Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 7

Subtract corresponding entries in the two matrices:Both matrices have shape $2\times 3$, so this subtraction is defined.

Using row-column matrix multiplication, the second row of the first matrix is $(1,6)$. Therefore

The row-column product is the dot product $4\cdot 6+8\cdot 3+1\cdot 0$, giving $\boxed{48}$. Its shape is $1\times 1$.

The first matrix product gives the column $(37,37,37{)}^T$. Its product with the final row is an outer product, not a scalar:

Row reduction, or the cofactor formula for a matrix inverse, givesMultiplying by the right-hand-side column $(5,13,35{)}^T$ gives, at $\lambda =3$,For general $\lambda$, the coefficient matrix has determinant $(\lambda -1)(\lambda -2)$. Thus other parameter values give a unique solution. If $\lambda =1$, subtracting the first equation from the second gives $y=8$, while subtracting it from the third gives $3y=30$, a contradiction. If $\lambda =2$, subtraction gives $y+z=8$ and $3y+3z=30$, again inconsistent. Hence