Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 1 / 5

Direct matrix multiplication gives $MN=\left(\begin{array}{cc}5& 18\\ 7& 25\end{array}\right)$. The three determinants are $\det M=8-9=-1$, $\det N=4-3=1$, and $\det (MN)=125-126=-1$. Hence $\boxed{\det (MN)=(-1)(1)=\det M\det N}$, as required.

The two-by-two matrix inverse formula givesMultiplication in the reversed order givesThe order reversal is needed because applying the inverse undoes the original two linear maps in reverse sequence.

Take $A=B=I_2$, which are invertible matrices. Then $\det (A+B)=\det (2I_2)=4$, whereas $\det A+\det B=2$. This proves that the determinant is not additive. Being a multilinear map of separate columns does not imply additivity in the entire matrix.

Again take $A=B=I_2$. Their sum is also invertible, but $(A+B{)}^{-1}=I_2/2$, while $A^{-1}+B^{-1}=2I_2$. Thus matrix inversion is not additive, even when every inverse in the proposed identity exists.

Use $A=M$ and $B=N$ from part (a). The proposed unreversed product iswhich differs from the actual inverse $\left(\begin{array}{cc}-25& 18\\ 7& -5\end{array}\right)$. Both original matrices are invertible matrices, so this is a valid counterexample. The universally valid order is $\boxed{(AB{)}^{-1}=B^{-1}{A}^{-1}}$.

The product of a $3\times n$ matrix with another $3\times n$ matrix requires its inner dimensions $n$ and three to match. Thus $\boxed{n=3}$ is necessary and sufficient for $M^2$ to be defined.

The transpose has shape $n\times 3$. Multiplying two copies requires $3=n$, so $\boxed{n=3}$.

The shapes in $M{M}^T$ are $(3\times n)(n\times 3)$, whose inner dimensions always match. Hence every positive integer $n$ is allowed, and the resulting matrix is $3\times 3$.

Matrix addition requires equal shapes. The shapes $3\times n$ and $n\times 3$ agree exactly when $\boxed{n=3}$.

A two-sided matrix inverse requires a square matrix, so $\boxed{n=3}$ is necessary. It exists only if that square matrix is also an invertible matrix, equivalently $\det M\ne 0$; choosing the dimension alone does not ensure invertibility.