Answer on Question #62621 – Math – Linear Algebra
A. Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination.
Question
1. { − x + y + 2 z = 1 2 x + 3 y + z = − 2 5 x + 4 y + 2 z = 4 \begin{cases} -x + y + 2z = 1 \\ 2x + 3y + z = -2 \\ 5x + 4y + 2z = 4 \end{cases} ⎩ ⎨ ⎧ − x + y + 2 z = 1 2 x + 3 y + z = − 2 5 x + 4 y + 2 z = 4
Solution
This system can be represented by the coefficient matrix: ( − 1 1 2 ∣ 1 2 3 1 ∣ − 2 5 4 2 ∣ 4 ) \begin{pmatrix} -1 & 1 & 2 & | & 1 \\ 2 & 3 & 1 & | & -2 \\ 5 & 4 & 2 & | & 4 \end{pmatrix} ⎝ ⎛ − 1 2 5 1 3 4 2 1 2 ∣ ∣ ∣ 1 − 2 4 ⎠ ⎞
Henceforth R 2 ← R 2 − 2 R 1 R2 \leftarrow R2 - 2R1 R 2 ← R 2 − 2 R 1
( − 1 1 2 ∣ 1 2 3 1 ∣ − 2 5 4 2 ∣ 4 ) → R 1 ← R 1 × ( − 1 ) ( 1 − 1 − 2 ∣ − 1 2 3 1 ∣ − 2 5 4 2 ∣ 4 ) → R 2 ← R 2 − 2 R 1 R 3 ← R 3 − 5 R 1 → ( 1 − 1 − 2 ∣ − 1 0 5 5 ∣ 0 0 9 12 ∣ 9 ) → R 3 ← 1 3 ⋅ R 3 ( 1 − 1 − 2 ∣ − 1 0 1 1 ∣ 0 0 0 1 ∣ 3 ) → R 3 ← R 3 − 3 R 2 → ( 1 − 1 − 2 ∣ − 1 0 1 1 ∣ 0 0 0 1 ∣ 3 ) ⇒ ⇒ { x − y − 2 z = − 1 y + z = 0 z = 3 ⇒ { x = − 1 + y + 2 z y = − z z = 3 ⇒ { x = − 1 + z y = − z z = 3 ⇒ { x = 2 y = − 3 z = 3 \begin{aligned}
&\left( \begin{array}{cccc|cc}
-1 & 1 & 2 & | & 1 \\
2 & 3 & 1 & | & -2 \\
5 & 4 & 2 & | & 4
\end{array} \right) \xrightarrow{R1 \leftarrow R1 \times (-1)} \left( \begin{array}{cccc|cc}
1 & -1 & -2 & | & -1 \\
2 & 3 & 1 & | & -2 \\
5 & 4 & 2 & | & 4
\end{array} \right) \xrightarrow{\frac{R2 \leftarrow R2 - 2R1}{R3 \leftarrow R3 - 5R1}} \\
&\rightarrow \left( \begin{array}{cccc|cc}
1 & -1 & -2 & | & -1 \\
0 & 5 & 5 & | & 0 \\
0 & 9 & 12 & | & 9
\end{array} \right) \xrightarrow{R3 \leftarrow \frac{1}{3} \cdot R3} \left( \begin{array}{cccc|cc}
1 & -1 & -2 & | & -1 \\
0 & 1 & 1 & | & 0 \\
0 & 0 & 1 & | & 3
\end{array} \right) \xrightarrow{R3 \leftarrow R3 - 3R2} \\
&\rightarrow \left( \begin{array}{cccc}
1 & -1 & -2 & | & -1 \\
0 & 1 & 1 & | & 0 \\
0 & 0 & 1 & | & 3
\end{array} \right) \Rightarrow \\
\Rightarrow \left\{ \begin{array}{l}
x - y - 2z = -1 \\
y + z = 0 \\
z = 3
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = -1 + y + 2z \\
y = -z \\
z = 3
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = -1 + z \\
y = -z \\
z = 3
\end{array} \right. \\
\Rightarrow \left\{ \begin{array}{l}
x = 2 \\
y = -3 \\
z = 3
\end{array} \right.
