Multiply a vector by a matrix, treat 2 × 2 matrices as transformations of the plane, and read the determinant as an area factor

Learning goal: Multiply a vector, regarded as a matrix with one column, by a matrix of suitable dimensions to produce another vector; work with matrices as transformations of vectors; work with 2 × 2 matrices as transformations of the plane; and interpret the absolute value of the determinant in terms of area.

Scope note. N.VM.11 and N.VM.12 are marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate standards "for students who are planning on taking advanced courses". They are taught here in full, as extension content for students on that path.

A vector as a column matrix. The vector ⟨5, −1⟩ can be written as a 2 × 1 matrix, a single column with entries 5 and −1. Regarded this way, it can be multiplied on the left by any matrix with two columns. A 2 × 2 matrix A times a 2 × 1 column gives a 2 × 1 column, another vector in the plane. A 2 × 3 matrix times a 3 × 1 column also gives a 2 × 1 column, so a matrix can take a vector from three-dimensional space to the plane. The rule is the ordinary matrix product: the dimensions must be suitable, with the number of columns of the matrix equal to the number of entries in the vector.

Computing Av. Let A have rows [1 2] and [3 4] and let v = [5; −1]. Row 1 of A with the column: 1(5) + 2(−1) = 5 − 2 = 3. Row 2: 3(5) + 4(−1) = 15 − 4 = 11. So Av = [3; 11]. The matrix has taken the vector ⟨5, −1⟩ to the vector ⟨3, 11⟩. For a 2 × 3 example, the matrix with rows [1 0 2] and [0 1 −1] times [3; 4; 5] gives [1(3) + 0(4) + 2(5); 0(3) + 1(4) − 1(5)] = [13; −1].

Matrices as transformations. Because Av is a vector for every v, the matrix A defines a rule that moves every point of the plane: the point with position vector v goes to the point with position vector Av. That rule is a transformation of the plane. Two properties make matrix transformations special. They send the origin to itself, since A[0; 0] = [0; 0]. And they respect vector arithmetic: A(u + w) = Au + Aw and A(cv) = cAv, which follow from the distributive rules for matrix products. Consequently, when the determinant is not zero, straight lines go to straight lines, parallel lines stay parallel, and the image of a parallelogram is a parallelogram; when the determinant is zero, the whole plane is squashed onto a line or a point, so a line can shrink to a single point.

Reading a matrix from its columns. Multiply A by the two basic vectors e₁ = [1; 0] and e₂ = [0; 1]. Ae₁ picks out the first column of A, and Ae₂ picks out the second. So the columns of a 2 × 2 matrix tell you where the two unit vectors go, and since every vector is a combination xe₁ + ye₂, that fixes everything. Some matrices to know: rows [0 −1], [1 0] send e₁ to ⟨0, 1⟩ and e₂ to ⟨−1, 0⟩, a rotation of 90° counterclockwise about the origin. Rows [1 0], [0 −1] send e₂ to ⟨0, −1⟩, a reflection across the x-axis. Rows [2 0], [0 2] double every vector, a dilation with factor 2. Rows [1 1], [0 1] leave e₁ alone and tilt e₂ to ⟨1, 1⟩, a shear that slides each horizontal line sideways by an amount equal to its height.

A fully worked model: the unit square. Let T have rows [2 1] and [0 3]. Apply it to the four corners of the unit square, (0, 0), (1, 0), (1, 1) and (0, 1), written as columns. T[0; 0] = [0; 0]. T[1; 0] = [2; 0]. T[1; 1] = [2 + 1; 0 + 3] = [3; 3]. T[0; 1] = [1; 3]. The square with area 1 has become the parallelogram with vertices (0, 0), (2, 0), (3, 3), (1, 3). (The shoelace formula: list the vertices (x₁, y₁), …, (xₖ, yₖ) in order around the figure, repeat the first vertex at the end, add xⱼyⱼ₊₁ − xⱼ₊₁yⱼ over each consecutive pair, and take half the absolute value.) Its area, by the shoelace formula, is ½|0·0 − 2·0 + 2·3 − 3·0 + 3·3 − 1·3 + 1·0 − 0·3| = ½|0 + 6 + 6 + 0| = 6. Now compute det T = 2(3) − 1(0) = 6. The area of the image is the determinant. This is not a coincidence of the example: for any 2 × 2 matrix, the image of the unit square is the parallelogram spanned by the two columns, and the area of that parallelogram is |ad − bc|.

The determinant as an area factor. Because the transformation respects addition and scaling, every region is stretched by the same factor as the unit square. So for any figure, area of image = |det A| × area of original. A triangle of area 4 sent through a matrix with determinant 5 becomes a triangle of area 20, and one sent through a matrix with determinant −1 keeps its area, 4, because |−1| = 1. The absolute value is needed since a determinant can be negative while an area cannot. A negative determinant means the transformation reverses orientation: a figure whose vertices read counterclockwise has an image whose vertices read clockwise, as happens with reflections. The reflection across the x-axis, rows [1 0], [0 −1], has determinant −1: areas unchanged, orientation flipped. A rotation has determinant +1: areas unchanged, orientation kept. The dilation with rows [2 0], [0 2] has determinant 4, and doubling every length does indeed multiply every area by 4, not by 2.

