Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Digital systems, linear algebra, introduction to formal proof resit / 4

Subtract the first vertex from the other two, giving side vectors $(2,1)$ and $(3,4)$. Their determinant is $2\cdot 4-3\cdot 1=5$, the signed area of the corresponding parallelogram. The triangle occupies half that area, hence its area is $5/2$. Taking the absolute value would remove any sign change caused by reversing the vertex order. This is the determinant formula for triangle area.

Let $c_1=(1,1,1{)}^T$ and $c_2=(2,-2,6{)}^T$. The remaining columns are $3c_1$ and $2c_2$, so these two vectors span the column space. They are linearly independent: the determinant of their first two coordinates is $-4\ne 0$. Therefore $\{(1,1,1{)}^T,(2,-2,6{)}^T\}$ is a basis of the range, and the matrix rank is two.

Equivalently the range is the plane $y_3=2y_1-y_2$: both displayed basis vectors satisfy this equation, and their two-dimensional span fills that plane.

The independent-column representation gives $T(x)=c_1(x_1+3x_3)+c_2(x_2+2x_4)$. Thus a vector lies in the kernel of a linear map represented by this matrix precisely when $x_1=-3x_3$ and $x_2=-2x_4$. Writing the two free coordinates separately givesThe two displayed vectors are linearly independent, so they form a basis of the kernel. Its dimension two also agrees with the rank-nullity theorem, $4=2+2$.

If $A$ is $m\times n$, then $Y$ is $n\times m$. The equation $AY=I_m$ makes every standard basis vector of ${\mathbb{R}}^m$ a linear combination of the columns of $A$: take the corresponding column of $Y$ as coefficients. Hence the column space is all of ${\mathbb{R}}^m$, giving $\mathrm{rank}(A)=m$. In particular, $n\ge m$ and $A$ has full row rank.

The system is consistent for every $b\in {\mathbb{R}}^m$: take $x=Yb$, since $A(Yb)=(AY)b=b$. Thus $Y$ is a right inverse giving an explicit solution for every right-hand side.

All solutions are $Yb+u$ with $u$ in the kernel of a linear map represented by $A$. The rank-nullity theorem gives kernel dimension $n-m$; the solution is unique when $n=m$, and is nonunique when $n>m$.

True. Transposing the inverse identities shows $(X^{-1}{)}^T=(X^T{)}^{-1}$. If $X^T=X$, this becomes $(X^{-1}{)}^T=X^{-1}$, exactly the symmetric matrix condition. The reasoning uses the matrix inverse and transpose identities, without assuming that different matrix factors commute.

True. Using the same transpose identity and $X^T=-X$ givesThe final equality follows by multiplying $-X$ and $-X^{-1}$ in either order and obtaining the identity. Thus the inverse is a skew-symmetric matrix as well. The invertibility hypothesis is retained throughout.

True for these square matrices. The determinant multiplication identity gives $\det (XY)=\det X\det Y$. Since $XY$ is invertible, its determinant is nonzero; both factors are therefore nonzero, so both matrices have matrix inverses.

Equivalently, $Y$ must be injective and $X$ must be surjective, and either property implies invertibility for an endomorphism of a finite-dimensional vector space. The equal square dimensions matter: a rectangular matrix and a right inverse can have identity product without being mutually invertible square matrices.