Question

A. Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination.

1.-x+y+2z=1
2x+3y+z=-2
5x+4y+2z=4

2. 2x+3y+z=10
2x-3y-3z=22
4x-2y+3z=-2

B. Finding the Inverse of a Matrix.
Find the inverse of the matrix.

1. [2 0; 0 3]
2. [-1 1; 3 -3]
3. [1 2; 3 7]

C. Finding the inverse of the Square of a Matrix.
Direction: Compute A^-2.

1. A = [0 -2; -1 3]
2. A = [2 7; -5 6]
3. A = [-2 0 0; 0 1 0; 0 0 3]

Answer 1

Answer on Question #62621 – Math – Linear Algebra

A. Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination.

Question

1. $\begin{cases} -x + y + 2z = 1 \\ 2x + 3y + z = -2 \\ 5x + 4y + 2z = 4 \end{cases}$

Solution

This system can be represented by the coefficient matrix: $\begin{pmatrix} -1 & 1 & 2 & | & 1 \\ 2 & 3 & 1 & | & -2 \\ 5 & 4 & 2 & | & 4 \end{pmatrix}$

Henceforth $R2 \leftarrow R2 - 2R1$ means “the second row equals the second row subtracting two times the first row”.

$$\begin{aligned} &\left( \begin{array}{cccc|cc} -1 & 1 & 2 & | & 1 \\ 2 & 3 & 1 & | & -2 \\ 5 & 4 & 2 & | & 4 \end{array} \right) \xrightarrow{R1 \leftarrow R1 \times (-1)} \left( \begin{array}{cccc|cc} 1 & -1 & -2 & | & -1 \\ 2 & 3 & 1 & | & -2 \\ 5 & 4 & 2 & | & 4 \end{array} \right) \xrightarrow{\frac{R2 \leftarrow R2 - 2R1}{R3 \leftarrow R3 - 5R1}} \\ &\rightarrow \left( \begin{array}{cccc|cc} 1 & -1 & -2 & | & -1 \\ 0 & 5 & 5 & | & 0 \\ 0 & 9 & 12 & | & 9 \end{array} \right) \xrightarrow{R3 \leftarrow \frac{1}{3} \cdot R3} \left( \begin{array}{cccc|cc} 1 & -1 & -2 & | & -1 \\ 0 & 1 & 1 & | & 0 \\ 0 & 0 & 1 & | & 3 \end{array} \right) \xrightarrow{R3 \leftarrow R3 - 3R2} \\ &\rightarrow \left( \begin{array}{cccc} 1 & -1 & -2 & | & -1 \\ 0 & 1 & 1 & | & 0 \\ 0 & 0 & 1 & | & 3 \end{array} \right) \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} x - y - 2z = -1 \\ y + z = 0 \\ z = 3 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = -1 + y + 2z \\ y = -z \\ z = 3 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = -1 + z \\ y = -z \\ z = 3 \end{array} \right. \\ \Rightarrow \left\{ \begin{array}{l} x = 2 \\ y = -3 \\ z = 3 \end{array} \right. \end{aligned}$$

Answer: $x = 2, y = -3, z = 3$.

Question

2. $\begin{cases} 2x + 3y + z = 10 \\ 2x - 3y - 3z = 22 \\ 4x - 2y + 3z = -2 \end{cases}$

Solution

This system can be represented by the matrix: $\begin{pmatrix} 2 & 3 & 1 & | & 10 \\ 2 & -3 & -3 & | & 22 \\ 4 & -2 & 3 & | & -2 \end{pmatrix}$

