10.018 Modelling Space and Systems — Weeks 8–13 Final Exam Prep
Solve the following system of linear equations using Gauss-Jordan elimination:
\[\begin{cases} x + 2y + z = 9 \\ 2x + 5y + 3z = 23 \\ x + 3y + 2z = 14 \end{cases}\]Step 1: Write the augmented matrix.
\[\left[\begin{array}{ccc|c} 1 & 2 & 1 & 9 \\ 2 & 5 & 3 & 23 \\ 1 & 3 & 2 & 14 \end{array}\right]\]Step 2: \(R_2 \leftarrow R_2 - 2R_1\), \(R_3 \leftarrow R_3 - R_1\)
\[\left[\begin{array}{ccc|c} 1 & 2 & 1 & 9 \\ 0 & 1 & 1 & 5 \\ 0 & 1 & 1 & 5 \end{array}\right]\]Step 3: \(R_3 \leftarrow R_3 - R_2\)
\[\left[\begin{array}{ccc|c} 1 & 2 & 1 & 9 \\ 0 & 1 & 1 & 5 \\ 0 & 0 & 0 & 0 \end{array}\right]\]Step 4: Back-substitute. \(R_1 \leftarrow R_1 - 2R_2\)
\[\left[\begin{array}{ccc|c} 1 & 0 & -1 & -1 \\ 0 & 1 & 1 & 5 \\ 0 & 0 & 0 & 0 \end{array}\right]\]Step 5: Express solution.
Let \(z = t\) (free variable). Then:
\[x = -1 + t, \quad y = 5 - t, \quad z = t, \quad t \in \mathbb{R}\]In vector form:
\[\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -1 \\ 5 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\]For what values of the parameter \(a\) does the following system have (i) no solution, (ii) a unique solution, (iii) infinitely many solutions?
\[\begin{cases} x + 2y + z = 3 \\ 2x + 5y + az = a + 3 \\ x + 3y + (a-1)z = 2 \end{cases}\]Step 1: Form augmented matrix.
\[\left[\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 2 & 5 & a & a+3 \\ 1 & 3 & a-1 & 2 \end{array}\right]\]Step 2: \(R_2 \leftarrow R_2 - 2R_1\), \(R_3 \leftarrow R_3 - R_1\)
\[\left[\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 0 & 1 & a-2 & a-3 \\ 0 & 1 & a-2 & -1 \end{array}\right]\]Step 3: \(R_3 \leftarrow R_3 - R_2\)
\[\left[\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 0 & 1 & a-2 & a-3 \\ 0 & 0 & 0 & -1-(a-3) \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 0 & 1 & a-2 & a-3 \\ 0 & 0 & 0 & 2-a \end{array}\right]\]Step 4: Classify.
Case 1: If \(a \neq 2\), then row 3 reads \(0 = 2 - a \neq 0\), which is a contradiction. No solution.
Case 2: If \(a = 2\), row 3 becomes \(0 = 0\). The system reduces to:
\[\left[\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right]\]With \(a=2\): Row 2 gives \(y + 0z = -1\), so \(y = -1\). Let \(z = t\). Row 1: \(x = 3 - 2(-1) - t = 5 - t\).
Infinitely many solutions when \(a = 2\).
There is no value of \(a\) giving a unique solution (the coefficient matrix always has rank at most 2 when consistent).
Find the general solution of the homogeneous system:
\[\begin{cases} x_1 + 2x_2 - x_3 + 3x_4 = 0 \\ 2x_1 + 4x_2 + x_3 + 3x_4 = 0 \\ 3x_1 + 6x_2 + 2x_4 = 0 \end{cases}\]Step 1: Row reduce the coefficient matrix.
\[\left[\begin{array}{cccc} 1 & 2 & -1 & 3 \\ 2 & 4 & 1 & 3 \\ 3 & 6 & 0 & 2 \end{array}\right]\]\(R_2 \leftarrow R_2 - 2R_1\), \(R_3 \leftarrow R_3 - 3R_1\):
\[\left[\begin{array}{cccc} 1 & 2 & -1 & 3 \\ 0 & 0 & 3 & -3 \\ 0 & 0 & 3 & -7 \end{array}\right]\]Step 2: \(R_3 \leftarrow R_3 - R_2\)
\[\left[\begin{array}{cccc} 1 & 2 & -1 & 3 \\ 0 & 0 & 3 & -3 \\ 0 & 0 & 0 & -4 \end{array}\right]\]Step 3: Scale rows. \(R_2 \leftarrow \frac{1}{3}R_2\), \(R_3 \leftarrow -\frac{1}{4}R_3\)
\[\left[\begin{array}{cccc} 1 & 2 & -1 & 3 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \end{array}\right]\]Step 4: Back-substitute to RREF. \(R_2 \leftarrow R_2 + R_3\), \(R_1 \leftarrow R_1 - 3R_3\), then \(R_1 \leftarrow R_1 + R_2\):
\[\left[\begin{array}{cccc} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]\]Step 5: Read off solution.
Pivots in columns 1, 3, 4. Free variable: \(x_2 = s\).
\[x_1 = -2s, \quad x_2 = s, \quad x_3 = 0, \quad x_4 = 0\]General solution:
\[\mathbf{x} = s\begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \quad s \in \mathbb{R}\]The augmented matrix of a system has been reduced to:
\[\left[\begin{array}{ccc|c} 1 & 3 & -2 & 4 \\ 0 & k-2 & 1 & 3 \\ 0 & 0 & k(k-2) & k-2 \end{array}\right]\]For what values of \(k\) does the system have (i) a unique solution, (ii) infinitely many solutions, (iii) no solution?
Step 1: Identify when rows become zero or contradictory.
Row 3: \(k(k-2)x_3 = k - 2\).
Row 2 requires \(k - 2 \neq 0\) for a pivot, i.e., \(k \neq 2\).
Step 2: Case \(k \neq 2\) and \(k \neq 0\).
All three pivots exist (row 2 has pivot since \(k-2 \neq 0\), row 3 has pivot since \(k(k-2) \neq 0\)).
Unique solution.
