From grade 10: solve \(x + 2y = 5\) and \(3x - y = 1\). Multiply the second by 2, add, kill a variable:
That is 100% of linear algebra's core move. Everything in this course is doing exactly this to bigger systems, faster, with better bookkeeping (matrices), and then asking deeper questions (determinants, invertibility, vectors).
Same system: solve one equation for one variable and sub it in. \(3x - y = 1\) β \(y = 3x - 1\). Put into the first: \(x + 2(3x-1) = 5\) β \(7x = 7\) β \(x=1, y=2\) β same answer. Substitution works but scales terribly β 8 variables Γ 8 equations would be torture. Elimination scales. That's why Gauss exists.
a) Solve: \(2x + 3y = 12\), \(x - y = 1\)
b) Are \((2,3)\) and \((-4,-6)\) on the same line through the origin?
c) Simplify: \(\frac{a^2b^{-3}}{ab^{-1}}\)
a) sub \(x = 1+y\): \(2(1+y)+3y=12\) β \(5y=10\) β \(\boxed{x=3,y=2}\)
b) slopes: \(3/2\) vs \(-6/-4 = 1.5\) β yes, both on \(y = 1.5x\) (one is a multiple of the other β that's "linear dependence" hiding in grade 9)
c) \(= a^1 b^{-2} = \frac{a}{b^2}\)
The Oct 5 announcement says Ch 3 "is partly high school review from your Calc+Vectors course β going through it very quickly." So vectors (dot product, length, projections) are the part where you should already be in shape. The exam-weighted NEW material is Chapters 1β2 β that's where the marks are.
\(a_1x_1 + a_2x_2 + \dots + a_nx_n = b\) β variables only to power 1, no \(x_1x_2\), no \(\sqrt{x}\), no \(\sin(x)\). Coefficients \(a_i\) and constant \(b\) can be any real number (including 0 and negatives).
Tell-tale test: if you can't write it in that form, it's nonlinear. \(9x_1x_3 - 3x_2 = 6x_4\) is NOT linear (product of two variables). \(7x_1 + \pi x_2 = x_3 - 9\) IS linear (\(\pi\) is just a coefficient).
Theorem: every linear system has no solution, exactly one solution, or infinitely many. Never 2, never 17.
| Case | 2-var picture | 3-var picture |
|---|---|---|
| Unique solution | lines cross once | planes meet at one point |
| No solution (inconsistent) | parallel lines | planes with no common point |
| Infinitely many | same line | planes share a line/plane |
Examiner's favorite: "A system with exactly 2 solutions exists." FALSE β that middle option list is exhaustive, and the proof (Ax = b lecture 3) shows two solutions βΉ infinitely many via the free-parameter mechanism.
Rows = equations. Columns = variables + the last column = constants. The bar (written or imagined) separates A from b.
$$\begin{cases}x_1 - 2x_2 + 3x_3 = 2\\ 3x_1 + x_2 = 4\\ x_2 - 2x_3 = -1\end{cases} \;\longrightarrow\; \begin{bmatrix}1 & -2 & 3 & 2\\ 3 & 1 & 0 & 4\\ 0 & 1 & -2 & -1\end{bmatrix}$$
Lecture-1 skill: transcribe every system into this form fast. Missing variables get a 0 entry (the \(x_3\) in eq 2, the \(x_1\) in eq 3).
These never change the solution set β they only make it easier to see. (For determinants later, note: these same moves DO change det β different rules apply there!)
$$\begin{bmatrix}1 & 2 & 8 & 3\\ 0 & 1 & 5 & -2\\ 0 & 0 & 1 & 7\end{bmatrix}$$
This "solve bottom-up" move is called back-substitution. REF matrices (next section) always allow it.
Is \((1, 1, 1)\) a solution of \(x_1 - 2x_2 + 3x_3 = 2\), \(3x_1 + x_2 = 4\), \(x_2 - 2x_3 = -1\)?
Eq 1: \(1 - 2 + 3 = 2\) β Eq 2: \(3 + 1 = 4\) β Eq 3: \(1 - 2 = -1\) β β all three satisfied β YES, a solution.
Verify = plug into EVERY equation. Satisfying 2 of 3 equations means nothing.
a) Write as augmented matrix: \(2x_1 + x_3 = 4\), \(x_1 - x_2 + x_3 = 0\), \(3x_2 = 1\)
b) Which are linear? (i) \(5x_1 - 6x_2 + 10x_3 = 8 + x_4\) (ii) \(x_1^2 + x_2 = 3\) (iii) \(\frac{x_1}{2} + x_2 = 7\)
Answers: a) \(\begin{bmatrix}2&0&1&4\\1&-1&1&0\\0&3&0&1\end{bmatrix}\) Β· b) (i) YES (move \(x_4\) left: linear), (ii) NO β squared variable, (iii) YES β \(\frac{1}{2}\) is just a coefficient.