$$\begin{array}{l} \left( \begin{array}{cccccc} 2 & 3 & 1 & | & 10 \\ 2 & -3 & -3 & | & 22 \\ 4 & -2 & 3 & | & -2 \end{array} \right) \xrightarrow{R2 \leftarrow R2 - R1} \left( \begin{array}{cccccc} 2 & 3 & 1 & | & 10 \\ 0 & -6 & -4 & | & 12 \\ 0 & -8 & 1 & | & -22 \end{array} \right) \xrightarrow{R2 \leftarrow -\frac{1}{2} R2} \xrightarrow{R3 \leftarrow -R3} \\ \rightarrow \left( \begin{array}{cccccc} 2 & 3 & 1 & | & 10 \\ 0 & 3 & 2 & | & -6 \\ 0 & 8 & -1 & | & 22 \end{array} \right) \xrightarrow{R3 \leftarrow R3 - \frac{8}{3} R2} \left( \begin{array}{cccccc} 2 & 3 & 1 & | & 10 \\ 0 & 3 & 2 & | & -6 \\ 0 & 0 & -\frac{19}{3} & | & 38 \end{array} \right) \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} 2x + 3y + z = 10 \\ 3y + 2z = -6 \\ -\frac{19}{3}z = 38 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = \frac{10 - 3y - z}{2} \\ y = \frac{-6 - 2z}{3} \\ z = -6 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = \frac{16 + z}{2} \\ y = \frac{-6 - 2z}{3} \\ z = -6 \end{array} \right. \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} x = 5 \\ y = 2 \\ z = -6 \end{array} \right. \end{array}$$

Answer: $x = 5, y = 2, z = -6$.

B. Find the inverse of the matrix.

**Question**

1. $\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$

**Solution**

$$\left[ \begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array} \right]^{-1} = \frac{1}{\operatorname*{det}(\left[ \begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array} \right])} \left[ \begin{array}{cc} 3 & 0 \\ 0 & 2 \end{array} \right] = \frac{1}{2 \cdot 3 - 0 \cdot 0} \left[ \begin{array}{cc} 3 & 0 \\ 0 & 2 \end{array} \right] = \frac{1}{6} \left[ \begin{array}{cc} 3 & 0 \\ 0 & 2 \end{array} \right] = \left[ \begin{array}{cc} \frac{3}{6} & \frac{0}{6} \\ \frac{0}{6} & \frac{2}{6} \end{array} \right] = \left[ \begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \end{array} \right].$$

Answer: $\begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \end{bmatrix}$.

**Question**

2. $\begin{bmatrix} -1 & 1 \\ 3 & -3 \end{bmatrix}$

**Solution**

$$\operatorname*{det}\left(\left[ \begin{array}{cc} -1 & 1 \\ 3 & -3 \end{array} \right]\right) = (-1) \cdot (-3) - 3 \cdot 1 = 3 - 3 = 0$$

It means that the inverse of the matrix doesn't exist. Such a matrix is called "singular".

Answer: it doesn't exist.

**Question**

3. $\begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix}$

**Solution**

$$\left[ \begin{array}{cc} 1 & 2 \\ 3 & 7 \end{array} \right]^{-1} = \frac{1}{\operatorname*{det}(\left[ \begin{array}{cc} 1 & 2 \\ 3 & 7 \end{array} \right])} \left[ \begin{array}{cc} 7 & -2 \\ -3 & 1 \end{array} \right] = \frac{1}{1 \cdot 7 - 2 \cdot 3} \left[ \begin{array}{cc} 7 & -2 \\ -3 & 1 \end{array} \right] = \left[ \begin{array}{cc} 7 & -2 \\ -3 & 1 \end{array} \right]$$

Answer: $\left[ \begin{array}{cc}7 & -2\\ -3 & 1 \end{array} \right]$

C. Finding the inverse of the Square of a Matrix.

Direction: Compute A^-2

Question

1. $A = \begin{bmatrix} 0 & -2 \\ -1 & 3 \end{bmatrix}$

Solution

$$\begin{aligned} A^2 &= \begin{bmatrix} 0 & -2 \\ -1 & 3 \end{bmatrix}^2 = \begin{bmatrix} 0 & -2 \\ -1 & 3 \end{bmatrix} \cdot \begin{bmatrix} 0 & -2 \\ -1 & 3 \end{bmatrix} = \\ &= \begin{bmatrix} 0 \cdot 0 + (-2)(-1) & 0 \cdot (-2) + (-2) \cdot 3 \\ (-1) \cdot 0 + 3 \cdot (-1) & (-1) \cdot (-2) + 3 \cdot 3 \end{bmatrix} = \begin{bmatrix} 2 & -6 \\ -3 & 11 \end{bmatrix} \\ A^{-2} &= (A^2)^{-1} = \begin{bmatrix} 2 & -6 \\ -3 & 11 \end{bmatrix}^{-1} = \frac{1}{\det\left(\begin{bmatrix} 2 & -6 \\ -3 & 11 \end{bmatrix}\right)} \begin{bmatrix} 11 & 6 \\ 3 & 2 \end{bmatrix} \\ &= \frac{1}{2 \cdot 11 - (-3) \cdot (-6)} \begin{bmatrix} 11 & 6 \\ 3 & 2 \end{bmatrix} = \\ &= \frac{1}{4} \begin{bmatrix} 11 & 6 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} \frac{11}{4} & \frac{3}{2} \\ \frac{3}{4} & \frac{1}{2} \end{bmatrix} \end{aligned}$$