\end{aligned} ⇒ ⎩ ⎨ ⎧ x − y − 2 z = − 1 y + z = 0 z = 3 ⇒ ⎩ ⎨ ⎧ x = − 1 + y + 2 z y = − z z = 3 ⇒ ⎩ ⎨ ⎧ x = − 1 + z y = − z z = 3 ⇒ ⎩ ⎨ ⎧ x = 2 y = − 3 z = 3 ⎝ ⎛ − 1 2 5 1 3 4 2 1 2 ∣ ∣ ∣ 1 − 2 4 ⎠ ⎞ R 1 ← R 1 × ( − 1 ) ⎝ ⎛ 1 2 5 − 1 3 4 − 2 1 2 ∣ ∣ ∣ − 1 − 2 4 ⎠ ⎞ R 3 ← R 3 − 5 R 1 R 2 ← R 2 − 2 R 1 → ⎝ ⎛ 1 0 0 − 1 5 9 − 2 5 12 ∣ ∣ ∣ − 1 0 9 ⎠ ⎞ R 3 ← 3 1 ⋅ R 3 ⎝ ⎛ 1 0 0 − 1 1 0 − 2 1 1 ∣ ∣ ∣ − 1 0 3 ⎠ ⎞ R 3 ← R 3 − 3 R 2 → ⎝ ⎛ 1 0 0 − 1 1 0 − 2 1 1 ∣ ∣ ∣ − 1 0 3 ⎠ ⎞ ⇒
Answer: x = 2 , y = − 3 , z = 3 x = 2, y = -3, z = 3 x = 2 , y = − 3 , z = 3
Question
2. { 2 x + 3 y + z = 10 2 x − 3 y − 3 z = 22 4 x − 2 y + 3 z = − 2 \begin{cases} 2x + 3y + z = 10 \\ 2x - 3y - 3z = 22 \\ 4x - 2y + 3z = -2 \end{cases} ⎩ ⎨ ⎧ 2 x + 3 y + z = 10 2 x − 3 y − 3 z = 22 4 x − 2 y + 3 z = − 2
Solution
This system can be represented by the matrix: ( 2 3 1 ∣ 10 2 − 3 − 3 ∣ 22 4 − 2 3 ∣ − 2 ) \begin{pmatrix} 2 & 3 & 1 & | & 10 \\ 2 & -3 & -3 & | & 22 \\ 4 & -2 & 3 & | & -2 \end{pmatrix} ⎝ ⎛ 2 2 4 3 − 3 − 2 1 − 3 3 ∣ ∣ ∣ 10 22 − 2 ⎠ ⎞
( 2 3 1 ∣ 10 2 − 3 − 3 ∣ 22 4 − 2 3 ∣ − 2 ) → R 2 ← R 2 − R 1 ( 2 3 1 ∣ 10 0 − 6 − 4 ∣ 12 0 − 8 1 ∣ − 22 ) → R 2 ← − 1 2 R 2 → R 3 ← − R 3 → ( 2 3 1 ∣ 10 0 3 2 ∣ − 6 0 8 − 1 ∣ 22 ) → R 3 ← R 3 − 8 3 R 2 ( 2 3 1 ∣ 10 0 3 2 ∣ − 6 0 0 − 19 3 ∣ 38 ) ⇒ ⇒ { 2 x + 3 y + z = 10 3 y + 2 z = − 6 − 19 3 z = 38 ⇒ { x = 10 − 3 y − z 2 y = − 6 − 2 z 3 z = − 6 ⇒ { x = 16 + z 2 y = − 6 − 2 z 3 z = − 6 ⇒ ⇒ { x = 5 y = 2 z = − 6 \begin{array}{l}
\left( \begin{array}{cccccc}
2 & 3 & 1 & | & 10 \\
2 & -3 & -3 & | & 22 \\
4 & -2 & 3 & | & -2
\end{array} \right)
\xrightarrow{R2 \leftarrow R2 - R1} \left( \begin{array}{cccccc}
2 & 3 & 1 & | & 10 \\
0 & -6 & -4 & | & 12 \\
0 & -8 & 1 & | & -22
\end{array} \right)
\xrightarrow{R2 \leftarrow -\frac{1}{2} R2} \xrightarrow{R3 \leftarrow -R3} \\
\rightarrow \left( \begin{array}{cccccc}
2 & 3 & 1 & | & 10 \\
0 & 3 & 2 & | & -6 \\
0 & 8 & -1 & | & 22
\end{array} \right)
\xrightarrow{R3 \leftarrow R3 - \frac{8}{3} R2} \left( \begin{array}{cccccc}
2 & 3 & 1 & | & 10 \\
0 & 3 & 2 & | & -6 \\
0 & 0 & -\frac{19}{3} & | & 38
\end{array} \right)
\Rightarrow \\
\Rightarrow \left\{
\begin{array}{l}
2x + 3y + z = 10 \\
3y + 2z = -6 \\
-\frac{19}{3}z = 38
\end{array}
\right.
