [ 1 − 2 0 1 ] × [ 1 2 0 1 ] × [ 1 − 2 0 1 ] = [ 1 − 2 0 1 ] × [ 1 0 0 1 ]
Simplify the previous formula and obtain the LU Factorization of the matrix A = [ 1 0 − 2 1 ] A = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} A = [ 1 − 2 0 1 ]
A = [ 1 0 − 2 1 ] = [ 1 0 − 2 1 ] × [ 1 0 0 1 ] = L U , where L = [ 1 0 − 2 1 ] , U = [ 1 0 0 1 ] . A = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = LU, \text{ where } L = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}, U = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. A = [ 1 − 2 0 1 ] = [ 1 − 2 0 1 ] × [ 1 0 0 1 ] = LU , where L = [ 1 − 2 0 1 ] , U = [ 1 0 0 1 ] .
Answer: [ 1 0 − 2 1 ] = [ 1 0 − 2 1 ] × [ 1 0 0 1 ] \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} [ 1 − 2 0 1 ] = [ 1 − 2 0 1 ] × [ 1 0 0 1 ]
Question
5.
B = [ − 2 1 − 6 4 ] B = \begin{bmatrix} -2 & 1 \\ -6 & 4 \end{bmatrix} B = [ − 2 − 6 1 4 ] Solution
Add ( − 3 × row1 ) (-3 \times \text{row1}) ( − 3 × row1 ) row2 \text{row2} row2
[ 1 0 − 3 1 ] × [ − 2 1 − 6 4 ] = [ − 2 1 0 1 ] ( ∗ ∗ ) \begin{bmatrix} 1 & 0 \\ -3 & 1 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \\ -6 & 4 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 0 & 1 \end{bmatrix} \quad (**) [ 1 − 3 0 1 ] × [ − 2 − 6 1 4 ] = [ − 2 0 1 1 ] ( ∗ ∗ )
Besides, the inverse of [ 1 0 − 3 1 ] \begin{bmatrix} 1 & 0 \\ -3 & 1 \end{bmatrix} [ 1 − 3 0 1 ] [ 1 0 3 1 ] \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} [ 1 3 0 1 ]
Let's multiply both sides of the equation ( ∗ ∗ ) (^{**}) ( ∗∗ ) [ 1 0 3 1 ] \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} [ 1 3 0 1 ]
[ 1 0 3 1 ] × [ 1 0 − 3 1 ] × [ − 2 1 − 6 4 ] = [ 1 0 3 1 ] × [ − 2 1 0 1 ] \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ -3 & 1 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \\ -6 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \\ 0 & 1 \end{bmatrix} [ 1 3 0 1 ] × [ 1 − 3 0 1 ] × [ − 2 − 6 1 4 ] = [ 1 3 0 1 ] × [ − 2 0 1 1 ]
Simplify the previous formula and obtain the LU Factorization of the matrix B = [ − 2 1 − 6 4 ] B = \begin{bmatrix} -2 & 1 \\ -6 & 4 \end{bmatrix} B = [ − 2 − 6 1 4 ]
B = [ − 2 1 − 6 4 ] = [ 1 0 3 1 ] × [ − 2 1 0 1 ] = L U , where L = [ 1 0 3 1 ] , U = [ − 2 1 0 1 ] . B = \begin{bmatrix} -2 & 1 \\ -6 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \\ 0 & 1 \end{bmatrix} = LU, \text{ where } L = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}, U = \begin{bmatrix} -2 & 1 \\ 0 & 1 \end{bmatrix}. B = [ − 2 − 6 1 4 ] = [ 1 3 0 1 ] × [ − 2 0 1 1 ] = LU , where L = [ 1 3 0 1 ] , U = [ − 2 0 1 1 ] .
Answer: [ − 2 1 − 6 4 ] = [ 1 0 3 1 ] × [ − 2 1 0 1 ] \begin{bmatrix} -2 & 1 \\ -6 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \\ 0 & 1 \end{bmatrix} [ − 2 − 6 1 4 ] = [ 1 3 0 1 ] × [ − 2 0 1 1 ]
Question
6.