Answer: $A^{-2} = \begin{bmatrix} \frac{11}{4} & \frac{3}{2} \\ \frac{3}{4} & \frac{1}{2} \end{bmatrix}$

Question

2. $A = \begin{bmatrix} 2 & 7 \\ -5 & 6 \end{bmatrix}$

Solution

$$\begin{aligned} A^2 &= \begin{bmatrix} 2 & 7 \\ -5 & 6 \end{bmatrix}^2 = \begin{bmatrix} 2 & 7 \\ -5 & 6 \end{bmatrix} \cdot \begin{bmatrix} 2 & 7 \\ -5 & 6 \end{bmatrix} = \\ &= \begin{bmatrix} 2 \cdot 2 + 7 \cdot (-5) & 2 \cdot 7 + 7 \cdot 6 \\ (-5) \cdot 2 + 6 \cdot (-5) & (-5) \cdot 7 + 6 \cdot 6 \end{bmatrix} = \begin{bmatrix} -31 & 56 \\ -40 & 1 \end{bmatrix} \\ A^{-2} &= (A^2)^{-1} = \begin{bmatrix} -31 & 56 \\ -40 & 1 \end{bmatrix}^{-1} = \frac{1}{\det\left(\begin{bmatrix} -31 & 56 \\ -40 & 1 \end{bmatrix}\right)} \begin{bmatrix} 1 & -56 \\ 40 & -31 \end{bmatrix} = \\ &= \frac{1}{(-31) \cdot 1 - 56 \cdot (-40)} \begin{bmatrix} 1 & -56 \\ 40 & -31 \end{bmatrix} = \frac{1}{2209} \begin{bmatrix} 1 & -56 \\ 40 & -31 \end{bmatrix} = \begin{bmatrix} \frac{1}{2209} & \frac{-56}{2209} \\ \frac{40}{2209} & \frac{-31}{2209} \end{bmatrix} \end{aligned}$$

Answer: $A^{-2} = \begin{bmatrix} \frac{1}{2209} & \frac{-56}{2209} \\ \frac{40}{2209} & \frac{-31}{2209} \end{bmatrix}$

Question

3. $A = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix}$

Solution

$$\begin{aligned} A^2 &= \begin{bmatrix} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix}^2 = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} \cdot \begin{bmatrix} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} = \\ &= \begin{bmatrix} (-2) \cdot (-2) + 0 \cdot 1 + 0 \cdot 3 & (-2) \cdot 0 + 0 \cdot 1 + 0 \cdot 0 & (-2) \cdot 0 + 0 \cdot 0 + 0 \cdot 3 \\ 0 \cdot (-2) + 1 \cdot 0 + 0 \cdot 0 & 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 & 0 \cdot 0 + 1 \cdot 0 + 0 \cdot 3 \\ 0 \cdot (-2) + 0 \cdot 0 + 3 \cdot 0 & 0 \cdot 0 + 0 \cdot 1 + 3 \cdot 0 & 0 \cdot 0 + 0 \cdot 0 + 3 \cdot 3 \end{bmatrix} = \\ &= \begin{bmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 9 \end{bmatrix} \end{aligned}$$ $$\begin{aligned} &\left[ \begin{array}{cccccccc} 4 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 9 & 0 & 0 & 1 \end{array} \right] \sim |R1 \leftarrow \frac{R1}{4}, R3 \leftarrow \frac{R3}{9}| \sim \begin{vmatrix} 1 & 0 & 0 & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & \frac{1}{9} \end{vmatrix} \end{aligned}$$ $$A^{-2} = (A^2)^{-1} = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 9 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{1}{4} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{9} \end{bmatrix}$$

Answer: $A^{-2} = \begin{bmatrix} \frac{1}{4} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{9} \end{bmatrix}$.

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