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1.3-1.4 Vector Equations & The Matrix Equation Ax = b

Theory: Vector Equations & The Matrix Equation Ax=bA\mathbf{x} = \mathbf{b}

Vectors in Rn\mathbb{R}^n & Linear Combinations

A column vector u∈Rn\mathbf{u} \in \mathbb{R}^n is an ordered list of nn real numbers:

u=[u1u2⋮un]\mathbf{u} = \begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_n \end{bmatrix}

Given vectors v1,v2,…,vp∈Rn\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_p \in \mathbb{R}^n and scalars c1,c2,…,cpc_1, c_2, \dots, c_p, the vector:

y=c1v1+c2v2+⋯+cpvp\mathbf{y} = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_p \mathbf{v}_p

is called a linear combination of v1,…,vp\mathbf{v}_1, \dots, \mathbf{v}_p with weights c1,…,cpc_1, \dots, c_p.

The set of all linear combinations is denoted by:

Span⁡{v1,…,vp}={c1v1+⋯+cpvp:c1,…,cp∈R}\operatorname{Span}\{\mathbf{v}_1, \dots, \mathbf{v}_p\} = \{c_1 \mathbf{v}_1 + \cdots + c_p \mathbf{v}_p : c_1, \dots, c_p \in \mathbb{R}\}

The Matrix Equation Ax=bA\mathbf{x} = \mathbf{b}

If AA is an m×nm \times n matrix with columns a1,…,an\mathbf{a}_1, \dots, \mathbf{a}_n, and x∈Rn\mathbf{x} \in \mathbb{R}^n, then the product AxA\mathbf{x} is defined as the linear combination of the columns of AA using the corresponding entries in x\mathbf{x} as weights:

Ax=[a1a2⋯an][x1x2⋮xn]=x1a1+x2a2+⋯+xnanA\mathbf{x} = \begin{bmatrix} \mathbf{a}_1 & \mathbf{a}_2 & \cdots & \mathbf{a}_n \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = x_1 \mathbf{a}_1 + x_2 \mathbf{a}_2 + \cdots + x_n \mathbf{a}_n

[!IMPORTANT] Key Equivalence: The matrix equation Ax=bA\mathbf{x} = \mathbf{b}, the vector equation x1a1+⋯+xnan=bx_1 \mathbf{a}_1 + \cdots + x_n \mathbf{a}_n = \mathbf{b}, and the linear system with augmented matrix [a1  a2  ⋯  an∣b][\mathbf{a}_1 \; \mathbf{a}_2 \; \cdots \; \mathbf{a}_n \mid \mathbf{b}] have the exact same solution set. Therefore, b∈Span⁡{a1,…,an}\mathbf{b} \in \operatorname{Span}\{\mathbf{a}_1, \dots, \mathbf{a}_n\} if and only if Ax=bA\mathbf{x} = \mathbf{b} is consistent.

Theorem 4: When Does Ax=bA\mathbf{x} = \mathbf{b} Have a Solution for Every b\mathbf{b}?

Let AA be an m×nm \times n matrix. The following four statements are logically equivalent (either all true or all false):

  1. For each b∈Rm\mathbf{b} \in \mathbb{R}^m, the equation Ax=bA\mathbf{x} = \mathbf{b} has a solution.
  2. Each b∈Rm\mathbf{b} \in \mathbb{R}^m is a linear combination of the columns of AA.
  3. The columns of AA span Rm\mathbb{R}^m (Span⁡{a1,…,an}=Rm\operatorname{Span}\{\mathbf{a}_1, \dots, \mathbf{a}_n\} = \mathbb{R}^m).
  4. AA has a pivot position in every row.

Solved Examples (Textbook Questions)

Exercise 1

(Adapted from Lay §1.3, Exercise 11)

Let a1=[1−20]\mathbf{a}_1 = \begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix}, a2=[012]\mathbf{a}_2 = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}, a3=[5−68]\mathbf{a}_3 = \begin{bmatrix} 5 \\ -6 \\ 8 \end{bmatrix}, and b=[2−16]\mathbf{b} = \begin{bmatrix} 2 \\ -1 \\ 6 \end{bmatrix}.

Determine whether b\mathbf{b} is in Span⁡{a1,a2,a3}\operatorname{Span}\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3\}. If it is, express b\mathbf{b} as a linear combination of a1,a2,a3\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3.