C = [ 3 0 1 6 1 1 − 3 1 0 ] C = \begin{bmatrix} 3 & 0 & 1 \\ 6 & 1 & 1 \\ -3 & 1 & 0 \end{bmatrix} C = ⎣ ⎡ 3 6 − 3 0 1 1 1 1 0 ⎦ ⎤ Solution
Add ( − 2 × row1 ) (-2 \times \text{row1}) ( − 2 × row1 ) row2 \text{row2} row2
[ 1 0 0 − 2 1 0 0 0 1 ] × [ 3 0 1 6 1 1 − 3 1 0 ] = [ 3 0 1 0 1 − 1 − 3 1 0 ] \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 3 & 0 & 1 \\ 6 & 1 & 1 \\ -3 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 1 \\ 0 & 1 & -1 \\ -3 & 1 & 0 \end{bmatrix} ⎣ ⎡ 1 − 2 0 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 6 − 3 0 1 1 1 1 0 ⎦ ⎤ = ⎣ ⎡ 3 0 − 3 0 1 1 1 − 1 0 ⎦ ⎤
Add ( 1 × row1 ) (1 \times \text{row1}) ( 1 × row1 ) row3 \text{row3} row3
[ 1 0 0 0 1 0 1 0 1 ] × [ 1 0 0 − 2 1 0 0 0 1 ] × [ 3 0 1 6 1 1 − 3 1 0 ] = [ 3 0 1 0 1 − 1 0 1 1 ] \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 6 & 1 & 1 \\ -3 & 1 & 0 \end{array} \right] = \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 1 & 1 \end{array} \right] ⎣ ⎡ 1 0 1 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 1 − 2 0 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 6 − 3 0 1 1 1 1 0 ⎦ ⎤ = ⎣ ⎡ 3 0 0 0 1 1 1 − 1 1 ⎦ ⎤
Add ( − 1 × r o w 2 ) (-1 \times row2) ( − 1 × ro w 2 )
[ 1 0 0 0 1 0 0 − 1 1 ] × [ 1 0 0 0 1 0 1 0 1 ] × [ 1 0 0 − 2 1 0 0 0 1 ] × [ 3 0 1 6 1 1 − 3 1 0 ] = [ 3 0 1 0 1 − 1 0 0 2 ] ( ∗ ∗ ∗ ) \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 6 & 1 & 1 \\ -3 & 1 & 0 \end{array} \right] = \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{array} \right] \quad (***) ⎣ ⎡ 1 0 0 0 1 − 1 0 0 1 ⎦ ⎤ × ⎣ ⎡ 1 0 1 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 1 − 2 0 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 6 − 3 0 1 1 1 1 0 ⎦ ⎤ = ⎣ ⎡ 3 0 0 0 1 0 1 − 1 2 ⎦ ⎤ ( ∗ ∗ ∗ )
The inverse of [ 1 0 0 0 1 0 0 − 1 1 ] \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} ⎣ ⎡ 1 0 0 0 1 − 1 0 0 1 ⎦ ⎤ [ 1 0 0 0 1 0 0 1 1 ] \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix} ⎣ ⎡ 1 0 0 0 1 1 0 0 1 ⎦ ⎤
Let's multiply both sides of the equation (***) by this inverse (that is, by [ 1 0 0 0 1 0 0 1 1 ] \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix} ⎣ ⎡ 1 0 0 0 1 1 0 0 1 ⎦ ⎤
[ 1 0 0 0 1 0 0 1 1 ] × [ 1 0 0 0 1 0 0 − 1 1 ] × [ 1 0 0 0 1 0 1 0 1 ] × [ 1 0 0 − 2 1 0 0 0 1 ] × [ 3 0 1 6 1 1 − 3 1 0 ] = = [ 1 0 0 0 1 0 0 1 1 ] × [ 3 0 1 0 1 − 1 0 0 2 ] . \begin{array}{l}
\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 6 & 1 & 1 \\ -3 & 1 & 0 \end{array} \right] = \\
= \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{array} \right].
