Past exam of the mathematics course of the University of Oxford / 2014 / Prelims / Mathematics I / 3

An elementary row operation is one of three operations: interchange two rows, multiply one row by a nonzero scalar, or add a scalar multiple of one row to a different row. Each operation is reversed by another operation of the same kind.

Performing such an operation on an $m\times n$ matrix $A$ is equivalent to left multiplication by the $m\times m$ elementary matrix obtained by applying that operation to the identity matrix. Each elementary matrix is invertible, with inverse corresponding to the reverse operation. Thus a sequence of operations givesFor square matrices, if $A$ were invertible, then $B=PA$ would have inverse $A^{-1}{P}^{-1}$. Hence if $B$ is singular, $A$ must be singular as well. In fact the two are invertible together, because $A=P^{-1}B$ gives the converse.

Solve $Ax=b$ for a general right-hand side $b=(b_1,b_2,b_3,b_4{)}^T$. Subtracting the fourth equation from the second gives $x_3=b_2-b_4$. The first and fourth then giveSubstitution into the third equation reduces the remaining condition toIf $t\ne 2$, every right-hand side therefore has a unique solution, and collecting its four coefficients yields the matrix inverseFor $t=2$, the nonzero vector $(-1,-1,0,1{)}^T$ lies in the kernel of a linear map represented by $A$, so $A$ is singular. Hence $A$ is invertible exactly when $t\ne 2$. The same calculation gives $\det A=t-2$; the formula above can also be checked by multiplying its numerator by $A$ to obtain $(t-2)I$.