Determinant zero. If det A = 0, every area becomes 0: the whole plane is flattened onto a line (or onto a single point, for the zero matrix). The matrix with rows [1 2], [2 4] sends the unit square's corners to (0, 0), (1, 2), (3, 6), (2, 4), all of which lie on the line y = 2x. The image has no area. This is the same matrix that has no inverse; a transformation that collapses the plane cannot be undone, because many different points have been sent to the same place.

The misconception. Students often expect a matrix with "2s in it" to double areas, or expect the determinant itself to be the area of the image of any figure. Neither is right. The determinant is a factor, applied to the original area: the image of a region of area 3 under T above has area 6 × 3 = 18, not 6. And the factor depends on all four entries through ad − bc, not on the size of any single entry: the shear with rows [1 1], [0 1] moves points a long way but has determinant 1 and changes no areas at all, since the unit square becomes a parallelogram with the same base 1 and the same height 1.

Worked, a matrix times a vector: A = [1 2; 3 4], v = [5; −1]. Row 1: 1(5) + 2(−1) = 3. Row 2: 3(5) + 4(−1) = 11. Av = [3; 11], so the transformation sends the point (5, −1) to (3, 11). For a 2 × 3 matrix with rows [1 0 2], [0 1 −1] and the 3 × 1 vector [3; 4; 5]: entries 3 + 0 + 10 = 13 and 0 + 4 − 5 = −1, giving [13; −1], a vector in the plane produced from one in space.

Worked, a rotation: R = [0 −1; 1 0]. The triangle with vertices (1, 0), (3, 0), (3, 2): R[1; 0] = [0; 1], R[3; 0] = [0; 3], R[3; 2] = [0(3) − 1(2); 1(3) + 0(2)] = [−2; 3]. The image has vertices (0, 1), (0, 3), (−2, 3): the triangle turned a quarter turn counterclockwise about the origin. det R = 0(0) − (−1)(1) = 1, so the area, ½ × 2 × 2 = 2, is unchanged, and the vertex order still reads the same way round.

Worked, area scale factor: M = [3 1; 1 2], det M = 3(2) − 1(1) = 5. Triangle with vertices (0, 0), (4, 0), (0, 2) has area ½ × 4 × 2 = 4. Images: M[0; 0] = [0; 0]; M[4; 0] = [12; 4]; M[0; 2] = [2; 4]. Shoelace on (0, 0), (12, 4), (2, 4): ½|0 + (12·4 − 2·4) + 0| = ½ × 40 = 20. And |det M| × 4 = 5 × 4 = 20. The determinant is the factor by which the area was multiplied.

Worked, a reflection and orientation: F = [1 0; 0 −1], det F = −1. Unit square corners go to (0, 0), (1, 0), (1, −1), (0, −1): a square of area 1 = |−1| × 1 below the axis. The original corners (0, 0), (1, 0), (1, 1), (0, 1) read counterclockwise; the images (0, 0), (1, 0), (1, −1), (0, −1) read clockwise. The negative sign records the reversed orientation; the absolute value gives the area.

Worked, determinant zero: N = [1 2; 2 4], det N = 1(4) − 2(2) = 0. Unit square corners: N[1; 0] = [1; 2], N[1; 1] = [3; 6], N[0; 1] = [2; 4]. All four image points, including (0, 0), satisfy y = 2x, so the image is a segment on that line with area 0. The plane has been flattened, which is why N has no inverse.

Recording task: On grid paper draw the unit square and its image under T = [2 1; 0 3], the parallelogram with vertices (0, 0), (2, 0), (3, 3), (1, 3). Record the four matrix-times-column products that give the image vertices, then compute the parallelogram's area two ways: by the shoelace formula (answer 6) and as base × height using the base from (0, 0) to (2, 0) and the height 3 (answer 6). Compute det T = 2(3) − 1(0) = 6 and write one sentence connecting the three results. Mark the direction in which the vertices are read on both figures and note that it is the same, matching the positive determinant.

Independent performance: Choose your own 2 × 2 matrix with all four entries nonzero and determinant not equal to 0, 1 or −1. (1) Multiply it by three vectors of your choice and by the two unit vectors, and explain how the columns of the matrix appear in the results. (2) Apply it to the triangle with vertices (0, 0), (3, 0), (0, 2) and to the unit square, plotting original and image. (3) Compute the area of each image by the shoelace formula and show that it equals |det| times the original area. (4) State whether orientation was preserved or reversed and connect that to the sign of the determinant. (5) Then modify one entry so that the determinant becomes 0, apply the new matrix to the same square, and describe the image. Success criteria: every product entry shows its sum; the two unit-vector images are the two columns; both shoelace areas match |det| × original area exactly; the orientation statement agrees with the sign of the determinant; and the zero-determinant image is correctly described as lying on a single line through the origin.