\end{array} ⎣ ⎡ 1 0 0 0 1 1 0 0 1 ⎦ ⎤ × ⎣ ⎡ 1 0 0 0 1 − 1 0 0 1 ⎦ ⎤ × ⎣ ⎡ 1 0 1 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 1 − 2 0 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 6 − 3 0 1 1 1 1 0 ⎦ ⎤ = = ⎣ ⎡ 1 0 0 0 1 1 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 0 0 0 1 0 1 − 1 2 ⎦ ⎤ .
Simplify:
[ 1 0 0 0 1 0 1 0 1 ] × [ 1 0 0 − 2 1 0 0 0 1 ] × [ 3 0 1 6 1 1 − 3 1 0 ] = [ 1 0 0 0 1 0 0 1 1 ] × [ 3 0 1 0 1 − 1 0 0 2 ] ( ∗ ∗ ∗ ) \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 6 & 1 & 1 \\ -3 & 1 & 0 \end{array} \right] = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{array} \right] \quad (***) ⎣ ⎡ 1 0 1 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 1 − 2 0 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 6 − 3 0 1 1 1 1 0 ⎦ ⎤ = ⎣ ⎡ 1 0 0 0 1 1 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 0 0 0 1 0 1 − 1 2 ⎦ ⎤ ( ∗ ∗ ∗ )
The inverse of [ 1 0 0 0 1 0 1 0 1 ] \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} ⎣ ⎡ 1 0 1 0 1 0 0 0 1 ⎦ ⎤ [ 1 0 0 0 1 0 − 1 0 1 ] \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} ⎣ ⎡ 1 0 − 1 0 1 0 0 0 1 ⎦ ⎤
Let's multiply both sides of the equation (***) by this inverse (that is, by
[ 1 0 0 0 1 0 − 1 0 1 ] ) : \begin{array}{l}
\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{array} \right] \quad):
\end{array} ⎣ ⎡ 1 0 − 1 0 1 0 0 0 1 ⎦ ⎤ ) : [ 1 0 0 0 1 0 − 1 0 1 ] × [ 1 0 0 0 1 0 1 0 1 ] × [ 1 0 0 − 2 1 0 0 0 1 ] × [ 3 0 1 6 1 1 − 3 1 0 ] = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 6 & 1 & 1 \\ -3 & 1 & 0 \end{array} \right] = ⎣ ⎡ 1 0 − 1 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 1 0 1 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 1 − 2 0 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 6 − 3 0 1 1 1 1 0 ⎦ ⎤ = = [ 1 0 0 0 1 0 − 1 0 1 ] × [ 1 0 0 0 1 0 0 1 1 ] × [ 3 0 1 0 1 − 1 0 0 2 ] = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{array} \right] = ⎣ ⎡ 1 0 − 1 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 1 0 0 0 1 1 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 0 0 0 1 0 1 − 1 2 ⎦ ⎤
Simplify:
[ 1 0 0 − 2 1 0 0 0 1 ] × [ 3 0 1 6 1 1 − 3 1 0 ] = [ 1 0 0 0 1 0 − 1 1 1 ] × [ 3 0 1 0 1 − 1 0 0 2 ] ( ∗ ∗ ∗ ) \left[ \begin{array}{ccc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 6 & 1 & 1 \\ -3 & 1 & 0 \end{array} \right] = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 1 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{array} \right] \quad (***) ⎣ ⎡ 1 − 2 0 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 6 − 3 0 1 1 1 1 0 ⎦ ⎤ = ⎣ ⎡ 1 0 − 1 0 1 1 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 0 0 0 1 0 1 − 1 2 ⎦ ⎤ ( ∗ ∗ ∗ )
The inverse of [ 1 0 0 − 2 1 0 0 0 1 ] \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} ⎣ ⎡ 1 − 2 0 0 1 0 0 0 1 ⎦ ⎤ [ 1 0 0 2 1 0 0 0 1 ] \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} ⎣ ⎡ 1 2 0 0 1 0 0 0 1 ⎦ ⎤