\Rightarrow
\left\{
\begin{array}{l}
x = \frac{10 - 3y - z}{2} \\
y = \frac{-6 - 2z}{3} \\
z = -6
\end{array}
\right.
\Rightarrow
\left\{
\begin{array}{l}
x = \frac{16 + z}{2} \\
y = \frac{-6 - 2z}{3} \\
z = -6
\end{array}
\right.
\Rightarrow \\
\Rightarrow \left\{
\begin{array}{l}
x = 5 \\
y = 2 \\
z = -6
\end{array}
\right.
\end{array} ⎝ ⎛ 2 2 4 3 − 3 − 2 1 − 3 3 ∣ ∣ ∣ 10 22 − 2 ⎠ ⎞ R 2 ← R 2 − R 1 ⎝ ⎛ 2 0 0 3 − 6 − 8 1 − 4 1 ∣ ∣ ∣ 10 12 − 22 ⎠ ⎞ R 2 ← − 2 1 R 2 R 3 ← − R 3 → ⎝ ⎛ 2 0 0 3 3 8 1 2 − 1 ∣ ∣ ∣ 10 − 6 22 ⎠ ⎞ R 3 ← R 3 − 3 8 R 2 ⎝ ⎛ 2 0 0 3 3 0 1 2 − 3 19 ∣ ∣ ∣ 10 − 6 38 ⎠ ⎞ ⇒ ⇒ ⎩ ⎨ ⎧ 2 x + 3 y + z = 10 3 y + 2 z = − 6 − 3 19 z = 38 ⇒ ⎩ ⎨ ⎧ x = 2 10 − 3 y − z y = 3 − 6 − 2 z z = − 6 ⇒ ⎩ ⎨ ⎧ x = 2 16 + z y = 3 − 6 − 2 z z = − 6 ⇒ ⇒ ⎩ ⎨ ⎧ x = 5 y = 2 z = − 6
Answer: x = 5 , y = 2 , z = − 6 x = 5, y = 2, z = -6 x = 5 , y = 2 , z = − 6
B. Find the inverse of the matrix.
**Question**
1. [ 2 0 0 3 ] \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} [ 2 0 0 3 ]
**Solution**
[ 2 0 0 3 ] − 1 = 1 det ( [ 2 0 0 3 ] ) [ 3 0 0 2 ] = 1 2 ⋅ 3 − 0 ⋅ 0 [ 3 0 0 2 ] = 1 6 [ 3 0 0 2 ] = [ 3 6 0 6 0 6 2 6 ] = [ 1 2 0 0 1 3 ] . \left[ \begin{array}{cc}
2 & 0 \\
0 & 3
\end{array} \right]^{-1} = \frac{1}{\operatorname*{det}(\left[ \begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array} \right])} \left[ \begin{array}{cc} 3 & 0 \\ 0 & 2 \end{array} \right] = \frac{1}{2 \cdot 3 - 0 \cdot 0} \left[ \begin{array}{cc} 3 & 0 \\ 0 & 2 \end{array} \right] = \frac{1}{6} \left[ \begin{array}{cc} 3 & 0 \\ 0 & 2 \end{array} \right] = \left[ \begin{array}{cc} \frac{3}{6} & \frac{0}{6} \\ \frac{0}{6} & \frac{2}{6} \end{array} \right] = \left[ \begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \end{array} \right]. [ 2 0 0 3 ] − 1 = det ( [ 2 0 0 3 ] ) 1 [ 3 0 0 2 ] = 2 ⋅ 3 − 0 ⋅ 0 1 [ 3 0 0 2 ] = 6 1 [ 3 0 0 2 ] = [ 6 3 6 0 6 0 6 2 ] = [ 2 1 0 0 3 1 ] .