Step 3: Case \(k = 0\).
Row 3 becomes: \(0 \cdot x_3 = 0 - 2 = -2\), i.e., \(0 = -2\). Contradiction.
No solution.
Step 4: Case \(k = 2\).
Row 2 becomes: \(0 \cdot x_2 + x_3 = 3\), so \(x_3 = 3\). Row 3 becomes: \(0 = 0\).
Only 2 pivots, with \(x_2\) free.
Infinitely many solutions.
Summary:
Solve the system:
\[\begin{cases} x + y + z + w = 4 \\ 2x + 3y + z + w = 7 \\ x + 2y + 2z + 3w = 9 \\ x + y + 3z + 4w = 10 \end{cases}\]Step 1: Augmented matrix.
\[\left[\begin{array}{cccc|c} 1 & 1 & 1 & 1 & 4 \\ 2 & 3 & 1 & 1 & 7 \\ 1 & 2 & 2 & 3 & 9 \\ 1 & 1 & 3 & 4 & 10 \end{array}\right]\]Step 2: \(R_2 \leftarrow R_2 - 2R_1\), \(R_3 \leftarrow R_3 - R_1\), \(R_4 \leftarrow R_4 - R_1\)
\[\left[\begin{array}{cccc|c} 1 & 1 & 1 & 1 & 4 \\ 0 & 1 & -1 & -1 & -1 \\ 0 & 1 & 1 & 2 & 5 \\ 0 & 0 & 2 & 3 & 6 \end{array}\right]\]Step 3: \(R_3 \leftarrow R_3 - R_2\)
\[\left[\begin{array}{cccc|c} 1 & 1 & 1 & 1 & 4 \\ 0 & 1 & -1 & -1 & -1 \\ 0 & 0 & 2 & 3 & 6 \\ 0 & 0 & 2 & 3 & 6 \end{array}\right]\]Step 4: \(R_4 \leftarrow R_4 - R_3\)
\[\left[\begin{array}{cccc|c} 1 & 1 & 1 & 1 & 4 \\ 0 & 1 & -1 & -1 & -1 \\ 0 & 0 & 2 & 3 & 6 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\]Step 5: RREF. \(R_3 \leftarrow \frac{1}{2}R_3\):
Row 3: \(z + \frac{3}{2}w = 3\). Let \(w = t\). Then \(z = 3 - \frac{3}{2}t\).
Row 2: \(y = -1 + z + w = -1 + 3 - \frac{3}{2}t + t = 2 - \frac{1}{2}t\).
Row 1: \(x = 4 - y - z - w = 4 - (2 - \frac{1}{2}t) - (3 - \frac{3}{2}t) - t = -1 + t\).
Step 6: Solution in vector form.
\[\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ 3 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ -1/2 \\ -3/2 \\ 1 \end{pmatrix}, \quad t \in \mathbb{R}\]Or multiplying the direction vector by 2:
\[\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ 3 \\ 0 \end{pmatrix} + s\begin{pmatrix} 2 \\ -1 \\ -3 \\ 2 \end{pmatrix}, \quad s \in \mathbb{R}\]The matrix \(A = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix}\) is written as a product of elementary matrices \(E_1 E_2 E_3\).
(a) Compute \(A\).
(b) Find \(A^{-1}\) using the factorisation.
Step 1: Identify the elementary matrices.
\(E_1 = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\) (add 3 times row 1 to row 2)
\(E_2 = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\) (scale row 1 by 2)
\(E_3 = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix}\) (add 4 times row 2 to row 1)
Step 2: Compute \(A = E_1 E_2 E_3\).
First: \(E_2 E_3 = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 8 \\ 0 & 1 \end{pmatrix}\)
Then: \(A = E_1 (E_2 E_3) = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 8 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 8 \\ 6 & 25 \end{pmatrix}\)
Step 3: Find \(A^{-1} = E_3^{-1} E_2^{-1} E_1^{-1}\).
\(E_1^{-1} = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}\), \(E_2^{-1} = \begin{pmatrix} 1/2 & 0 \\ 0 & 1 \end{pmatrix}\), \(E_3^{-1} = \begin{pmatrix} 1 & -4 \\ 0 & 1 \end{pmatrix}\)
\(E_2^{-1}E_1^{-1} = \begin{pmatrix} 1/2 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 1/2 & 0 \\ -3 & 1 \end{pmatrix}\)
\(A^{-1} = E_3^{-1}(E_2^{-1}E_1^{-1}) = \begin{pmatrix} 1 & -4 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1/2 & 0 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 25/2 & -4 \\ -3 & 1 \end{pmatrix}\)
Simplifying: \(A^{-1} = \frac{1}{2}\begin{pmatrix} 25 & -8 \\ -6 & 2 \end{pmatrix}\)
Verification:
\(\det(A) = 2(25) - 8(6) = 50 - 48 = 2\). So \(A^{-1} = \frac{1}{2}\begin{pmatrix} 25 & -8 \\ -6 & 2 \end{pmatrix}\). ✓
Find the inverse of \(A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 5 & 3 \\ 1 & 3 & 3 \end{pmatrix}\) using Gauss-Jordan elimination.
Step 1: Augment with identity.
\[\left[\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0 \\ 2 & 5 & 3 & 0 & 1 & 0 \\ 1 & 3 & 3 & 0 & 0 & 1 \end{array}\right]\]Step 2: \(R_2 \leftarrow R_2 - 2R_1\), \(R_3 \leftarrow R_3 - R_1\)
\[\left[\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & -2 & 1 & 0 \\ 0 & 1 & 2 & -1 & 0 & 1 \end{array}\right]\]Step 3: \(R_3 \leftarrow R_3 - R_2\)
\[\left[\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & -2 & 1 & 0 \\ 0 & 0 & 1 & 1 & -1 & 1 \end{array}\right]\]Step 4: Back-substitute. \(R_2 \leftarrow R_2 - R_3\), \(R_1 \leftarrow R_1 - R_3\)
\[\left[\begin{array}{ccc|ccc} 1 & 2 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & -3 & 2 & -1 \\ 0 & 0 & 1 & 1 & -1 & 1 \end{array}\right]\]Step 5: \(R_1 \leftarrow R_1 - 2R_2\)
\[\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 6 & -3 & 1 \\ 0 & 1 & 0 & -3 & 2 & -1 \\ 0 & 0 & 1 & 1 & -1 & 1 \end{array}\right]\]Result:
\[A^{-1} = \begin{pmatrix} 6 & -3 & 1 \\ -3 & 2 & -1 \\ 1 & -1 & 1 \end{pmatrix}\]Given \(A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) and \(B = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\), solve for \(X\) in the equation \(AXB = C\) where \(C = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}\).