Show solution ↓
Solution

The question asks whether there exist scalars x1,x2,x3x_1, x_2, x_3 such that:

x1a1+x2a2+x3a3=bx_1 \mathbf{a}_1 + x_2 \mathbf{a}_2 + x_3 \mathbf{a}_3 = \mathbf{b}

Step 1: Form the augmented matrix.

[1052−21−6−10286]\left[\begin{array}{ccc|c} 1 & 0 & 5 & 2 \\ -2 & 1 & -6 & -1 \\ 0 & 2 & 8 & 6 \end{array}\right]

Step 2: Row reduce to RREF.

  • R2←R2+2R1R_2 \leftarrow R_2 + 2R_1: [−2,1,−6,−1]+2[1,0,5,2]=[0,1,4,3][-2, 1, -6, -1] + 2[1, 0, 5, 2] = [0, 1, 4, 3]

The matrix becomes:

[105201430286]\left[\begin{array}{ccc|c} 1 & 0 & 5 & 2 \\ 0 & 1 & 4 & 3 \\ 0 & 2 & 8 & 6 \end{array}\right]
  • R3←R3−2R2R_3 \leftarrow R_3 - 2R_2: [0,2,8,6]−2[0,1,4,3]=[0,0,0,0][0, 2, 8, 6] - 2[0, 1, 4, 3] = [0, 0, 0, 0]

The RREF is:

[105201430000]\left[\begin{array}{ccc|c} 1 & 0 & 5 & 2 \\ 0 & 1 & 4 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right]

Step 3: Conclusion. There is no pivot in the augmented column, so the system is consistent. Thus, b∈Span⁡{a1,a2,a3}\mathbf{b} \in \operatorname{Span}\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3\}.

The solution equations are:

x1+5x3=2  ⟹  x1=2−5x3x2+4x3=3  ⟹  x2=3−4x3x3 is free\begin{aligned} x_1 + 5x_3 &= 2 \implies x_1 = 2 - 5x_3 \\ x_2 + 4x_3 &= 3 \implies x_2 = 3 - 4x_3 \\ x_3 &\text{ is free} \end{aligned}

Choosing x3=0x_3 = 0, we get weights x1=2x_1 = 2 and x2=3x_2 = 3:

b=2a1+3a2+0a3=2a1+3a2\mathbf{b} = 2\mathbf{a}_1 + 3\mathbf{a}_2 + 0\mathbf{a}_3 = 2\mathbf{a}_1 + 3\mathbf{a}_2

Verification:

2[1−20]+3[012]=[2+0−4+30+6]=[2−16]=b✓2\begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix} + 3\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 + 0 \\ -4 + 3 \\ 0 + 6 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 6 \end{bmatrix} = \mathbf{b} \quad \checkmark
Exercise 2

(Adapted from Lay §1.3, Exercise 17)

For what value(s) of hh is y\mathbf{y} in the plane generated by v1\mathbf{v}_1 and v2\mathbf{v}_2, where:

v1=[14−2],v2=[−2−37],y=[41h]\mathbf{v}_1 = \begin{bmatrix} 1 \\ 4 \\ -2 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} -2 \\ -3 \\ 7 \end{bmatrix}, \quad \mathbf{y} = \begin{bmatrix} 4 \\ 1 \\ h \end{bmatrix}
Show solution ↓
Solution

The vector y\mathbf{y} lies in the plane Span⁡{v1,v2}\operatorname{Span}\{\mathbf{v}_1, \mathbf{v}_2\} if and only if the vector equation x1v1+x2v2=yx_1 \mathbf{v}_1 + x_2 \mathbf{v}_2 = \mathbf{y} is consistent.

Step 1: Set up the augmented matrix.

[1−244−31−27h]\left[\begin{array}{cc|c} 1 & -2 & 4 \\ 4 & -3 & 1 \\ -2 & 7 & h \end{array}\right]

Step 2: Row reduce to echelon form.

  • R2←R2−4R1R_2 \leftarrow R_2 - 4R_1: [4,−3,1]−4[1,−2,4]=[0,5,−15][4, -3, 1] - 4[1, -2, 4] = [0, 5, -15]
  • R3←R3+2R1R_3 \leftarrow R_3 + 2R_1: [−2,7,h]+2[1,−2,4]=[0,3,h+8][-2, 7, h] + 2[1, -2, 4] = [0, 3, h + 8]

The matrix is:

[1−2405−1503h+8]\left[\begin{array}{cc|c} 1 & -2 & 4 \\ 0 & 5 & -15 \\ 0 & 3 & h + 8 \end{array}\right]

Scale row 2 by 15\frac{1}{5} (R2←15R2R_2 \leftarrow \frac{1}{5}R_2):