A matrix is in row echelon form (REF) if:
Add the 4th condition and it's RREF:
RREF is UNIQUE for every matrix. REF is not unique β two different people row-reducing can get different REFs, but the same RREF. (They love this as T/F.)
Gaussian = stop at REF + back-substitute. Gauss-Jordan = go all the way to RREF and read the answers off. Same answers, Jordan does more row-ops up front, no back-sub at the end.
Solve: \(2x_1 + 3x_2 - x_3 = -3\), \(\;x_1 + x_2 = 0\), \(\;3x_1 + x_3 = 11\)
Note the trick: swap first to get a 1 β saves you fraction arithmetic everywhere. Always scan for an easy 1 before dividing rows.
Row-reduce, then read the LAST rows:
| You see | Meaning |
|---|---|
| a row like \([0\;0\;0\;|\;c]\), \(c \neq 0\) | 0 = 5? contradiction β NO solutions |
| every variable has a leading 1, no junk rows | exactly one solution |
| β₯ 1 variable has NO leading 1 (a free variable) | infinitely many β set free vars = parameters \(t, s\) |
Free variable = a column with no pivot. Number of parameters = number of free variables.
$$\begin{bmatrix}1 & 2 & -1 & 5 & -7 & 1\\ 0 & 1 & 5 & 0 & 0 & 2\end{bmatrix}$$ (system of 2 eqs, 5 unknowns, already RREF-ish)
a) Is \(\begin{bmatrix}1&0&2&0&1\\0&1&0&1&0\\0&0&1&0&0\\0&0&0&0&0\end{bmatrix}\) REF? RREF? (check each of the 4 properties)
b) A system row-reduces to \(\begin{bmatrix}1&0&3&|&0\\0&1&-2&|&0\\0&0&0&|&0\end{bmatrix}\). How many solutions? Write the general solution.
Answers: a) It has REF structure (leading 1s staircase, zero row at bottom) but column 3's leading 1 has a nonzero (the 2) above it β property 4 fails β REF only, not RREF.
b) Zero row = consistent. \(x_3\) is free (no pivot). Infinitely many: \(x_3 = t\) β \(\boxed{x_1 = -3t,\; x_2 = 2t,\; x_3 = t}\). Also note: all constants were 0 β this was a homogeneous system β this is exactly its nontrivial solution family.
An \(m \times n\) matrix has \(m\) rows, \(n\) columns. Entry \(a_{ij}\) = row \(i\), column \(j\). Column vector = one column; row vector = one row.
Note: the exam writes \(m \times n\) as "rows Γ columns". A 3Γ2 matrix is NOT the same as 2Γ3 β transpose is the operation between them: \((A^T)_{ij} = a_{ji}\), i.e. flip rows and columns.
Find \(3A_1 - 5A_2 + 2A_3\) for \(A_1 = \begin{bmatrix}2&0\\-1&3\end{bmatrix}\), \(A_2 = \begin{bmatrix}3&5\\2&-1\end{bmatrix}\), \(A_3 = \begin{bmatrix}1&-1\\0&2\end{bmatrix}\).
If \(A\) is \(m \times r\) and \(B\) is \(r \times n\), then \(AB\) is \(m \times n\) with: $$(AB)_{ij} = (\text{row } i \text{ of } A) \cdot (\text{column } j \text{ of } B) = a_{i1}b_{1j} + a_{i2}b_{2j} + \dots + a_{ir}b_{rj}$$
Size check FIRST: columns of A must equal rows of B. "Inner dimensions must match, outer dimensions are the answer's size."
A great sanity phrase: \(AB\) means "A acts on B" β read right-to-left like function composition.
$$\begin{bmatrix}4 & 7\\ -1 & 3\end{bmatrix}\begin{bmatrix}2 & 8\\ 5 & -2\end{bmatrix}$$
Dot-product connection (from lecture 9): \((AB)_{ij} = r_i \cdot c_j\) β this is why dot products are reviewed with matrices.