Step 1: Isolate X.
\(AXB = C \implies X = A^{-1}CB^{-1}\)
Step 2: Find \(A^{-1}\).
\(A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\), \(\det(A) = 1\).
\(A^{-1} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}\)
Step 3: Find \(B^{-1}\).
\(B = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\), \(\det(B) = 1\).
\(B^{-1} = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}\)
Step 4: Compute \(A^{-1}C\).
\[A^{-1}C = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\]Step 5: Compute \(X = (A^{-1}C)B^{-1}\).
\[X = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\]Verification:
\(AXB = A \cdot I \cdot B = AB = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix} = C\). ✓
Suppose \(A\) is an \(n \times n\) matrix satisfying \(A^2 - 3A + 2I = 0\). Show that \(A\) is invertible and express \(A^{-1}\) in terms of \(A\).
Step 1: Rearrange the equation to isolate \(I\).
\[A^2 - 3A + 2I = 0\] \[2I = 3A - A^2\] \[2I = A(3I - A)\]Step 2: Factor to reveal the inverse.
\[I = A \cdot \frac{1}{2}(3I - A)\]This shows that \(A\) has a right inverse: \(A^{-1} = \frac{1}{2}(3I - A)\).
Step 3: Verify it is also a left inverse.
\[\frac{1}{2}(3I - A) \cdot A = \frac{1}{2}(3A - A^2) = \frac{1}{2}(3A - (3A - 2I)) = \frac{1}{2}(2I) = I \; \checkmark\](We used \(A^2 = 3A - 2I\) from the given equation.)
Answer:
\[\boxed{A^{-1} = \frac{1}{2}(3I - A)}\]Given that elementary row operations \(E_3 E_2 E_1 A = I\) where:
(a) Express \(A\) in terms of \(E_1, E_2, E_3\).
(b) Express \(A^{-1}\) in terms of \(E_1, E_2, E_3\).
(c) Find \(\det(A)\).
(a) From \(E_3 E_2 E_1 A = I\):
\[A = E_1^{-1} E_2^{-1} E_3^{-1}\](b) \(A^{-1} = E_3 E_2 E_1\)
(c) Compute \(\det(A)\).
\(\det(E_1) = -1\) (row swap)
\(\det(E_2) = \frac{1}{2}\) (row scaling)
\(\det(E_3) = 1\) (row addition)
From \(E_3 E_2 E_1 A = I\):
\[\det(E_3)\det(E_2)\det(E_1)\det(A) = 1\] \[(1)\left(\frac{1}{2}\right)(-1)\det(A) = 1\] \[\det(A) = -2\]Prove that if \(A\) is a symmetric invertible matrix (i.e., \(A^T = A\)), then \(A^{-1}\) is also symmetric.
Proof:
We want to show that \((A^{-1})^T = A^{-1}\).
Recall the property: \((A^{-1})^T = (A^T)^{-1}\).
Since \(A\) is symmetric, \(A^T = A\).
Therefore: \((A^{-1})^T = (A^T)^{-1} = A^{-1}\).
Hence \(A^{-1}\) is symmetric. \(\square\)
Justification of the property used:
To prove \((A^{-1})^T = (A^T)^{-1}\): Start from \(AA^{-1} = I\). Transpose both sides:
\[(AA^{-1})^T = I^T\] \[(A^{-1})^T A^T = I\]This shows \((A^{-1})^T\) is the inverse of \(A^T\), i.e., \((A^{-1})^T = (A^T)^{-1}\). \(\square\)
Compute the determinant of:
\[A = \begin{pmatrix} 2 & 0 & 1 & 3 \\ 1 & 2 & 0 & 1 \\ 0 & 3 & 1 & 0 \\ 1 & 0 & 2 & 1 \end{pmatrix}\]Step 1: Expand along Row 1 (has a zero, reduces work).
\[\det(A) = 2 \cdot C_{11} + 0 \cdot C_{12} + 1 \cdot C_{13} + 3 \cdot C_{14}\]where \(C_{ij} = (-1)^{i+j} M_{ij}\).
Step 2: Compute \(M_{11}\) (delete row 1, col 1).
\[M_{11} = \det\begin{pmatrix} 2 & 0 & 1 \\ 3 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix}\]Expand along row 1: \(= 2(1 \cdot 1 - 0 \cdot 2) - 0 + 1(3 \cdot 2 - 1 \cdot 0) = 2(1) + 1(6) = 8\)
\(C_{11} = (+1)(8) = 8\)
Step 3: Compute \(M_{13}\) (delete row 1, col 3).
\[M_{13} = \det\begin{pmatrix} 1 & 2 & 1 \\ 0 & 3 & 0 \\ 1 & 0 & 1 \end{pmatrix}\]Expand along row 2 (two zeros): \(= 3 \cdot (-1)^{2+2} \det\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = 3(0) = 0\)
\(C_{13} = (+1)(0) = 0\)
Step 4: Compute \(M_{14}\) (delete row 1, col 4).
\[M_{14} = \det\begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 1 \\ 1 & 0 & 2 \end{pmatrix}\]Expand along column 1: \(= 1 \cdot \det\begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix} - 0 + 1 \cdot (-1)^{3+1}\det\begin{pmatrix} 2 & 0 \\ 3 & 1 \end{pmatrix}\)
\(= 1(6 - 0) + 1(2 - 0) = 6 + 2 = 8\)
\(C_{14} = (-1)^{1+4}(8) = -8\)
Step 5: Combine.
\[\det(A) = 2(8) + 0 + 1(0) + 3(-8) = 16 - 24 = -8\]The matrix \(A\) is reduced to upper triangular form \(U\) via the following operations:
The resulting upper triangular matrix is \(U = \begin{pmatrix} 3 & 1 & 2 \\ 0 & -1 & 4 \\ 0 & 0 & 2 \end{pmatrix}\).
