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9231 · 1.4

Matrices flashcards

Revision flashcards for Cambridge 9231 Matrices (syllabus 1.4). Flip, recall, then mark a real past-paper question.

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    What is a singular matrix?

    A square matrix whose determinant is zero. A singular matrix does not have an inverse.

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    What is a non-singular matrix?

    A square matrix whose determinant is non-zero. A non-singular matrix always has an inverse.

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    What is the condition for the matrix product AB to be defined?

    The number of columns in the first matrix, A, must be equal to the number of rows in the second matrix, B.

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    Is matrix multiplication commutative?

    No. In general, for two matrices A and B, the product AB is not equal to the product BA. The order matters.

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    What is the identity matrix, I?

    A square matrix with 1s on the main diagonal (top-left to bottom-right) and 0s everywhere else. For any compatible matrix A, AI = IA = A.

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    What is the formula for the inverse of a 2x2 matrix $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$?

    $\mathbf{A}^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. This is only valid if the determinant, $ad-bc$, is not zero.

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    What is the adjugate (or adjoint) of a matrix?

    The transpose of the matrix of cofactors. It's a key intermediate step in finding the inverse of a 3x3 matrix by hand.

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    How do you use matrices to solve the system $\mathbf{A}\mathbf{x} = \mathbf{b}$?

    If A is non-singular, you can pre-multiply both sides by the inverse of A. This gives the unique solution $\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}$.

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    What does a determinant of 0 for a system of equations $\mathbf{A}\mathbf{x} = \mathbf{b}$ imply?

    It means there is no unique solution. The system will either have no solutions or infinitely many solutions. You cannot find $\mathbf{A}^{-1}$.

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    What is the 'matrix of cofactors'?

    It's a matrix of the same size as the original, where each element $a_{ij}$ is replaced by its cofactor. The cofactor is the minor of the element multiplied by $(-1)^{i+j}$.