[1−2401−303h+8]\left[\begin{array}{cc|c} 1 & -2 & 4 \\ 0 & 1 & -3 \\ 0 & 3 & h + 8 \end{array}\right]

Eliminate entry in row 3 (R3←R3−3R2R_3 \leftarrow R_3 - 3R_2):

[0,3,h+8]−3[0,1,−3]=[0,0,h+8−(−9)]=[0,0,h+17][0, 3, h + 8] - 3[0, 1, -3] = [0, 0, h + 8 - (-9)] = [0, 0, h + 17]

The echelon form is:

[1−2401−300h+17]\left[\begin{array}{cc|c} 1 & -2 & 4 \\ 0 & 1 & -3 \\ 0 & 0 & h + 17 \end{array}\right]

Step 3: Enforce consistency. By Theorem 2, the system is consistent if and only if the last row does not have a pivot in the augmented column:

h+17=0  ⟹  h=−17h + 17 = 0 \implies h = -17

Conclusion: y∈Span⁡{v1,v2}\mathbf{y} \in \operatorname{Span}\{\mathbf{v}_1, \mathbf{v}_2\} if and only if h=−17h = -17.

Exercise 3

(Adapted from Lay §1.4, Exercise 17)

Let A=[1303−1−1−110−42−8203−1]A = \begin{bmatrix} 1 & 3 & 0 & 3 \\ -1 & -1 & -1 & 1 \\ 0 & -4 & 2 & -8 \\ 2 & 0 & 3 & -1 \end{bmatrix}.

Do the columns of AA span R4\mathbb{R}^4? Does the equation Ax=bA\mathbf{x} = \mathbf{b} have a solution for each b∈R4\mathbf{b} \in \mathbb{R}^4? Justify your answer using Theorem 4.

Show solution ↓
Solution

According to Theorem 4, the columns of AA span R4\mathbb{R}^4 if and only if AA has a pivot position in every row. Since AA has 4 rows, it must have 4 pivots.

Step 1: Row reduce AA to echelon form.

A=[1303−1−1−110−42−8203−1]A = \begin{bmatrix} 1 & 3 & 0 & 3 \\ -1 & -1 & -1 & 1 \\ 0 & -4 & 2 & -8 \\ 2 & 0 & 3 & -1 \end{bmatrix}
  • R2←R2+R1R_2 \leftarrow R_2 + R_1: [−1,−1,−1,1]+[1,3,0,3]=[0,2,−1,4][-1, -1, -1, 1] + [1, 3, 0, 3] = [0, 2, -1, 4]
  • R4←R4−2R1R_4 \leftarrow R_4 - 2R_1: [2,0,3,−1]−2[1,3,0,3]=[0,−6,3,−7][2, 0, 3, -1] - 2[1, 3, 0, 3] = [0, -6, 3, -7]

Matrix:

[130302−140−42−80−63−7]\begin{bmatrix} 1 & 3 & 0 & 3 \\ 0 & 2 & -1 & 4 \\ 0 & -4 & 2 & -8 \\ 0 & -6 & 3 & -7 \end{bmatrix}
  • R3←R3+2R2R_3 \leftarrow R_3 + 2R_2: [0,−4,2,−8]+2[0,2,−1,4]=[0,0,0,0][0, -4, 2, -8] + 2[0, 2, -1, 4] = [0, 0, 0, 0]
  • R4←R4+3R2R_4 \leftarrow R_4 + 3R_2: [0,−6,3,−7]+3[0,2,−1,4]=[0,0,0,5][0, -6, 3, -7] + 3[0, 2, -1, 4] = [0, 0, 0, 5]

Interchange R3R_3 and R4R_4 to place the zero row at the bottom (R3↔R4R_3 \leftrightarrow R_4):

[130302−1400050000]\begin{bmatrix} 1 & 3 & 0 & 3 \\ 0 & 2 & -1 & 4 \\ 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 \end{bmatrix}

Step 2: Count the pivots.

  • Pivot in row 1 (column 1: entry 11)
  • Pivot in row 2 (column 2: entry 22)
  • Pivot in row 3 (column 4: entry 55)
  • Row 4 is all zeros: no pivot in row 4.

Conclusion: Matrix AA has only 3 pivot positions, not 4. By Theorem 4:

  1. The columns of AA do not span R4\mathbb{R}^4.
  2. The equation Ax=bA\mathbf{x} = \mathbf{b} does not have a solution for every b∈R4\mathbf{b} \in \mathbb{R}^4.