Find: (a) \(\det(A)\), (b) \(\det(3A)\), (c) \(\det(A^{-1})\), (d) \(\det(A^2)\).
Step 1: \(\det(U) = 3 \times (-1) \times 2 = -6\).
Step 2: Track effect of row operations on the determinant.
So \(\det(U) = (-1) \cdot \det(A)\), hence \(\det(A) = -\det(U) = -(-6) = 6\).
(a) \(\det(A) = 6\)
(b) \(\det(3A) = 3^3 \det(A) = 27 \times 6 = 162\)
(For an \(n \times n\) matrix: \(\det(cA) = c^n \det(A)\))
(c) \(\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{6}\)
(d) \(\det(A^2) = (\det A)^2 = 36\)
Determine which of the following sets are subspaces of \(\mathbb{R}^3\). Justify your answer.
(a) \(S_1\): YES, it is a subspace.
(b) \(S_2\): NO, not a subspace.
The zero vector \((0,0,0)\) does not satisfy \(0 + 0 - 0 = 1\). ✗
(c) \(S_3\): NO, not a subspace.
Not closed under addition. Counterexample: \((1, 0, 0) \in S_3\) (since \(1 \cdot 0 = 0\)) and \((0, 0, 1) \in S_3\) (since \(0 \cdot 1 = 0\)), but \((1, 0, 0) + (0, 0, 1) = (1, 0, 1)\) and \(1 \cdot 1 = 1 \neq 0\). ✗
(d) \(S_4\): YES, it is a subspace.
\(S_4 = \text{span}\{(2, -1, 3)\}\), which is the span of a single vector, hence a subspace (a line through the origin). ✓
Determine if the following vectors are linearly independent:
\[\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix} 2 \\ 3 \\ 0 \end{pmatrix}, \quad \mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\]Step 1: Form the matrix with these vectors as columns and row reduce (or compute the determinant).
\[\det\begin{pmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 0 & -1 \end{pmatrix}\]Step 2: Expand along row 1.
\[= 1\det\begin{pmatrix} 3 & 1 \\ 0 & -1 \end{pmatrix} - 2\det\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} + 1\det\begin{pmatrix} 2 & 3 \\ 1 & 0 \end{pmatrix}\] \[= 1(-3 - 0) - 2(-2 - 1) + 1(0 - 3)\] \[= -3 - 2(-3) + (-3)\] \[= -3 + 6 - 3 = 0\]Step 3: Conclusion.
Since \(\det = 0\), the vectors are linearly dependent.
To find the relation: note that \(\mathbf{v}_3 = \mathbf{v}_1 - \mathbf{v}_2 + 0\)... Let's check: \((1,2,1) - (2,3,0) = (-1,-1,1) = -\mathbf{v}_3\). So \(\mathbf{v}_1 - \mathbf{v}_2 + \mathbf{v}_3 = (0,0,0)\).
Express \(\mathbf{b} = \begin{pmatrix} 5 \\ 3 \\ 7 \end{pmatrix}\) as a linear combination of \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\), \(\mathbf{v}_2 = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\), \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\), or show it is impossible.
Step 1: Solve \(c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = \mathbf{b}\).
\[\left[\begin{array}{ccc|c} 1 & 2 & 1 & 5 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 1 & 7 \end{array}\right]\]Step 2: \(R_2 \leftarrow R_2 - R_1\), \(R_3 \leftarrow R_3 - 2R_1\).
\[\left[\begin{array}{ccc|c} 1 & 2 & 1 & 5 \\ 0 & -1 & -1 & -2 \\ 0 & -3 & -1 & -3 \end{array}\right]\]Step 3: \(R_3 \leftarrow R_3 - 3R_2\).
\[\left[\begin{array}{ccc|c} 1 & 2 & 1 & 5 \\ 0 & -1 & -1 & -2 \\ 0 & 0 & 2 & 3 \end{array}\right]\]Step 4: Back-substitute.
Row 3: \(2c_3 = 3 \implies c_3 = \frac{3}{2}\)
Row 2: \(-c_2 - c_3 = -2 \implies c_2 = 2 - c_3 = \frac{1}{2}\)
Row 1: \(c_1 + 2c_2 + c_3 = 5 \implies c_1 = 5 - 1 - \frac{3}{2} = \frac{5}{2}\)
Answer:
\[\mathbf{b} = \frac{5}{2}\mathbf{v}_1 + \frac{1}{2}\mathbf{v}_2 + \frac{3}{2}\mathbf{v}_3\]Verification: \(\frac{5}{2}(1,1,2) + \frac{1}{2}(2,1,1) + \frac{3}{2}(1,0,1) = (\frac{5}{2}+1+\frac{3}{2}, \frac{5}{2}+\frac{1}{2}, 5+\frac{1}{2}+\frac{3}{2}) = (5, 3, 7)\). ✓
Prove that there are no \(2 \times 2\) real matrices \(A\) and \(B\) satisfying \(AB - BA = I\).
Proof using the trace:
Suppose for contradiction that \(AB - BA = I\) for some \(2 \times 2\) matrices \(A, B\).
Take the trace of both sides:
\[\text{tr}(AB - BA) = \text{tr}(I)\] \[\text{tr}(AB) - \text{tr}(BA) = 2\]Key property:
By the cyclic property of trace: \(\text{tr}(AB) = \text{tr}(BA)\) for any matrices \(A, B\).
Therefore the left side equals \(\text{tr}(AB) - \text{tr}(AB) = 0\).
Contradiction:
We get \(0 = 2\), which is a contradiction.
Hence no such matrices \(A, B\) exist. \(\square\)
Remark:
This argument works for any \(n \times n\) matrices (the RHS would be \(n\)), so \(AB - BA = I\) has no solution in \(M_{n \times n}(\mathbb{R})\) for any \(n\).