\(A: 5\times 4\), \(B: 2\times 3\), \(C: 5\times 1\), \(D: 7\times 2\). Do these exist, and what size?
| Product | Inner check | Result |
|---|---|---|
| \(BD\) | 3 vs 7 β | undefined |
| \(DB\) | 2 vs 2 β | \(7 \times 3\) |
| \(A^TC\) | \(A^T\) is \(4\times5\), C is \(5\times1\) β | \(4 \times 1\) |
| Law | Says |
|---|---|
| Associative (mult) | \(A(BC) = (AB)C\) |
| Distributive L/R | \(A(B+C) = AB+AC\), \((B+C)A = BA+CA\) |
| Scalar laws | \(a(bC) = (ab)C\), \(a(BC) = (aB)C = B(aC)\) |
| Transpose laws | \((A^T)^T = A\), \((A\pm B)^T = A^T \pm B^T\), \((kA)^T = kA^T\), \((AB)^T = B^TA^T\) (reverses order!) |
Inversion also reverses order: \((AB)^{-1} = B^{-1}A^{-1}\). Same pattern as transpose β "socks and shoes": to undo socks-then-shoes you remove shoes-then-socks.
$$\begin{cases}2x_1 - 3x_2 + 5x_3 = 5\\ x_1 - 2x_2 = 9\end{cases} \;\longrightarrow\; \begin{bmatrix}2 & -3 & 5\\ 1 & -2 & 0\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}5\\9\end{bmatrix}$$
Missing variables contribute a 0 in A. If A is invertible: \(x = A^{-1}b\) β the one-solution guarantee, next section.
a) \(tr\left(\begin{bmatrix}9&1\\-3&3\end{bmatrix}\right)\)?
b) \(A\) is \(2\times3\), \(B\) is \(3\times3\). What is \(AB\)? What is \(BA\)?
c) T/F: if \(AB = 0\) then \(A = 0\) or \(B = 0\).
d) Write \(x_1 + 2x_2 = 5,\; 4x_1 - x_2 = 1\) as \(Ax=b\).
Answers: a) \(9 + 3 = 12\) Β· b) \(AB\): \(2\times3\) β; \(BA\): 3 vs 2 inner mismatch β undefined Β· c) FALSE β no-cancellation counterexamples exist; d) \(\begin{bmatrix}1&2\\4&-1\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}5\\1\end{bmatrix}\)
\(\frac{1}{a}\) undoes multiplication by \(a\): \(a \cdot \frac1a = 1\). The matrix inverse \(A^{-1}\) does the same: \(AA^{-1} = A^{-1}A = I\), where \(I\) is the identity matrix (1s on diagonal, 0s elsewhere β acts like the number 1).
Singular = no inverse exists (like \(a=0\) for numbers). Invertible = one exists, and it's UNIQUE (lecture proved \(B = C\) if both satisfy \(AB = I\)).
$$A = \begin{bmatrix}a & b\\ c & d\end{bmatrix} \;\; \Rightarrow \;\; A^{-1} = \frac{1}{ad - bc}\begin{bmatrix}d & -b\\ -c & a\end{bmatrix} \quad \text{iff } ad - bc \neq 0$$
\(ad - bc\) β swap the diagonal, negate the off-diagonal, divide by the cross-difference. That cross-difference IS the determinant (Chapter 2 preview!).
\(A = \begin{bmatrix}7 & -1\\ 2 & 3\end{bmatrix}\)
Verify (they demand it): \(AA^{-1} = \frac{1}{23}\begin{bmatrix}21+2 & 7-7\\ 6+6 & -2+21\end{bmatrix} = \frac{1}{23}\begin{bmatrix}23&0\\0&23\end{bmatrix} = I\) β
For any size: put \(A\) and \(I\) side by side, row-reduce the LEFT block to \(I\); the right block becomes \(A^{-1}\).
Why it works: elementary matrices. Each row op = left-multiplying by an elementary matrix E. Doing all of them to reach I means \(E_k\cdots E_1 A = I\), so \(E_k \cdots E_1 = A^{-1}\) β and that product is exactly what collects on the right side.
\(A = \begin{bmatrix}1 & 2 & 0\\ 0 & 1 & 0\\ 2 & 4 & 1\end{bmatrix}\)
Check: \(AA^{-1} = \begin{bmatrix}1Β·1+2Β·0+0Β·(-2) & ...\end{bmatrix}\) β entry (1,1): \(1\), entry (1,2): \(-2+2 = 0\) β, entry (2,Β·): row 2 of A is (0,1,0) β picks out row 2 of \(A^{-1}\) = (0,1,0) β. Looks right.
\(x_1 + 2x_2 + 3x_3 = 5\), \(2x_1 + 5x_2 + 3x_3 = 3\), \(x_1 + 8x_3 = 17\) β with \(A^{-1} = \begin{bmatrix}-40&16&9\\13&-5&-3\\5&-2&-1\end{bmatrix}\) (given).