Answer: [ 1 2 0 0 1 3 ] \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \end{bmatrix} [ 2 1 0 0 3 1 ]
**Question**
2. [ − 1 1 3 − 3 ] \begin{bmatrix} -1 & 1 \\ 3 & -3 \end{bmatrix} [ − 1 3 1 − 3 ]
**Solution**
det ( [ − 1 1 3 − 3 ] ) = ( − 1 ) ⋅ ( − 3 ) − 3 ⋅ 1 = 3 − 3 = 0 \operatorname*{det}\left(\left[ \begin{array}{cc} -1 & 1 \\ 3 & -3 \end{array} \right]\right) = (-1) \cdot (-3) - 3 \cdot 1 = 3 - 3 = 0 det ( [ − 1 3 1 − 3 ] ) = ( − 1 ) ⋅ ( − 3 ) − 3 ⋅ 1 = 3 − 3 = 0
It means that the inverse of the matrix doesn't exist. Such a matrix is called "singular".
Answer: it doesn't exist.
**Question**
3. [ 1 2 3 7 ] \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix} [ 1 3 2 7 ]
**Solution**
[ 1 2 3 7 ] − 1 = 1 det ( [ 1 2 3 7 ] ) [ 7 − 2 − 3 1 ] = 1 1 ⋅ 7 − 2 ⋅ 3 [ 7 − 2 − 3 1 ] = [ 7 − 2 − 3 1 ] \left[ \begin{array}{cc} 1 & 2 \\ 3 & 7 \end{array} \right]^{-1} = \frac{1}{\operatorname*{det}(\left[ \begin{array}{cc} 1 & 2 \\ 3 & 7 \end{array} \right])} \left[ \begin{array}{cc} 7 & -2 \\ -3 & 1 \end{array} \right] = \frac{1}{1 \cdot 7 - 2 \cdot 3} \left[ \begin{array}{cc} 7 & -2 \\ -3 & 1 \end{array} \right] = \left[ \begin{array}{cc} 7 & -2 \\ -3 & 1 \end{array} \right] [ 1 3 2 7 ] − 1 = det ( [ 1 3 2 7 ] ) 1 [ 7 − 3 − 2 1 ] = 1 ⋅ 7 − 2 ⋅ 3 1 [ 7 − 3 − 2 1 ] = [ 7 − 3 − 2 1 ]
Answer: [ 7 − 2 − 3 1 ] \left[ \begin{array}{cc}7 & -2\\ -3 & 1 \end{array} \right] [ 7 − 3 − 2 1 ]
C. Finding the inverse of the Square of a Matrix.
Direction: Compute A^-2
Question
1. A = [ 0 − 2 − 1 3 ] A = \begin{bmatrix} 0 & -2 \\ -1 & 3 \end{bmatrix} A = [ 0 − 1 − 2 3 ]
Solution
A 2 = [ 0 − 2 − 1 3 ] 2 = [ 0 − 2 − 1 3 ] ⋅ [ 0 − 2 − 1 3 ] = = [ 0 ⋅ 0 + ( − 2 ) ( − 1 ) 0 ⋅ ( − 2 ) + ( − 2 ) ⋅ 3 ( − 1 ) ⋅ 0 + 3 ⋅ ( − 1 ) ( − 1 ) ⋅ ( − 2 ) + 3 ⋅ 3 ] = [ 2 − 6 − 3 11 ] A − 2 = ( A 2 ) − 1 = [ 2 − 6 − 3 11 ] − 1 = 1 det ( [ 2 − 6 − 3 11 ] ) [ 11 6 3 2 ] = 1 2 ⋅ 11 − ( − 3 ) ⋅ ( − 6 ) [ 11 6 3 2 ] = = 1 4 [ 11 6 3 2 ] = [ 11 4 3 2 3 4 1 2 ] \begin{aligned}
