Class prep · TJ Linear Algebra

Understand the structure,then solve with confidence.

A complete first course in Linear Algebra, from systems and matrices through vector spaces, eigenvalues, orthogonality and quadratic forms. Learn what each calculation means, then practice exact computation, geometric reasoning and clear proofs with original problems. Every problem has a full worked solution. Built by Neil Kothari, a TJ alumnus.

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$395 covers the full school year. The first lesson is free.

Try one. This is a test-level question from the course.

Problem 1

Test level
Multiple choice
No calculator
4 pts

Which statement correctly describes the three planes $x+y+2z=4,\qquad 2x-y+z=1,\qquad 4x+y+5z=6?$

Answer

No point lies on all three planes; each pair meets in a line, and the three lines are parallel.

Full solution

Row reduce the augmented matrix $\left[\begin{array}{ccc|c}1&1&2&4\\2&-1&1&1\\4&1&5&6\end{array}\right]$ . $R_2-2R_1$ gives $[0\;-3\;-3\,|\,-7]$ and $R_3-4R_1$ gives $[0\;-3\;-3\,|\,-10]$ . Then $R_3-R_2$ gives $[0\;0\;0\,|\,-3]$ .
The row $[0\;0\;0\,|\,-3]$ says $0=-3$ , so the system is inconsistent: no point lies on all three planes. That rules out (A) and (B).
Is any pair parallel? Normals $(1,1,2)$ , $(2,-1,1)$ , $(4,1,5)$ : no one is a scalar multiple of another, so no two planes are parallel and each pair meets in a line. That rules out (C) and (D).
Direction of those lines: the line of planes 1 and 2 runs along $(1,1,2)\times(2,-1,1)=(3,3,-3)$ , a multiple of $(1,1,-1)$ . This is perpendicular to the third normal too: $4+1-5=0$ . So all three pairwise lines run in the direction $(1,1,-1)$ , like the edges of a triangular prism. That is (E).
Why (B) is tempting: the left sides do satisfy $2(x+y+2z)+(2x-y+z)=4x+y+5z$ . But then the constants would have to match too, $2(4)+1=9$ , and the third constant is $6$ , not $9$ . Same left-side combination with a different constant means the third plane is parallel to the common line, not through it.
Why (D) is wrong: no solution does not mean parallel planes. Parallel planes need proportional normals, and none of these are. The coefficient matrix having only two pivots says the three normals lie in one plane, which is what lines up the three intersection lines; whether they share a point is decided by the augmented matrix, because only it can see the constants.

Where students lose points

Deciding the picture from the left sides only. Spotting that row 3's coefficients are a combination of rows 1 and 2 tells you the planes are not in general position; you still have to check the same combination of the constants. Matching constants give a common line (infinitely many), mismatched constants give the prism (none).

original problems, each with a worked answer: 546; original problems, each with a worked answer
at test level or beyond: 378; at test level or beyond
short lessons, one for every topic: 42; short lessons, one for every topic
units, all open now: 7; units, all open now

Free lesson

Lesson 1.1: Linear Systems, Augmented Matrices, and Their Geometry

How to turn a word statement into a linear system, store it as an augmented matrix, and read off whether it has no solution, one solution, or infinitely many, with the picture of lines and planes behind each case.

Start with a system of equations and learn how its algebra connects to its geometry.

Problem 2

Core
True or false
No calculator
2 pts

True or False: A linear system with more equations than unknowns (for example, three equations in $x$ and $y$ ) can never have exactly one solution. Justify your answer.

Answer

False. The system $2x+y=2$ , $x-3y=-13$ , $5x+2y=3$ has three equations in two unknowns and exactly one solution, $(x,y)=(-1,4)$ : all three lines pass through that single point.

Full solution

A claim with the word never is false as soon as you find one counterexample, so try to build a system with three equations in two unknowns that has exactly one solution.
Geometrically, each equation in $x$ and $y$ is a line. Exactly one solution means all three lines pass through one common point and no other point is shared. So pick a point, say $(-1,4)$ , and write three different lines through it.
Choose any coefficients and let the point fix the constant: $2(-1)+4=2$ , $(-1)-3(4)=-13$ , $5(-1)+2(4)=3$ . That gives $2x+y=2$ , $x-3y=-13$ , $5x+2y=3$ .
Check that the solution is unique. From the first equation, $y=2-2x$ . Substitute into the second: $x-3(2-2x)=-13$ , so $7x-6=-13$ , $7x=-7$ and $x=-1$ , then $y=2-2(-1)=4$ . Two lines with different slopes ( $-2$ and $\frac13$ ) meet in exactly one point, so the first two equations already force $(x,y)=(-1,4)$ .
The third equation is satisfied by that point: $5(-1)+2(4)=3$ . Adding an equation can only remove solutions, never create new ones, so the system still has exactly one solution.
In matrix language: the augmented matrix $\left[\begin{array}{cc|c}2&1&2\\1&-3&-13\\5&2&3\end{array}\right]$ row reduces to $\left[\begin{array}{cc|c}1&0&-1\\0&1&4\\0&0&0\end{array}\right]$ . A zero row is fine; it just says the third equation was redundant. There is a pivot in each variable column and no row of the form $[0\;0\,|\,c]$ with $c\neq0$ , so the solution is unique.
The true statement hiding behind the false one: more equations than unknowns makes inconsistency likely for randomly chosen numbers, but all three cases (none, one, infinitely many) are still possible.