Sanity check: sub into eq 3: \(1 + 8(2) = 17\) β
Given \(Ax = b_1\), \(Ax = b_2\), \(Ax = b_3\): two methods:
Worked in lecture: \(A = \begin{bmatrix}1&2\\1&1\end{bmatrix}\), three b's β answers (3,β1), (β2,2), (β7,6).
a) Invert \(\begin{bmatrix}3&-1\\2&3\end{bmatrix}\)
b) T/F: \((AB)^{-1} = A^{-1}B^{-1}\)
c) If \(AB\) is invertible, must \(A\) be?
d) When row-reducing \([A|I]\) you hit \(\begin{bmatrix}1&3&-2&|&1&0&0\\0&1&5&|&0&1&0\\0&0&0&|&-3&0&1\end{bmatrix}\). Conclusion?
Answers: a) det \(= 9+2 = 11\): \(A^{-1} = \frac{1}{11}\begin{bmatrix}3&1\\-2&3\end{bmatrix}\) Β· b) FALSE β reverses to \(B^{-1}A^{-1}\) Β· c) YES (theorem 1.4/2.3) Β· d) left block has a zero row β A is singular, no inverse exists.
An elementary matrix = I with ONE row operation applied. Left-multiplying \(E\) onto \(A\) performs that same row operation on A:
$$\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}3&7\\-5&16\end{bmatrix} = \begin{bmatrix}-5&16\\3&7\end{bmatrix}$$ (row swap applied instantly)
Zeros everywhere except the main diagonal. Powers and inverses are entry-wise β the easiest matrices alive:
$$D = \begin{bmatrix}2&0&0\\0&1&0\\0&0&-5\end{bmatrix} \;\Rightarrow\; D^3 = \begin{bmatrix}8&0&0\\0&1&0\\0&0&-125\end{bmatrix}, \;\; D^{-1} = \begin{bmatrix}1/2&0&0\\0&1&0\\0&0&-1/5\end{bmatrix}$$
Upper triangular: zeros BELOW the diagonal. Lower: zeros ABOVE.
| Fact | Statement |
|---|---|
| Transpose flips | \((\text{upper})^T = \text{lower}\) |
| Product keeps type | upper Γ upper = upper |
| Invertibility | triangular invertible βΊ ALL diagonal entries β 0 |
| Inverse type | inverse of upper is upper |
| Determinant (ch 2!) | \(\det = \) product of diagonal entries |
\(A^T = A\). Like adjacency matrices of friendships, stiffness/flexibility matrices.
\(A = \begin{bmatrix}1 & 2 & 0\\ 0 & -1 & 3\end{bmatrix}\)
a) Classify: \(\begin{bmatrix}2&-1\\0&3\end{bmatrix}\), \(\begin{bmatrix}3&0&0\\1&0&0\\1&2&7\end{bmatrix}\)
b) T/F: the product of two symmetric matrices is symmetric.
c) \(a_{ij} = 5i^2j\) β is this matrix symmetric? (compute \(a_{12}\) vs \(a_{21}\))
Answers: a) first is upper triangular (not diagonal β off-diagonal \(-1\) β 0); second is lower triangular, not upper Β· b) FALSE in general (though it's true if they happen to commute) Β· c) \(a_{12} = 5(1)(2) = 10\), \(a_{21} = 5(4)(1) = 20\) β not symmetric.
$$\begin{matrix}a_{11}x_1 + \dots + a_{1n}x_n = 0\\ \vdots\end{matrix}$$
Memorize the 4 T/F statements from lecture:
$$\begin{cases}x + ky = 3\\ 2x + k^2y = k + 4\end{cases}$$ β find all \(k\) giving (i) no solution (ii) exactly one (iii) infinitely many.
The method: row-reduce keeping k symbolic, factor (never divide by an expression containing k!), then case-split on the values that zero-out pivots.
The deadly sin: dividing by \(k(k-2)\) mid-reduction silently assumes \(k \neq 0,2\) and loses cases. Keep it as a factor, split at the end.
What must \(b_1, b_2, b_3\) satisfy for this to be consistent?
$$x_1 + x_2 + x_3 = b_1,\quad -x_1 - 2x_3 = b_2,\quad x_2 - x_3 = b_3$$
a) \(x_1 + 2x_2 - x_3 = 0\), \(x_1 + 5x_2 - 7x_3 = 0\) β why infinitely many solutions without row-reducing?
b) For \(\begin{cases}2x + ky = k+4\\ x + ky = 3\end{cases}\), which k gives no solutions?