A^2 &= \begin{bmatrix} 0 & -2 \\ -1 & 3 \end{bmatrix}^2 = \begin{bmatrix} 0 & -2 \\ -1 & 3 \end{bmatrix} \cdot \begin{bmatrix} 0 & -2 \\ -1 & 3 \end{bmatrix} = \\
&= \begin{bmatrix} 0 \cdot 0 + (-2)(-1) & 0 \cdot (-2) + (-2) \cdot 3 \\ (-1) \cdot 0 + 3 \cdot (-1) & (-1) \cdot (-2) + 3 \cdot 3 \end{bmatrix} = \begin{bmatrix} 2 & -6 \\ -3 & 11 \end{bmatrix} \\
A^{-2} &= (A^2)^{-1} = \begin{bmatrix} 2 & -6 \\ -3 & 11 \end{bmatrix}^{-1} = \frac{1}{\det\left(\begin{bmatrix} 2 & -6 \\ -3 & 11 \end{bmatrix}\right)} \begin{bmatrix} 11 & 6 \\ 3 & 2 \end{bmatrix} \\
&= \frac{1}{2 \cdot 11 - (-3) \cdot (-6)} \begin{bmatrix} 11 & 6 \\ 3 & 2 \end{bmatrix} = \\
&= \frac{1}{4} \begin{bmatrix} 11 & 6 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} \frac{11}{4} & \frac{3}{2} \\ \frac{3}{4} & \frac{1}{2} \end{bmatrix}
\end{aligned} A 2 A − 2 = [ 0 − 1 − 2 3 ] 2 = [ 0 − 1 − 2 3 ] ⋅ [ 0 − 1 − 2 3 ] = = [ 0 ⋅ 0 + ( − 2 ) ( − 1 ) ( − 1 ) ⋅ 0 + 3 ⋅ ( − 1 ) 0 ⋅ ( − 2 ) + ( − 2 ) ⋅ 3 ( − 1 ) ⋅ ( − 2 ) + 3 ⋅ 3 ] = [ 2 − 3 − 6 11 ] = ( A 2 ) − 1 = [ 2 − 3 − 6 11 ] − 1 = det ( [ 2 − 3 − 6 11 ] ) 1 [ 11 3 6 2 ] = 2 ⋅ 11 − ( − 3 ) ⋅ ( − 6 ) 1 [ 11 3 6 2 ] = = 4 1 [ 11 3 6 2 ] = [ 4 11 4 3 2 3 2 1 ]
Answer: A − 2 = [ 11 4 3 2 3 4 1 2 ] A^{-2} = \begin{bmatrix} \frac{11}{4} & \frac{3}{2} \\ \frac{3}{4} & \frac{1}{2} \end{bmatrix} A − 2 = [ 4 11 4 3 2 3 2 1 ]
Question
2. A = [ 2 7 − 5 6 ] A = \begin{bmatrix} 2 & 7 \\ -5 & 6 \end{bmatrix} A = [ 2 − 5 7 6 ]
Solution
A 2 = [ 2 7 − 5 6 ] 2 = [ 2 7 − 5 6 ] ⋅ [ 2 7 − 5 6 ] = = [ 2 ⋅ 2 + 7 ⋅ ( − 5 ) 2 ⋅ 7 + 7 ⋅ 6 ( − 5 ) ⋅ 2 + 6 ⋅ ( − 5 ) ( − 5 ) ⋅ 7 + 6 ⋅ 6 ] = [ − 31 56 − 40 1 ] A − 2 = ( A 2 ) − 1 = [ − 31 56 − 40 1 ] − 1 = 1 det ( [ − 31 56 − 40 1 ] ) [ 1 − 56 40 − 31 ] = = 1 ( − 31 ) ⋅ 1 − 56 ⋅ ( − 40 ) [ 1 − 56 40 − 31 ] = 1 2209 [ 1 − 56 40 − 31 ] = [ 1 2209 − 56 2209 40 2209 − 31 2209 ] \begin{aligned}
A^2 &= \begin{bmatrix} 2 & 7 \\ -5 & 6 \end{bmatrix}^2 = \begin{bmatrix} 2 & 7 \\ -5 & 6 \end{bmatrix} \cdot \begin{bmatrix} 2 & 7 \\ -5 & 6 \end{bmatrix} = \\
