Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 12

A real square orthogonal matrix $Q$ satisfiesequivalently $Q^{-1}=Q^T$. Its columns, and also its rows, form an orthonormal basis. Consequently it preserves dot products and Euclidean norms.

To invert the matrix, row-reduce the augmented matrix $(M\mid I)$. Subtracting row one from rows two and three gives left rows $(0,-4,1)$ and $(0,0,2)$, so the pivots are nonzero and back-substitution is possible. The resulting matrix inverse isMultiplying by $M$ verifies the identity matrix. This inverse is not $M^T$; indeedThus $M$ is not orthogonal.

Use the order-reversal rule for the transpose of a product:Therefore the product of two orthogonal matrices is orthogonal. The argument does not require $A$ and $B$ to commute.

Collect the coefficients of the unknowns into the matrix and the constants into a column. The requested linear system isThe unknown coordinate $y$ here is a scalar component; the right-hand column is the vector denoted by the question's matrix-form notation.

At $a=2$, eliminate the first column by $R_2\leftarrow R_2-2R_1$ and $R_3\leftarrow R_3-2R_1$ in the augmented matrix. The last two rows give $-z=-2$ and $-y+z=-1$. Hence $z=2$, $y=3$, and the first row then gives $x=-2$. ThusThe coefficient determinant at this parameter is nonzero, so this solution is unique.

The same elimination at general $a$ gives the rowsThe coefficient determinant is $(a-3)(a-1)$. Only $a=1$ or $a=3$ can therefore fail to have a unique solution. At $a=3$, the second row reads $0=-2$, so there is no solution. At $a=1$, the last two rows are proportional and both specify $z=1$, leaving one free unknown; hence there are infinitely many solutions. Every other real $a$ gives a unique solution.

At $a=1$, the reduced system gives $z=1$ and $x+y=2$. Thus all solutions areGeometrically the solution set is an affine straight line, and substitution verifies all three original equations.