Let's multiply both sides of the equation (***) by this inverse (that is, by
[ 1 0 0 2 1 0 0 0 1 ] : [ 1 0 0 2 1 0 0 0 1 ] × [ 1 0 0 − 2 1 0 0 0 1 ] × [ 3 0 1 6 1 1 − 3 1 0 ] = = [ 1 0 0 2 1 0 0 0 1 ] × [ 1 0 0 0 1 0 − 1 1 1 ] × [ 3 0 1 0 1 − 1 0 0 2 ] \begin{array}{l}
\left[ \begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]: \\
\left[ \begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{lll} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{lll} 3 & 0 & 1 \\ 6 & 1 & 1 \\ -3 & 1 & 0 \end{array} \right] = \\
= \left[ \begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 1 & 1 \end{array} \right] \times \left[ \begin{array}{lll} 3 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{array} \right]
\end{array} ⎣ ⎡ 1 2 0 0 1 0 0 0 1 ⎦ ⎤ : ⎣ ⎡ 1 2 0 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 1 − 2 0 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 6 − 3 0 1 1 1 1 0 ⎦ ⎤ = = ⎣ ⎡ 1 2 0 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ 1 0 − 1 0 1 1 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 0 0 0 1 0 1 − 1 2 ⎦ ⎤
Simplify the previous formula and obtain the LU Factorization of the matrix
C = [ 3 0 1 6 1 1 − 3 1 0 ] : C = \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 6 & 1 & 1 \\ -3 & 1 & 0 \end{array} \right]: C = ⎣ ⎡ 3 6 − 3 0 1 1 1 1 0 ⎦ ⎤ : C = [ 3 0 1 6 1 1 − 3 1 0 ] = [ 1 0 0 2 1 0 − 1 1 1 ] × [ 3 0 1 0 1 − 1 0 0 2 ] = L U , C = \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 6 & 1 & 1 \\ -3 & 1 & 0 \end{array} \right] = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -1 & 1 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} 3 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{array} \right] = LU, C = ⎣ ⎡ 3 6 − 3 0 1 1 1 1 0 ⎦ ⎤ = ⎣ ⎡ 1 2 − 1 0 1 1 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 0 0 0 1 0 1 − 1 2 ⎦ ⎤ = LU ,
where L = [ 1 0 0 2 1 0 − 1 1 1 ] , U = [ 3 0 1 0 1 − 1 0 0 2 ] . L = \left[ \begin{array}{lll}1 & 0 & 0\\ 2 & 1 & 0\\ -1 & 1 & 1 \end{array} \right], U = \left[ \begin{array}{lll}3 & 0 & 1\\ 0 & 1 & -1\\ 0 & 0 & 2 \end{array} \right]. L = ⎣ ⎡ 1 2 − 1 0 1 1 0 0 1 ⎦ ⎤ , U = ⎣ ⎡ 3 0 0 0 1 0 1 − 1 2 ⎦ ⎤ .
Answer: [ 3 0 1 6 1 1 − 3 1 0 ] = [ 1 0 0 2 1 0 − 1 1 1 ] × [ 3 0 1 0 1 − 1 0 0 2 ] . \left[ \begin{array}{ccc}3 & 0 & 1\\ 6 & 1 & 1\\ -3 & 1 & 0 \end{array} \right] = \left[ \begin{array}{ccc}1 & 0 & 0\\ 2 & 1 & 0\\ -1 & 1 & 1 \end{array} \right]\times \left[ \begin{array}{ccc}3 & 0 & 1\\ 0 & 1 & -1\\ 0 & 0 & 2 \end{array} \right]. ⎣ ⎡ 3 6 − 3 0 1 1 1 1 0 ⎦ ⎤ = ⎣ ⎡ 1 2 − 1 0 1 1 0 0 1 ⎦ ⎤ × ⎣ ⎡ 3 0 0 0 1 0 1 − 1 2 ⎦ ⎤ .
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#340153 on Dec 2023
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