Answers: a) homogeneous + 3 unknowns > 2 equations β guaranteed free variable β infinitely many (includes trivial) Β· b) reduce: subtract: \(x = 1\)β then \(y = 2/k\)β¦ careful: from the second row \(x + ky = 3\): with \(x = 1\), \(ky = 2\); first: \(2 + ky = k + 4\) β \(ky = k + 2\) β \(2 = k+2\) β \(\boxed{k = 0}\) gives contradiction \(0\cdot y = 2\)... check: at k=0 eq1: \(2x = 4\), eq2: \(x = 3\) β \(x = 2\) AND \(x = 3\) β no solution β
\(\det(A)\) is a NUMBER, not a matrix (instant T/F trap). Notation: \(\det(A)\) or \(|A|\).
2Γ2: \(\begin{vmatrix}a & b\\ c & d\end{vmatrix} = ad - bc\)
3Γ3 trick (diagonals): \(a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33}\)
THE 3Γ3-TRICK TRAP (lecture's explicit warning): the diagonal-trick works ONLY for 3Γ3. Using it on 4Γ4 or 5Γ5 = "completely wrong, automatic 0." For 4Γ4+, use cofactor expansion or row reduction.
Minor \(M_{ij}\): delete row i, column j; take the det of what's left.
Cofactor \(C_{ij} = (-1)^{i+j}M_{ij}\) β checkerboard of signs:
$$\begin{bmatrix}+ & - & +\\ - & + & -\\ + & - & +\end{bmatrix}$$
Expansion along any row i: \(\det(A) = a_{i1}C_{i1} + a_{i2}C_{i2} + \dots + a_{in}C_{in}\). Same for any column. Every choice gives the same answer β so pick the row/column with the most zeros (less work).
$$A = \begin{bmatrix}0 & 8 & 0 & 1\\ 4 & 7 & 3 & -1\\ 3 & 0 & 0 & 2\\ 0 & -5 & 1 & 6\end{bmatrix}$$
The zeros did 90% of the labor. ALWAYS scan for the easiest row/column first.
Upper/lower/diagonal: \(\det = a_{11}a_{22}\cdots a_{nn}\) (product of diagonal).
Worked: \(\begin{vmatrix}5&0&0&0\\1&3&0&0\\2&7&-2&0\\0&2&9&1\end{vmatrix} = 5Β·3Β·(-2)Β·1 = \boxed{-30}\) β instantly, no expansion.
Corollary: \(\det(I_n) = 1\).
a) \(\begin{vmatrix}-1&2&3\\2&4&-5\\0&1&-3\end{vmatrix}\) by the 3Γ3 trick.
b) T/F: \(\det(A)\) is a matrix.
c) Find \(C_{12}\) for \(\begin{bmatrix}-1&5&4\\3&8&2\\4&-7&1\end{bmatrix}\)
Answers: a) \((-1)(4)(-3) + 2(-5)(0) + 3(2)(1) - 3(4)(0) - (-1)(-5)(1) - 2(2)(-3) = 12 + 0 + 6 - 0 - 5 + 12 = 25\) Β· b) FALSE β it's a number Β· c) \(M_{12} = \begin{vmatrix}3&2\\4&1\end{vmatrix} = 3 - 8 = -5\), sign \((-1)^{1+2} = -\) β \(C_{12} = 5\)
| Operation on A | Effect on det |
|---|---|
| Swap two rows | \(\det \to -\det\) |
| Multiply ONE row by k | \(\det \to k\det\) |
| Add multiple of a row to another | \(\det\) UNCHANGED |
| Multiply the WHOLE matrix by k | \(\det(kA) = k^n\det(A)\) β n factors! |
Contrast to remember: row ops don't change SOLUTIONS when solving systems, but they (mostly) DO change det values. Also: when computing det you may use COLUMN ops too (same rules) β which you can't do when solving systems.
\(\det(A^T) = \det(A)\). A zero row/column β det = 0. Two proportional rows/columns β det = 0 (inspection questions!)
$$\det\begin{bmatrix}1 & -1 & 2 & -1\\ 2 & -2 & 1 & -3\\ -1 & 1 & 4 & 6\\ 0 & 1 & 2 & -1\end{bmatrix}$$
If you swap rows to get a pivot: multiply your running det by β1. If you scale a row: divide it out again. Track every swap/scale in the margin.
a) \(\begin{vmatrix}-2&1&3\\9&-4.5&-13.5\\-4&2&6\end{vmatrix}\) β row 2 = \(-4.5\)Γrow 1, row 3 = 2Γrow 1 β proportional rows β \(\det = 0\) instantly.
b) \(\det(A) = 3\), B results from \(R_2 \to R_2 + 5R_3\), C from swapping rows: \(\det(B) = 3\), \(\det(C) = -3\).
c) \(\det(A) = -5\), A is 4Γ4: \(\det(2A) = 2^4(-5) = -80\).
d) \(\det AB\) for \(|A|=5, |B|=-2\): \(= -10\). \(\det(3B) = 3^3(-2) = -54\).