Let \(A = \begin{pmatrix} 1 & 2 & 0 & 1 \\ 2 & 4 & 1 & 3 \\ 3 & 6 & 1 & 4 \end{pmatrix}\) with RREF \(R = \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}\).
Find bases for: (a) \(\text{Row}(A)\), (b) \(\text{Col}(A)\), (c) \(\text{Null}(A)\). (d) Verify the rank-nullity theorem.
(a) Basis for Row(A):
The nonzero rows of the RREF form a basis for the row space:
\[\{(1, 2, 0, 1),\; (0, 0, 1, 1)\}\](b) Basis for Col(A):
Pivots are in columns 1 and 3. Take the corresponding columns of the original matrix \(A\):
\[\left\{\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix},\; \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\right\}\](c) Basis for Null(A):
Free variables: \(x_2 = s\), \(x_4 = t\). From RREF:
Row 2: \(x_3 = -t\)
Row 1: \(x_1 = -2s - t\)
\[\mathbf{x} = s\begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + t\begin{pmatrix} -1 \\ 0 \\ -1 \\ 1 \end{pmatrix}\]Basis for Null(A): \(\left\{\begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix},\; \begin{pmatrix} -1 \\ 0 \\ -1 \\ 1 \end{pmatrix}\right\}\)
(d) Rank-Nullity Theorem verification:
\(\text{rank}(A) + \text{nullity}(A) = 2 + 2 = 4 = n\) (number of columns). ✓
Determine whether \(S = \left\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\right\}\) is a basis for \(\mathbb{R}^3\).
Step 1: A set of 3 vectors is a basis for \(\mathbb{R}^3\) iff the vectors are linearly independent (equivalently, the matrix formed has nonzero determinant).
\[\det\begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}\]Step 2: Compute the determinant.
Expand along row 1:
\[= 1(1 \cdot 1 - 0 \cdot 1) - 0 + 1(1 \cdot 1 - 1 \cdot 0) = 1(1) + 1(1) = 2\]Conclusion:
Since \(\det = 2 \neq 0\), the vectors are linearly independent and hence form a basis for \(\mathbb{R}^3\).
Let \(B = \left\{\mathbf{b}_1, \mathbf{b}_2\right\}\) where \(\mathbf{b}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\), \(\mathbf{b}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\).
Given \(\mathbf{x} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}\) (in standard coordinates), find \([\mathbf{x}]_B\).
Step 1: We need \(c_1, c_2\) such that \(c_1 \mathbf{b}_1 + c_2 \mathbf{b}_2 = \mathbf{x}\).
\[\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}\]Step 2: Solve. The change-of-basis matrix is \(P = [\mathbf{b}_1 | \mathbf{b}_2]\). We need \(P^{-1}\mathbf{x}\).
\(\det(P) = -1 - 1 = -2\).
\[P^{-1} = \frac{1}{-2}\begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}\]Step 3: Compute \([\mathbf{x}]_B = P^{-1}\mathbf{x}\).
\[[\mathbf{x}]_B = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}\begin{pmatrix} 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}\]Verification:
\(4\begin{pmatrix} 1 \\ 1 \end{pmatrix} + 1\begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}\). ✓
Let \(\mathcal{A} = \{\mathbf{a}_1, \mathbf{a}_2\}\) where \(\mathbf{a}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\), \(\mathbf{a}_2 = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\), and let \(\mathcal{B} = \{\mathbf{b}_1, \mathbf{b}_2\}\) where \(\mathbf{b}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\), \(\mathbf{b}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\).
If \([\mathbf{x}]_\mathcal{A} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\), find \([\mathbf{x}]_\mathcal{B}\).
Step 1: Find \(\mathbf{x}\) in standard coordinates.
\[\mathbf{x} = 2\mathbf{a}_1 + 1\mathbf{a}_2 = 2\begin{pmatrix} 1 \\ 2 \end{pmatrix} + 1\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}\]Step 2: Find \([\mathbf{x}]_\mathcal{B}\) by solving \(P_B [\mathbf{x}]_\mathcal{B} = \mathbf{x}\) where \(P_B = [\mathbf{b}_1 | \mathbf{b}_2]\).
\[\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}\]Step 3: Solve by back-substitution.
Row 2: \(c_2 = 5\).
Row 1: \(c_1 + 5 = 5 \implies c_1 = 0\).
Answer:
\[[\mathbf{x}]_\mathcal{B} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}\]Verification: \(0\begin{pmatrix} 1 \\ 0 \end{pmatrix} + 5\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}\). ✓
Alternative (transition matrix method):
\(P_{\mathcal{A} \to \mathcal{B}} = P_B^{-1} P_A\) where \(P_A = \begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix}\), \(P_B^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}\).
\[P_{\mathcal{A} \to \mathcal{B}} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 2 & 1 \end{pmatrix}\] \[[\mathbf{x}]_\mathcal{B} = \begin{pmatrix} -1 & 2 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \end{pmatrix} \; \checkmark\]Check whether the following set is (i) orthogonal, (ii) orthonormal:
\[S = \left\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix}\right\}\]Step 1: Check orthogonality (all dot products between distinct vectors must be zero).
\(\mathbf{v}_1 \cdot \mathbf{v}_2 = (1)(-1) + (1)(1) + (0)(0) = -1 + 1 + 0 = 0\) ✓
\(\mathbf{v}_1 \cdot \mathbf{v}_3 = (1)(0) + (1)(0) + (0)(2) = 0\) ✓
\(\mathbf{v}_2 \cdot \mathbf{v}_3 = (-1)(0) + (1)(0) + (0)(2) = 0\) ✓
The set is orthogonal.
Step 2: Check orthonormality (each vector must have unit length).
\(\|\mathbf{v}_1\| = \sqrt{1 + 1 + 0} = \sqrt{2} \neq 1\)
The set is not orthonormal.
