GCSEFurther MathsAQAFlashcards5. Matrix TransformationsMatrix Multiplication

Matrix Multiplication (AQA GCSE Further Maths): Flashcards

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  • FrontIntroduction to Matrices

    Define a matrix.

Cards in this collection (17)

  • Define a matrix.

    A matrix is a rectangular grid of elements, which may be numbers or letters, arranged in rows and columns.

    The plural is matrices, and a bold capital letter is normally used to name one.

  • What is meant by the order of a matrix?

    The order is the number of rows followed by the number of columns.

    So \begin{pmatrix} a & b \\ c & d \end{pmatrix} is a 2 \times 2 matrix and \begin{pmatrix} a \\ b \end{pmatrix} is a 2 \times 1 matrix.

  • Complete the names for these matrices:

    A 2 \times 1 matrix is called a \_\_\_\_\_\_ matrix.

    A 2 \times 2 matrix is called a \_\_\_\_\_\_ matrix.

    The completed names are:

    A 2 \times 1 matrix is called a column matrix.

    A 2 \times 2 matrix is called a square matrix.

    A column matrix is also called a column vector.

  • How is the point \left(3 , 5\right) written as a matrix?

    The point is written as the column matrix \begin{pmatrix} 3 \\ 5 \end{pmatrix}, with the x-coordinate on top.

    Writing coordinates this way is what allows a matrix to transform a point.

  • How do you multiply a matrix by a scalar?

    Multiply every element by that scalar.

    So 2\begin{pmatrix} 5 & 2 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} 10 & 4 \\ 0 & 8 \end{pmatrix}.

  • True or False?

    In k\mathbf{A}, the k is a scalar rather than a matrix.

    True.

    Lower-case letters are used for scalars and bold capitals for matrices.

    So k\mathbf{A} is a matrix of the same order as \mathbf{A}, not a bigger one.

  • Complete the product:

    \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ \_\_\_\_\_\_ \end{pmatrix}

    The completed product is:

    \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}

    Each entry comes from a row of the first matrix paired with the column of the second.

  • Work out \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 10 \\ 20 \end{pmatrix}.

    Pair each row with the column: 1 \times 10 + 2 \times 20 = 50 and 3 \times 10 + 4 \times 20 = 110.

    So the answer is \begin{pmatrix} 50 \\ 110 \end{pmatrix}.

  • True or False?

    A 2 \times 2 matrix multiplied by a 2 \times 1 matrix gives a 2 \times 1 matrix.

    True.

    The answer keeps the rows of the first matrix and the columns of the second, so it is 2 \times 1.

    That is why transforming a point, written as a 2 \times 1 column, gives back another point.

  • True or False?

    To square a matrix you square each of its elements.

    False.

    You have to write out the multiplication in full, so \mathbf{P}^{2} = \mathbf{P} \times \mathbf{P}.

    For \mathbf{P} = \begin{pmatrix} 2 & 4 \\ 1 & -3 \end{pmatrix} that gives \begin{pmatrix} 8 & -4 \\ -1 & 13 \end{pmatrix}, which contains negatives even though it is a square.

  • Work out \begin{pmatrix} 3 & 1 \\ -2 & 0 \end{pmatrix}\begin{pmatrix} 5 & -5 \\ 4 & 2 \end{pmatrix}.

    Each entry pairs a row of the first with a column of the second, so the top-left is 3 \times 5 + 1 \times 4 = 19.

    Completing all four entries gives \begin{pmatrix} 19 & -13 \\ -10 & 10 \end{pmatrix}.

  • Does \mathbf{AB} always equal \mathbf{BA} for matrices?

    Matrix multiplication is not commutative, so \mathbf{AB} and \mathbf{BA} are usually different.

    \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 5 & 1 \end{pmatrix} = \begin{pmatrix} 10 & 3 \\ 20 & 7 \end{pmatrix}, but the other way round gives \begin{pmatrix} 3 & 4 \\ 8 & 14 \end{pmatrix}.

  • What does it mean that matrix multiplication is associative?

    Being associative means \left(\mathbf{AB}\right)\mathbf{C} = \mathbf{A}\left(\mathbf{BC}\right), so with three matrices you may group them either way.

    What you must not do is change their order, which is a different thing entirely.

  • Complete the identity matrix:

    \mathbf{I} = \begin{pmatrix} 1 & 0 \\ \_\_\_\_\_\_ & \_\_\_\_\_\_ \end{pmatrix}

    The completed identity matrix is:

    \mathbf{I} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

    Multiplying any 2 \times 2 matrix by \mathbf{I} leaves it unchanged either way round, so \mathbf{AI} = \mathbf{IA} = \mathbf{A}.

  • If \mathbf{A} = \begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix}, show that \mathbf{A}^{2} = 4\mathbf{I}.

    Multiplying out gives \mathbf{A}^{2} = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix}.

    Taking out the factor 4 leaves 4\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, which is 4\mathbf{I}.

  • How do you find unknowns in an equation between two matrices?

    Multiply out any products first, then equate corresponding elements.

    Each position gives its own equation, and harder questions produce simultaneous equations.

  • If \begin{pmatrix} a & 3 \\ 9 & b \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 3 & 0 \end{pmatrix} = k\mathbf{I}, find a, b and k.

    The product is \begin{pmatrix} 2a + 9 & a \\ 18 + 3b & 9 \end{pmatrix}, while k\mathbf{I} = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}.

    Equating gives a = 0 from the top-right, then k = 9 and b = -6.

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