&= \begin{bmatrix} 2 \cdot 2 + 7 \cdot (-5) & 2 \cdot 7 + 7 \cdot 6 \\ (-5) \cdot 2 + 6 \cdot (-5) & (-5) \cdot 7 + 6 \cdot 6 \end{bmatrix} = \begin{bmatrix} -31 & 56 \\ -40 & 1 \end{bmatrix} \\
A^{-2} &= (A^2)^{-1} = \begin{bmatrix} -31 & 56 \\ -40 & 1 \end{bmatrix}^{-1} = \frac{1}{\det\left(\begin{bmatrix} -31 & 56 \\ -40 & 1 \end{bmatrix}\right)} \begin{bmatrix} 1 & -56 \\ 40 & -31 \end{bmatrix} = \\
&= \frac{1}{(-31) \cdot 1 - 56 \cdot (-40)} \begin{bmatrix} 1 & -56 \\ 40 & -31 \end{bmatrix} = \frac{1}{2209} \begin{bmatrix} 1 & -56 \\ 40 & -31 \end{bmatrix} = \begin{bmatrix} \frac{1}{2209} & \frac{-56}{2209} \\ \frac{40}{2209} & \frac{-31}{2209} \end{bmatrix}
\end{aligned} A 2 A − 2 = [ 2 − 5 7 6 ] 2 = [ 2 − 5 7 6 ] ⋅ [ 2 − 5 7 6 ] = = [ 2 ⋅ 2 + 7 ⋅ ( − 5 ) ( − 5 ) ⋅ 2 + 6 ⋅ ( − 5 ) 2 ⋅ 7 + 7 ⋅ 6 ( − 5 ) ⋅ 7 + 6 ⋅ 6 ] = [ − 31 − 40 56 1 ] = ( A 2 ) − 1 = [ − 31 − 40 56 1 ] − 1 = det ( [ − 31 − 40 56 1 ] ) 1 [ 1 40 − 56 − 31 ] = = ( − 31 ) ⋅ 1 − 56 ⋅ ( − 40 ) 1 [ 1 40 − 56 − 31 ] = 2209 1 [ 1 40 − 56 − 31 ] = [ 2209 1 2209 40 2209 − 56 2209 − 31 ]
Answer: A − 2 = [ 1 2209 − 56 2209 40 2209 − 31 2209 ] A^{-2} = \begin{bmatrix} \frac{1}{2209} & \frac{-56}{2209} \\ \frac{40}{2209} & \frac{-31}{2209} \end{bmatrix} A − 2 = [ 2209 1 2209 40 2209 − 56 2209 − 31 ]
Question
3. A = [ − 2 0 0 0 1 0 0 0 3 ] A = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} A = ⎣ ⎡ − 2 0 0 0 1 0 0 0 3 ⎦ ⎤
Solution
A 2 = [ − 2 0 0 0 1 0 0 0 3 ] 2 = [ − 2 0 0 0 1 0 0 0 3 ] ⋅ [ − 2 0 0 0 1 0 0 0 3 ] = = [ ( − 2 ) ⋅ ( − 2 ) + 0 ⋅ 1 + 0 ⋅ 3 ( − 2 ) ⋅ 0 + 0 ⋅ 1 + 0 ⋅ 0 ( − 2 ) ⋅ 0 + 0 ⋅ 0 + 0 ⋅ 3 0 ⋅ ( − 2 ) + 1 ⋅ 0 + 0 ⋅ 0 0 ⋅ 0 + 1 ⋅ 1 + 0 ⋅ 0 0 ⋅ 0 + 1 ⋅ 0 + 0 ⋅ 3 0 ⋅ ( − 2 ) + 0 ⋅ 0 + 3 ⋅ 0 0 ⋅ 0 + 0 ⋅ 1 + 3 ⋅ 0 0 ⋅ 0 + 0 ⋅ 0 + 3 ⋅ 3 ] = = [ 4 0 0 0 1 0 0 0 9 ] \begin{aligned}
A^2 &= \begin{bmatrix} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix}^2 = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} \cdot \begin{bmatrix} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} = \\