Step 3: To make it orthonormal, normalise each vector:
\[\hat{\mathbf{v}}_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \quad \hat{\mathbf{v}}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}, \quad \hat{\mathbf{v}}_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\]Let \(A = \begin{pmatrix} 1 & 3 & 2 \\ 2 & 6 & 5 \\ 1 & 3 & 3 \end{pmatrix}\) with RREF: \(\begin{pmatrix} 1 & 3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}\).
Find the rank, nullity, and bases for all four fundamental subspaces of \(A\).
Rank and Nullity:
\(\text{rank}(A) = 2\) (two pivots), \(\text{nullity}(A) = 3 - 2 = 1\).
1. Row Space \(\text{Row}(A) \subseteq \mathbb{R}^3\):
Nonzero rows of RREF: \(\{(1, 3, 0),\; (0, 0, 1)\}\). Dimension = 2.
2. Column Space \(\text{Col}(A) \subseteq \mathbb{R}^3\):
Pivots in columns 1 and 3. Take columns 1 and 3 of original \(A\):
\[\left\{\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix},\; \begin{pmatrix} 2 \\ 5 \\ 3 \end{pmatrix}\right\}\]Dimension = 2.
3. Null Space \(\text{Null}(A) \subseteq \mathbb{R}^3\):
Free variable: \(x_2 = t\). From RREF: \(x_3 = 0\), \(x_1 = -3t\).
\[\text{Null}(A) = \text{span}\left\{\begin{pmatrix} -3 \\ 1 \\ 0 \end{pmatrix}\right\}\]Dimension = 1.
4. Left Null Space \(\text{Null}(A^T) \subseteq \mathbb{R}^3\):
Dimension = \(m - \text{rank}(A) = 3 - 2 = 1\).
Solve \(A^T \mathbf{y} = \mathbf{0}\): row reduce \(A^T = \begin{pmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 5 & 3 \end{pmatrix}\).
\(R_2 \leftarrow R_2 - 3R_1\), \(R_3 \leftarrow R_3 - 2R_1\): \(\begin{pmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \end{pmatrix}\)
Swap \(R_2 \leftrightarrow R_3\): \(\begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}\)
Free: \(y_3 = t\). Then \(y_2 = -t\), \(y_1 = -2(-t) - t = t\).
\[\text{Null}(A^T) = \text{span}\left\{\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\right\}\]Find the eigenvalues and eigenvectors of \(A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}\).
Step 1: Characteristic equation \(\det(A - \lambda I) = 0\).
\[\det\begin{pmatrix} 4-\lambda & 2 \\ 1 & 3-\lambda \end{pmatrix} = (4-\lambda)(3-\lambda) - 2 = 0\] \[\lambda^2 - 7\lambda + 12 - 2 = 0\] \[\lambda^2 - 7\lambda + 10 = 0\] \[(\lambda - 5)(\lambda - 2) = 0\]Eigenvalues: \(\lambda_1 = 5\), \(\lambda_2 = 2\).
Step 2: Eigenvector for \(\lambda_1 = 5\).
\[A - 5I = \begin{pmatrix} -1 & 2 \\ 1 & -2 \end{pmatrix}\]Row reduce: \(R_2 \leftarrow R_2 + R_1\): \(\begin{pmatrix} -1 & 2 \\ 0 & 0 \end{pmatrix}\).
\(-x_1 + 2x_2 = 0 \implies x_1 = 2x_2\). Let \(x_2 = 1\).
\[\mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\]Step 3: Eigenvector for \(\lambda_2 = 2\).
\[A - 2I = \begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix}\]Row reduce: \(\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\).
\(x_1 + x_2 = 0 \implies x_1 = -x_2\). Let \(x_2 = 1\).
\[\mathbf{v}_2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}\]Find the eigenvalues and eigenspaces of \(A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 3 \\ 0 & 0 & 5 \end{pmatrix}\).
Step 1: For a triangular matrix, eigenvalues are the diagonal entries.
Eigenvalues: \(\lambda_1 = 2\) (algebraic multiplicity 2), \(\lambda_2 = 5\) (algebraic multiplicity 1).
Step 2: Eigenspace for \(\lambda = 2\).
\[A - 2I = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 3 \\ 0 & 0 & 3 \end{pmatrix} \xrightarrow{\text{RREF}} \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}\]From RREF: \(x_2 = 0\), \(x_3 = 0\), \(x_1\) is free.
\[E_2 = \text{span}\left\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\right\}\]Geometric multiplicity = 1.
Step 3: Eigenspace for \(\lambda = 5\).
\[A - 5I = \begin{pmatrix} -3 & 1 & 0 \\ 0 & -3 & 3 \\ 0 & 0 & 0 \end{pmatrix}\]From row 2: \(-3x_2 + 3x_3 = 0 \implies x_2 = x_3\). Let \(x_3 = 3\), then \(x_2 = 3\).
From row 1: \(-3x_1 + x_2 = 0 \implies x_1 = 1\).
\[E_5 = \text{span}\left\{\begin{pmatrix} 1 \\ 3 \\ 3 \end{pmatrix}\right\}\]Determine if \(A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & -2 & 3 \end{pmatrix}\) is diagonalisable. If yes, find \(P\) and \(D\) such that \(A = PDP^{-1}\).
Step 1: Characteristic polynomial.
\[\det(A - \lambda I) = (1-\lambda)\det\begin{pmatrix} 3-\lambda & -2 \\ -2 & 3-\lambda \end{pmatrix}\] \[= (1-\lambda)[(3-\lambda)^2 - 4] = (1-\lambda)(\lambda^2 - 6\lambda + 5)\] \[= (1-\lambda)(5-\lambda)(1-\lambda) = -(\ \lambda-1)^2(\lambda - 5)\]Eigenvalues: \(\lambda_1 = 1\) (mult. 2), \(\lambda_2 = 5\) (mult. 1).