\(\det(A) = -2\), 2Γ2:
\(A = \begin{bmatrix}2&3\\4&1\end{bmatrix}\), \(B = \begin{bmatrix}5&-3\\-2&2\end{bmatrix}\)
a) Row \(R_2 \to 6R_2 - R_1\) applied to A β what's det now (in terms of \(\det A\))? Careful, it's a compound op.
b) T/F: \(\det(A+B) = \det A + \det B\).
c) 5Γ5, \(|A| = 3\): find \(\det(-2A)\).
Answers: a) \(6R_2 - R_1\): think of \(6R_2\) first (det Γ6) then \(R_2 - R_1\) (det unchanged, since subtracting a multiple of ANOTHER row)... but careful: \(6R_2 - R_1\) means new row 2 = 6(row 2) β (row 1) β you can do it as two steps: \(R_2 \to R_2 - \frac16 R_1\) then scale \(R_2\) by 6: net effect det Γ 6. So \(6\det(A)\) Β· b) FALSE Β· c) \((-2)^5(3) = -96\)
If \(\det(A) \neq 0\) (n equations, n unknowns), the unique solution of \(Ax = b\) is:
$$x_1 = \frac{\det(A_1)}{\det(A)}, \quad x_2 = \frac{\det(A_2)}{\det(A)}, \quad \dots$$
where \(A_j\) = copy of A with its j-th column replaced by b.
Best use: you only need ONE variable (they explicitly ask "find xβ") β one determinant instead of a full solve.
You must know what \(A_1, A_2, \dots\) are β the formula alone is not the memorization ask. And it needs a SQUARE coefficient matrix with \(\det \neq 0\).
Find \(x_3\) only: \(x_1 + 2x_2 - 2x_3 = 2\), \(-x_1 + x_3 = -2\), \(2x_1 + 4x_2 - 5x_3 = 1\), given \(\det(A) = 2\).
\(\begin{bmatrix}-3&5\\2&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}4\\-6\end{bmatrix}\)
(Lecture's exact answers: \(x_1 = -34/13, x_2 = -10/13\))
For an \(n \times n\) matrix A, ALL of these are the same statement (all true or all false together):
Exam move: they give you ONE fact ("det(A) = 5" or "Ax = 0 has only the trivial solution") and ask which other statements follow. Everything follows β that's the point of "equivalent."
Watch direction: "If we know det(A) = 5" β can \(Ax=0\) have infinitely many solutions? NO (statement 1 says trivial only). "Ax = b must be consistent"? YES (statement 5). "A expressible as a product of elementary matrices"? YES (statement 4).
A is 7Γ7 and \(Ax = b\) is consistent for every b. True/False:
a) By Cramer: \(\begin{bmatrix}2&1\\1&3\end{bmatrix}x = \begin{bmatrix}5\\10\end{bmatrix}\) β xβ?
b) If RREF of A is \(\begin{bmatrix}1&3\\0&0\end{bmatrix}\), which equivalence statements hold?
c) T/F: if \(Ax = 0\) has infinitely many solutions, then C is not invertible.
Answers: a) \(\det = 5\); \(A_1 = \begin{bmatrix}5&1\\10&3\end{bmatrix}\), \(\det = 5\) β \(x_1 = 1\) (then xβ = 3, and check: \(2+3=5\) β) Β· b) NOT invertible β NONE of the 7 hold Β· c) TRUE β infinitely many nontrivial solutions to \(Ax=0\) βΉ statement 2 fails βΉ not invertible.
They flagged it explicitly: "The midterm covers Chapters 1 and 2 (incl. indep. study material on polynomial interpolation)." Do not skip this section.
Given n+1 points \((x_0, y_0), \dots, (x_n, y_n)\) with distinct x-values, there's a unique polynomial \(p(x) = a_0 + a_1x + \dots + a_nx^n\) of degree β€ n through ALL of them.
