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Vanier · 201-SN4-RE

201-SN4-RE Linear Algebra and Vector Geometry tutoring

Get one-to-one help with vectors, matrices and determinants for Vanier 201-SN4-RE. Try the short example below, or bring a question and the step you want to understand.

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Linear algebra and vector geometry at Vanier

Vanier’s Linear Algebra & Vector Geometry course introduces vectors, matrices, determinants and Euclidean vector geometry with applications. The example below connects these ideas in two dimensions: build a target vector from two given vectors, then explain what the determinant says about uniqueness and area.

Skills to work on

  • Adding vectors and multiplying them by real numbers
  • Reading matrix columns as the vectors in a linear combination
  • Calculating a determinant and interpreting its geometric meaning
  • Checking whether a target vector can be formed, and whether its coefficients are unique

Before trying the example

Review ordered pairs, multiplication by a scalar and solving two equations in two unknowns. A scalar is a number: multiply each component of a vector by that number, then add corresponding components. Keep the horizontal component first and the vertical component second.

Two steps worth checking

  • Keeping the coefficients of a vector combination separate from the components of the resulting vector
  • Checking the target vector before deciding what a zero determinant means for the number of solutions

Worked example: two vectors, one target and an area

In the coordinate plane, let u=(2,1)\mathbf u=(2,1), v=(1,3)\mathbf v=(1,3) and the target vector w=(8,9)\mathbf w=(8,9). Find real numbers aa and bb such that au+bv=wa\mathbf u+b\mathbf v=\mathbf w. Explain why this pair of coefficients is unique, and find the area of the parallelogram whose adjacent sides are u\mathbf u and v\mathbf v, drawn from the same point.

Show a hint

Match the horizontal and vertical components to obtain two equations. Place u\mathbf u and v\mathbf v in the columns of a matrix. Its determinant connects the two questions about uniqueness and area.

Show the reasoning

  1. 1. Translate the vector equation

    The numbers aa and bb scale the two given vectors. Matching components gives

    {2a+b=8,a+3b=9.\begin{cases}2a+b=8,\\a+3b=9.\end{cases}

    Call the matrix of their columns AA. The same equations can be written as

    A=[2113],A[ab]=[89].\begin{gathered}A=\begin{bmatrix}2&1\\1&3\end{bmatrix},\\A\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}8\\9\end{bmatrix}.\end{gathered}

    The first column belongs to the first coefficient. Changing the order of the columns without changing the coefficient order would change the vector combination.

  2. 2. Find and check the coefficients

    From the first equation, b=8−2ab=8-2a. Substitute into the second: a+3(8−2a)=9a+3(8-2a)=9, so −5a=−15-5a=-15. Therefore a=3a=3 and b=2b=2. Check the original vector equation:

    3(2,1)+2(1,3)=(6,3)+(2,6)=(8,9)=w.\begin{gathered}3(2,1)+2(1,3)\\=(6,3)+(2,6)\\=(8,9)=\mathbf w.\end{gathered}

    The coefficients are (3,2)(3,2); the resulting vector has components (8,9)(8,9). They describe different things.

  3. 3. Explain why the answer is unique

    For a two-by-two matrix, take the top-left entry times the bottom-right entry, then subtract the top-right entry times the bottom-left entry. The notation det⁡(A)\det(A) means the determinant of AA:

    det⁡(A)=2⋅3−1⋅1=5.\det(A)=2\cdot3-1\cdot1=5.

    This is nonzero, so the two column vectors are linearly independent: neither is a multiple of the other. They give a unique pair of coefficients for every vector in the plane. In this calculation, the equation −5a=−15-5a=-15 fixes aa, and then b=8−2ab=8-2a fixes bb.

  4. 4. Interpret the determinant as an area

    For two vectors in the plane, the absolute value of the determinant of their columns is the area of the parallelogram they form. Here that area is ∣5∣=5|5|=5 square units. The determinant’s sign records orientation; area is nonnegative.