Step 2: Eigenspace for \(\lambda = 1\).
\[A - I = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 2 & -2 \\ 0 & -2 & 2 \end{pmatrix} \xrightarrow{\text{RREF}} \begin{pmatrix} 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\]Free variables: \(x_1 = s\), \(x_3 = t\). Then \(x_2 = t\).
\[E_1 = \text{span}\left\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\right\}\]Geometric multiplicity = 2 = algebraic multiplicity. ✓
Step 3: Eigenspace for \(\lambda = 5\).
\[A - 5I = \begin{pmatrix} -4 & 0 & 0 \\ 0 & -2 & -2 \\ 0 & -2 & -2 \end{pmatrix} \xrightarrow{\text{RREF}} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}\]\(x_1 = 0\), \(x_2 = -x_3\). Let \(x_3 = 1\):
\[E_5 = \text{span}\left\{\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}\right\}\]Step 4: Since geometric mult. = algebraic mult. for all eigenvalues, \(A\) is diagonalisable.
\[P = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 1 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5 \end{pmatrix}\]Let \(A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}\). Compute \(A^5\) using diagonalisation.
Step 1: Eigenvalues (triangular matrix).
\(\lambda_1 = 3\), \(\lambda_2 = 2\).
Step 2: Eigenvectors.
For \(\lambda = 3\): \(A - 3I = \begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}\). So \(x_2 = 0\), \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\).
For \(\lambda = 2\): \(A - 2I = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\). So \(x_1 = -x_2\), \(\mathbf{v}_2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}\).
Step 3: \(A = PDP^{-1}\) where:
\[P = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}\] \[P^{-1} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\]Step 4: \(A^5 = PD^5P^{-1}\).
\[D^5 = \begin{pmatrix} 3^5 & 0 \\ 0 & 2^5 \end{pmatrix} = \begin{pmatrix} 243 & 0 \\ 0 & 32 \end{pmatrix}\] \[A^5 = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 243 & 0 \\ 0 & 32 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\]First: \(PD^5 = \begin{pmatrix} 243 & -32 \\ 0 & 32 \end{pmatrix}\)
Then: \(A^5 = \begin{pmatrix} 243 & -32 \\ 0 & 32 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 243 & 243 - 32 \\ 0 & 32 \end{pmatrix} = \begin{pmatrix} 243 & 211 \\ 0 & 32 \end{pmatrix}\]
A city has two weather states: Sunny (S) and Rainy (R). The transition probabilities are:
(a) Write the transition matrix. (b) Find the long-run (steady-state) probability distribution.
(a) Transition matrix (columns are probability vectors):
\[P = \begin{pmatrix} 0.8 & 0.4 \\ 0.2 & 0.6 \end{pmatrix}\](Column 1: from Sunny. Column 2: from Rainy.)
(b) Find steady state \(\mathbf{q}\) satisfying \(P\mathbf{q} = \mathbf{q}\), i.e., \((P - I)\mathbf{q} = \mathbf{0}\).
\[P - I = \begin{pmatrix} -0.2 & 0.4 \\ 0.2 & -0.4 \end{pmatrix}\]Row 1: \(-0.2q_1 + 0.4q_2 = 0 \implies q_1 = 2q_2\).
Step 3: Use constraint \(q_1 + q_2 = 1\).
\(2q_2 + q_2 = 1 \implies q_2 = \frac{1}{3}\), \(q_1 = \frac{2}{3}\).
Answer:
Steady-state distribution: \(\mathbf{q} = \begin{pmatrix} 2/3 \\ 1/3 \end{pmatrix}\).
In the long run, it is sunny \(\frac{2}{3}\) of the time and rainy \(\frac{1}{3}\) of the time.
A \(4 \times 4\) matrix \(B\) has characteristic polynomial \((\lambda - 1)^2(\lambda + 2)^2\). The eigenspace for \(\lambda = 1\) has dimension 1, and the eigenspace for \(\lambda = -2\) has dimension 2. Is \(B\) diagonalisable?
Step 1: Check the condition for diagonalisability.
A matrix is diagonalisable if and only if, for every eigenvalue, the geometric multiplicity equals the algebraic multiplicity.
Step 2: Compare multiplicities.
Conclusion:
\(B\) is NOT diagonalisable because the geometric multiplicity for \(\lambda = 1\) is less than its algebraic multiplicity.
Prove: If \(H^2 = H\) (i.e., \(H\) is idempotent), then the only possible eigenvalues of \(H\) are 0 and 1.
Proof:
Let \(\lambda\) be an eigenvalue of \(H\) with eigenvector \(\mathbf{v} \neq \mathbf{0}\), so \(H\mathbf{v} = \lambda \mathbf{v}\).
Step 1: Apply \(H\) to both sides.
\[H^2 \mathbf{v} = H(H\mathbf{v}) = H(\lambda \mathbf{v}) = \lambda H\mathbf{v} = \lambda^2 \mathbf{v}\]Step 2: Use \(H^2 = H\).
\[H^2 \mathbf{v} = H\mathbf{v} = \lambda \mathbf{v}\]Step 3: Equate the two expressions.
\[\lambda^2 \mathbf{v} = \lambda \mathbf{v}\] \[(\lambda^2 - \lambda)\mathbf{v} = \mathbf{0}\]Since \(\mathbf{v} \neq \mathbf{0}\):
\[\lambda^2 - \lambda = 0 \implies \lambda(\lambda - 1) = 0\] \[\lambda = 0 \text{ or } \lambda = 1\]Conclusion:
The only possible eigenvalues of an idempotent matrix are 0 and 1. \(\square\)
Determine which of the following functions are linear transformations:
(a) \(T_1\): YES, linear.
Can be written as matrix multiplication: \(T_1(\mathbf{x}) = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}\). Any function of the form \(T(\mathbf{x}) = A\mathbf{x}\) is linear.
(b) \(T_2\): NO, not linear.
\(T_2(\mathbf{0}) = T_2(0,0) = (1, 0) \neq (0,0)\). A linear transformation must map the zero vector to zero.
(c) \(T_3\): NO, not linear.
Check additivity: \(T_3(1,1) + T_3(1,1) = 1 + 1 = 2\), but \(T_3(2,2) = 4 \neq 2\). Fails.
Alternatively, \(T_3(c \cdot (1,1)) = T_3(c,c) = c^2 \neq c \cdot T_3(1,1) = c\) for general \(c\).
(d) \(T_4\): YES, linear.