Finding it = a linear system: plug each point into p(x), get one equation in the coefficients \(a_i\):
$$a_0 + a_1x_0 + a_1 ... \quad\text{per point: } a_0 + a_1x_i + a_2x_i^2 + \dots + a_nx_i^n = y_i$$
n+1 equations, n+1 unknowns (\(a_0, \dots, a_n\)) β solve with Gaussian elimination. Existence+uniqueness = coefficient matrix is invertible (distinct x's β Vandermonde, nonzero det).
Find the degree-β€2 polynomial through \((1, 4), (2, 3), (3, 6)\).
Points \((0, -2)\) and \((2, 6)\): \(p(x) = a_0 + a_1x\). \(a_0 = -2\) from the first; \(6 = -2 + 2a_1\) β \(a_1 = 4\). \(p(x) = 4x - 2\).
For equally-spaced x-values there are shortcuts (finite differences), but the safe exam method is always: set up the system, row-reduce, sanity-check every point.
Find the degree-β€2 polynomial through \((-1, 6), (0, 1), (1, 0)\).
\(a_0 = 1\) (from x=0). Then \(-a_1... \): eq1: \(a_0 - a_1 + a_2 = 6\) β \(-a_1 + a_2 = 5\); eq3: \(a_0 + a_1 + a_2 = 0\) β \(a_1 + a_2 = -1\). Subtract: \(-2a_1 = 6\) β \(a_1 = -3\), \(a_2 = 2\). \(\boxed{1 - 3x + 2x^2}\). Check: x=β1: \(1 + 3 + 2 = 6\) β; x=1: \(1 - 3 + 2 = 0\) β
Weekly update: "Chapter 3 is partly high school review... going through it very quickly." The midterm covers Ch 1β2, but lecture 9 material (dot product geometry, projections, cross product) lands right before the test β a light taste is possible. 15 minutes here buys insurance.
| Object | Formula | Read it as |
|---|---|---|
| Dot product | \(u \cdot v = u_1v_1 + \dots + u_nv_n\) (a NUMBER) | how aligned two vectors are |
| Norm | \(\|u\| = \sqrt{u \cdot u} = \sqrt{u_1^2 + \dots + u_n^2}\) | length (Pythagoras) |
| Distance | \(d(u,v) = \|u - v\|\) | straight-line gap |
| Angle | \(u \cdot v = \|u\|\|v\|\cos\theta\) β \(\theta = \cos^{-1}\frac{u\cdot v}{\|u\|\|v\|}\) | the geometry link |
Unit vector: divide by your length: \(\frac{u}{\|u\|}\) β same direction, length 1.
\(u = (2, -1, 3)\), \(a = (1, 0, 2)\). (a) dot? (b) angle? (c) projection of u onto a?
\(u = (3, 0, 2)\), \(v = (6, 1, 3)\). Find \(u \times v\) and the parallelogram area.
a) Unit vector in direction of \((1, -5, 3, 2)\) (note: 4D!)
b) Which pairs orthogonal: \(u = (2,-1,2,1)\), \(v = (1,1,-1,1)\), \(w = (1,1,3,1)\)?
c) Line through \((5, -2, 7)\) perpendicular to plane \(3x - y + 6z = 8\)?
Answers: a) \(\|(1,-5,3,2)\| = \sqrt{1+25+9+4} = \sqrt{39}\) β \(\frac{1}{\sqrt{39}}(1,-5,3,2)\) Β· b) \(uΒ·v = 2 -1 -2 +1 = 0\) β orthogonal; \(uΒ·w = 2 -1 + 6 + 1 \neq 0\), \(vΒ·w = 1 + 1 - 3 + 1 = 0\) β too β so BOTH pairs! Β· c) normal of the plane = direction of the line: \((x,y,z) = (5,-2,7) + t(3,-1,6)\)
Ch 1 heavy: Gaussian/Jordan elimination WILL be a long-answer (it's the course's first toolbelt). Inverse via [A|I] highly likely. Ch 2: cofactor expansion + row-reduction dets + Cramer for a single variable. Homogeneous systems + the k-value trifecta = classic MC cluster. Interpolation = flagged, expect at least one question. Equivalence theorem list = T/F machine.
3 ops: scale (β 0), swap, add-multiple.
REF: leading 1s staircase, zero rows bottom.
RREF adds: zeros above + below leading 1s (unique!).
Junk row \([0\;0\;|\;c\neq0]\) β no solution.
Free var = no pivot β parameters.
No solution / one / infinitely many β never 2.
Junk row β none. Full pivots β one.
Free vars β β (one param each).
Homogeneous: always consistent; more unknowns than eqs β β.
\(AB\): inner sizes match; result \(m\times n\).