&= \begin{bmatrix} (-2) \cdot (-2) + 0 \cdot 1 + 0 \cdot 3 & (-2) \cdot 0 + 0 \cdot 1 + 0 \cdot 0 & (-2) \cdot 0 + 0 \cdot 0 + 0 \cdot 3 \\ 0 \cdot (-2) + 1 \cdot 0 + 0 \cdot 0 & 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 & 0 \cdot 0 + 1 \cdot 0 + 0 \cdot 3 \\ 0 \cdot (-2) + 0 \cdot 0 + 3 \cdot 0 & 0 \cdot 0 + 0 \cdot 1 + 3 \cdot 0 & 0 \cdot 0 + 0 \cdot 0 + 3 \cdot 3 \end{bmatrix} = \\
&= \begin{bmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 9 \end{bmatrix}
\end{aligned} A 2 = ⎣ ⎡ − 2 0 0 0 1 0 0 0 3 ⎦ ⎤ 2 = ⎣ ⎡ − 2 0 0 0 1 0 0 0 3 ⎦ ⎤ ⋅ ⎣ ⎡ − 2 0 0 0 1 0 0 0 3 ⎦ ⎤ = = ⎣ ⎡ ( − 2 ) ⋅ ( − 2 ) + 0 ⋅ 1 + 0 ⋅ 3 0 ⋅ ( − 2 ) + 1 ⋅ 0 + 0 ⋅ 0 0 ⋅ ( − 2 ) + 0 ⋅ 0 + 3 ⋅ 0 ( − 2 ) ⋅ 0 + 0 ⋅ 1 + 0 ⋅ 0 0 ⋅ 0 + 1 ⋅ 1 + 0 ⋅ 0 0 ⋅ 0 + 0 ⋅ 1 + 3 ⋅ 0 ( − 2 ) ⋅ 0 + 0 ⋅ 0 + 0 ⋅ 3 0 ⋅ 0 + 1 ⋅ 0 + 0 ⋅ 3 0 ⋅ 0 + 0 ⋅ 0 + 3 ⋅ 3 ⎦ ⎤ = = ⎣ ⎡ 4 0 0 0 1 0 0 0 9 ⎦ ⎤ [ 4 0 0 1 0 0 0 1 0 0 1 0 0 0 9 0 0 1 ] ∼ ∣ R 1 ← R 1 4 , R 3 ← R 3 9 ∣ ∼ ∣ 1 0 0 1 4 0 0 0 1 0 0 1 0 0 0 1 0 0 1 9 ∣ \begin{aligned}
&\left[ \begin{array}{cccccccc}
4 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 9 & 0 & 0 & 1
\end{array} \right] \sim |R1 \leftarrow \frac{R1}{4}, R3 \leftarrow \frac{R3}{9}| \sim
\begin{vmatrix}
1 & 0 & 0 & \frac{1}{4} & 0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & \frac{1}{9}
\end{vmatrix}
\end{aligned} ⎣ ⎡ 4 0 0 0 1 0 0 0 9 1 0 0 0 1 0 0 0 1 ⎦ ⎤ ∼ ∣ R 1 ← 4 R 1 , R 3 ← 9 R 3 ∣ ∼ ∣ ∣ 1 0 0 0 1 0 0 0 1 4 1 0 0 0 1 0 0 0 9 1 ∣ ∣ A − 2 = ( A 2 ) − 1 = [ 4 0 0 0 1 0 0 0 9 ] − 1 = [ 1 4 0 0 0 1 0 0 0 1 9 ] A^{-2} = (A^2)^{-1} = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 9 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{1}{4} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{9} \end{bmatrix} A − 2 = ( A 2 ) − 1 = ⎣ ⎡ 4 0 0 0 1 0 0 0 9 ⎦ ⎤ − 1 = ⎣ ⎡ 4 1 0 0 0 1 0 0 0 9 1 ⎦ ⎤
Answer: A − 2 = [ 1 4 0 0 0 1 0 0 0 1 9 ] A^{-2} = \begin{bmatrix} \frac{1}{4} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{9} \end{bmatrix} A − 2 = ⎣ ⎡ 4 1 0 0 0 1 0 0 0 9 1 ⎦ ⎤
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#340153 on Dec 2023