\(T_4(\mathbf{x}) = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 2 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\). Matrix multiplication, so linear.
A linear transformation \(T: \mathbb{R}^3 \to \mathbb{R}^2\) is defined by \(T(x, y, z) = (x + 2y - z,\; 3x - y + 2z)\). Find the standard matrix \([T]\).
Step 1: The standard matrix has columns \(T(\mathbf{e}_1)\), \(T(\mathbf{e}_2)\), \(T(\mathbf{e}_3)\).
\(T(\mathbf{e}_1) = T(1,0,0) = (1, 3)\)
\(T(\mathbf{e}_2) = T(0,1,0) = (2, -1)\)
\(T(\mathbf{e}_3) = T(0,0,1) = (-1, 2)\)
Step 2: Form the matrix.
\[[T] = \begin{pmatrix} 1 & 2 & -1 \\ 3 & -1 & 2 \end{pmatrix}\]Verification:
\[[T]\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x + 2y - z \\ 3x - y + 2z \end{pmatrix} \; \checkmark\]Find the matrix representing the composite transformation: first reflect across the \(y\)-axis, then rotate \(60°\) counter-clockwise about the origin.
Step 1: Reflection across the \(y\)-axis.
\[M_{\text{ref}} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\]Step 2: Rotation by \(60°\) CCW.
\[R_{60°} = \begin{pmatrix} \cos 60° & -\sin 60° \\ \sin 60° & \cos 60° \end{pmatrix} = \begin{pmatrix} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix}\]Step 3: Composite = (second transformation) × (first transformation).
\[T = R_{60°} \cdot M_{\text{ref}} = \begin{pmatrix} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\] \[= \begin{pmatrix} -1/2 & -\sqrt{3}/2 \\ -\sqrt{3}/2 & 1/2 \end{pmatrix}\]Verification:
Check: \(\mathbf{e}_1 = (1,0) \xrightarrow{\text{reflect}} (-1,0) \xrightarrow{\text{rotate 60°}} (-\cos 60°, -\sin 60°) = (-1/2, -\sqrt{3}/2)\).
Column 1 of our matrix: \((-1/2, -\sqrt{3}/2)\). ✓
Let \(L\) be the line through the origin in direction \(\mathbf{u} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\).
(a) Find the projection matrix \(P\) that projects onto \(L\).
(b) Find the projection of \(\mathbf{b} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\) onto \(L\).
(c) Find the shortest distance from \(\mathbf{b}\) to \(L\).
(a) Projection matrix formula:
\[P = \frac{\mathbf{u}\mathbf{u}^T}{\mathbf{u}^T\mathbf{u}}\]\(\mathbf{u}^T\mathbf{u} = 1 + 4 = 5\)
\(\mathbf{u}\mathbf{u}^T = \begin{pmatrix} 1 \\ 2 \end{pmatrix}(1 \; 2) = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}\)
\[P = \frac{1}{5}\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}\](b) Projection of \(\mathbf{b}\):
\[\text{proj}_L \mathbf{b} = P\mathbf{b} = \frac{1}{5}\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 5 \\ 10 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\](c) Shortest distance = \(\|\mathbf{b} - \text{proj}_L \mathbf{b}\|\):
\[\mathbf{b} - \text{proj}_L \mathbf{b} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}\] \[\text{distance} = \sqrt{4 + 1} = \sqrt{5}\]Find the least squares line \(y = c_0 + c_1 x\) that best fits the data points: \((0, 1)\), \((1, 2)\), \((2, 4)\), \((3, 5)\).
Step 1: Set up the system \(A\mathbf{c} = \mathbf{y}\) (overdetermined).
\[A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 1 & 2 \\ 1 & 3 \end{pmatrix}, \quad \mathbf{y} = \begin{pmatrix} 1 \\ 2 \\ 4 \\ 5 \end{pmatrix}, \quad \mathbf{c} = \begin{pmatrix} c_0 \\ c_1 \end{pmatrix}\]Step 2: Form the normal equations \(A^T A \mathbf{c} = A^T \mathbf{y}\).
\[A^T A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 1 & 2 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 6 & 14 \end{pmatrix}\] \[A^T \mathbf{y} = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 12 \\ 25 \end{pmatrix}\]Step 3: Solve the \(2 \times 2\) system.
\[\begin{cases} 4c_0 + 6c_1 = 12 \\ 6c_0 + 14c_1 = 25 \end{cases}\]From equation 1: \(c_0 = 3 - \frac{3}{2}c_1\). Substitute into equation 2:
\[6(3 - \frac{3}{2}c_1) + 14c_1 = 25\] \[18 - 9c_1 + 14c_1 = 25\] \[5c_1 = 7 \implies c_1 = \frac{7}{5}\] \[c_0 = 3 - \frac{3}{2} \cdot \frac{7}{5} = 3 - \frac{21}{10} = \frac{9}{10}\]Answer:
\[\boxed{y = \frac{9}{10} + \frac{7}{5}x}\]Or equivalently: \(y = 0.9 + 1.4x\).
A linear transformation \(T: \mathbb{R}^2 \to \mathbb{R}^2\) satisfies:
\[T\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}, \quad T\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 7 \end{pmatrix}\]Find the standard matrix \([T]\).
Step 1: Use the formula \([T] = B \cdot A^{-1}\) where:
\(A\) = matrix of inputs (as columns): \(A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}\)
\(B\) = matrix of outputs (as columns): \(B = \begin{pmatrix} 3 & 4 \\ 5 & 7 \end{pmatrix}\)
Logic: \([T] \cdot A = B\), so \([T] = BA^{-1}\).
Step 2: Find \(A^{-1}\).
\(\det(A) = 2 - 1 = 1\).
\[A^{-1} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}\]Step 3: Compute \([T] = BA^{-1}\).
\[[T] = \begin{pmatrix} 3 & 4 \\ 5 & 7 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 6-4 & -3+4 \\ 10-7 & -5+7 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\]Verification:
\([T]\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}\) ✓
\([T]\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2+2 \\ 3+4 \end{pmatrix} = \begin{pmatrix} 4 \\ 7 \end{pmatrix}\) ✓