\(AB \neq BA\). No cancellation. \(AB=0 \nRightarrow A=0\).
\((AB)^T = B^TA^T\) and \((AB)^{-1} = B^{-1}A^{-1}\) β reverse order.
\(A^{-1} = \frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\), valid iff \(ad-bc\neq0\).
General: \([A|I] \to [I|A^{-1}]\).
Zero row in left block = singular.
Swap β flip sign. RowΓk β Γk. Add-multiple β unchanged.
\(\det(kA) = k^n\det(A)\). \(\det(AB) = \det A \det B\).
\(\det(A^{-1}) = 1/\det A\). \(\det A^T = \det A\).
Triangular β diagonal product. Det = number, not matrix.
3Γ3 diagonal trick works ONLY for 3Γ3.
Invertible βΊ trivial-only Ax=0 βΊ RREF = I βΊ product of elementary βΊ consistent βb βΊ unique βb βΊ detβ 0.
One true βΉ all true.
\(x_j = \det(A_j)/\det(A)\) where \(A_j\) = A with column j swapped for b. Needs square + detβ 0. Great for single variables.
n+1 points β degree-β€n poly. Plug each point in = one linear equation per coefficient. Solve, then verify ALL points.
\(uΒ·v\) = number. \(\|u\| = \sqrt{uΒ·u}\). \(d(u,v) = \|u-v\|\).
\(uβ₯v \iff uΒ·v = 0\). Unit: divide by norm.
\(\text{proj}_a u = \frac{uΒ·a}{\|a\|^2}a\).
\(uΓv\): vector β₯ both, anticommutes, \|uΓv\| = parallelogram area.
Plane: read normal off \(ax+by+cz=d\).
Q1: lecture's exact system β row-reduce to \([1\;1\;1\,|\,2;\;0\;1\;4\,|\,2... \) (add R1) β solve: subtract eq1 from eq3: \(-5x_3 = 0\)... full reduce: \(x = (1, 1, 0)\) β (lecture's stated answer)
Q2: FALSE β trivial-only is possible (e.g. \(x+y=0, 2x+2y=0\) is infinite, but \(x_1+x_2+x_3=0, x_2+x_3=0, x_3=0\) has only trivial). "Always" kills it. Infinitely many is guaranteed only when unknowns > equations.
Q3: FALSE β no cancellation law (counterexample exists with nonzero A).
Q4: det \(= 4+3 = 7\): \(A^{-1} = \frac17\begin{bmatrix}2&1\\-3&2\end{bmatrix}\). Verify \(AA^{-1} = \frac17\begin{bmatrix}7&0\\0&7\end{bmatrix} = I\) β
Q5: \(\frac{8}{5}(1,0,2) = (\frac85,0,\frac{16}{5})\); orthogonal part \((\frac25,-1,-\frac15)\)
Q6: \(X = (AB^{-1})^{-1}DC^{-T}\)... carefully: \(X = B A^{-1} D (C^T)^{-1} = BA^{-1}DC^{-T}\) (reverse order!)
Q7: k=0 β none; k=2 β β; else unique
Q8: R3 = 2ΓR1 (rows proportional) β det = 0, zero work needed
Q9: \(\det(2A) = 2^4(-5) = -80\); \(\det(A^T) = -5\); \(\det(A^{-1}) = -\frac15\)
Q10: \(A_2 = \begin{bmatrix}-3&4\\2&-6\end{bmatrix}\), det = 10 β \(x_2 = \frac{10}{-13} = -\frac{10}{13}\)
Q11: FALSE β consistency βb βΉ exactly one solution βb βΉ \(Ax=0\) trivial-only
Q12: \(a_0 = 2\), \(a_0 + a_1 + a_2 = 4\), \(a_0 + 2a_1 + 4a_2 = 8\) β \(a_1 + a_2 = 2\), \(2a_1 + 4a_2 = 6\) β \(a_2 = 1... \) solve: subtractΓ2: \(2a_1+4a_2 - 2(a_1+a_2) = 6-4 = 2a_2 = 2\)? β \(a_2 = 1\), \(a_1 = 1\). \(\boxed{2 + x + x^2}\) β check: 0β2 β 1β4 β 2β8 β
Q13: Yes β reduces cleanly to I (worked in section 4: \(A^{-1} = \begin{bmatrix}1&-2&0\\0&1&0\\-2&0&1\end{bmatrix}\))
Q14: \((-2, 3, 3)\), area \(= \sqrt{22}\)
Q15: (i) NO (product of variables) (ii) YES (Ο is a legit coefficient) (iii) NO (β of a variable)