    As a check, swapping the columns gives 1⋅1−2⋅3=−51\cdot1-2\cdot3=-5. It reverses their order but leaves the same parallelogram, so its area stays 55 square units.

Your turn: when the two directions collapse into one

Keep u=(2,1)\mathbf u=(2,1), but replace the second vector by q=(4,2)\mathbf q=(4,2). (1) Find the determinant of the matrix BB with columns u,q\mathbf u,\mathbf q, and describe the vectors that au+bqa\mathbf u+b\mathbf q can produce. (2) For target r=(6,3)\mathbf r=(6,3), give every real coefficient pair (a,b)(a,b). (3) Can the same two vectors produce target s=(6,4)\mathbf s=(6,4)? Explain.

Show a hint

First write q\mathbf q as a multiple of u\mathbf u. Factor au+bqa\mathbf u+b\mathbf q, then compare both components of each target with that single direction.

Check the three steps

  1. 1. Identify the reachable line

    q=2u\mathbf q=2\mathbf u, so the two vectors are linearly dependent. Their matrix and determinant are

    B=[2412],det⁡(B)=2⋅2−4⋅1=0.\begin{gathered}B=\begin{bmatrix}2&4\\1&2\end{bmatrix},\\\begin{aligned}\det(B)&=2\cdot2-4\cdot1\\&=0.\end{aligned}\end{gathered}

    Every combination is au+bq=(a+2b)ua\mathbf u+b\mathbf q=(a+2b)\mathbf u. Write c=a+2bc=a+2b; the possible vectors are (2c,c)(2c,c), where cc is any real number. Their endpoints lie on the line y=x/2y=x/2, with xx and yy the usual horizontal and vertical coordinates. The parallelogram has collapsed to a line segment and has zero area.

  2. 2. Give all coefficients for the first target

    Because r=(6,3)=3u\mathbf r=(6,3)=3\mathbf u, we need a+2b=3a+2b=3. Let tt be any real number and choose b=tb=t. Then every solution is

    (a,b)=(3−2t,t),t∈R.\begin{gathered}(a,b)=(3-2t,t),\\t\in\mathbb R.\end{gathered}

    For example, (3,0)(3,0) and (1,1)(1,1) both produce (6,3)(6,3). There are infinitely many coefficient pairs, but they all produce the same target vector.

  3. 3. Test the second target

    To produce s=(6,4)\mathbf s=(6,4), the horizontal component 2c=62c=6 would require c=3c=3, while the vertical component would require c=4c=4. Both cannot hold. The target is off the reachable line, so there is no solution.

    A zero determinant tells us that a unique coefficient pair is impossible. Checking the target distinguishes infinitely many solutions from no solution.

If your coefficients were wrong, check the component equations, then substitute the values you found for aa and bb into the original vector equation. If you concluded that a zero determinant always means no solution, revisit the two different targets in the self-check.

Work through the difficult step with a tutor

Your tutor can help you sketch the vectors, label the coefficients and turn a vector equation into component equations. You can compare a correct column setup with a misplaced entry and explain how the geometry changes.

After checking the calculation, your tutor can change one vector or the target and ask you to predict the result before solving. Together, you can connect a determinant to the available directions, check a proposed answer and write a complete family of solutions when needed.

Plan your review

From your section’s assigned topics, choose one linear system and one problem that combines two vectors in the plane. Check a proposed solution by substitution. For the vector problem, turn the components into equations and explain what the determinant tells you about uniqueness. If the determinant is zero, check whether the target is reachable before deciding how many solutions there are. Redo a corrected question without the solution, using only the tools your teacher permits.

Independent practice

Choose your next practice step

13 min

Practise matrices, vectors and determinants

Use the matrix, independence and determinant examples for more practice. This shared resource was written for Concordia MATH 204; choose the topics that match your Vanier work.

Vanier course information

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