Where students lose points

Counting equations and unknowns and stopping there. The number of solutions is decided by the pivots and by whether a row $[0\;\cdots\;0\,|\,c]$ with $c\neq0$ appears, not by whether the system is tall or wide. Always picture the lines (or planes) or row reduce.

Problem 3

Core
Multiple choice
No calculator
4 pts

Using the variable order $x$ , $y$ , $z$ , which matrix is the augmented matrix of this system? $\begin{aligned}2y-x&=6+3z\\5x+z&=-4\\y&=4x-7\end{aligned}$

Answer

$\left[\begin{array}{ccc|c}-1&2&-3&6\\5&0&1&-4\\-4&1&0&-7\end{array}\right]$

Full solution

Before writing any matrix, rewrite every equation in standard form: variables on the left in the order $x,y,z$ , constant on the right, and a $0$ written in for any missing variable.
Equation 1: $2y-x=6+3z$ . Subtract $3z$ from both sides and reorder: $-x+2y-3z=6$ . Row: $[-1\;\;2\;\;-3\,|\,6]$ . Choice (A) keeps $+3$ for $z$ , the sign error from moving $3z$ across the equals sign without changing its sign.
Equation 2: $5x+z=-4$ has no $y$ , so write $5x+0y+1z=-4$ . Row: $[5\;\;0\;\;1\,|\,-4]$ . Choice (C) slides the $1$ into the $y$ column because the placeholder $0$ was skipped; that turns the equation into $5x+y=-4$ , a different plane.
Equation 3: $y=4x-7$ . Subtract $4x$ : $-4x+y=-7$ , with $0z$ . Row: $[-4\;\;1\;\;0\,|\,-7]$ . Choice (E) has $+4$ for $x$ , which comes from moving $4x$ without changing its sign.
Choice (D) is the coefficient matrix: it is correct as far as it goes but has no column of constants, so it is $3\times3$ instead of $3\times4$ and cannot describe the system by itself.
Stacking the three rows gives (B). Check by reading each row back as an equation and rearranging it to the original: $-x+2y-3z=6$ becomes $2y-x=6+3z$ , $5x+0y+z=-4$ is $5x+z=-4$ , and $-4x+y+0z=-7$ becomes $y=4x-7$ .

Where students lose points

Copying coefficients straight off the page. Terms on the wrong side change sign when you move them, and a missing variable needs an explicit $0$ to hold its column. Always rewrite in standard form first, then copy.

Open the rest of this lesson, free

The full lesson and all 13 of its problems, with every worked answer.

How it works

Class prep that works alongside class

A complete Linear Algebra course for students in grades 9 to 12. Work alongside your school course or study the material in sequence. Each lesson starts with a concrete example, develops the idea, and shows how to check your reasoning before you try the practice.

Original problems in several formats: true or false, multiple choice, multi-part and free response.
Each one is marked core, test level or beyond.
Practice explains the structure behind each calculation, including proofs, geometric interpretations and applications.
A worked answer for every problem, ending with the mistake that most often costs points.
A lesson on every topic, with diagrams wherever a geometric picture makes the idea clear.

What is inside

1Linear Systems and Row Reduction6 lessons · 78 problems
2Matrix Algebra, Inverses and Factorizations6 lessons · 78 problems
3Determinants6 lessons · 78 problems
4Vector Spaces6 lessons · 78 problems
5Eigenvalues and Eigenvectors6 lessons · 78 problems
6Orthogonality and Least Squares6 lessons · 78 problems
7Symmetric Matrices and Quadratic Forms6 lessons · 78 problems

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Who built it

Neil Kothari

TJHSST alumnus · Co-founder, EduAvenues · 10+ years in TJ preparation

Neil is a TJ alumnus and has spent more than a decade working with TJ students and their families. He built this course so students can connect geometric pictures, exact calculations and clear mathematical arguments. Every problem is written for the course and every answer is checked before it goes in.

Tuition

$395for the full school year

One tuition covers every unit of the course. All 7 units are open now. The Linear Systems, Augmented Matrices, and Their Geometry lesson stays free for everyone.

All 7 units: 42 lessons and 546 original problems
A worked answer for every problem
Access through the end of the school year

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