Review : Matrices & Vectors — Question 1

Question 1 – Solution

Strategy. A rectangular matrix can have a right inverse without having a left inverse. Check what happens to vectors, not just products.

Step 1: Multiply with dimensions attached. The sizes are $A:2\times 3$ and $B:3\times 2$. Thus $$AB=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I_2, \qquad BA=\begin{pmatrix}1&0&1\\0&1&1\\0&0&0\end{pmatrix}\ne I_3.$$ The two products even act on spaces of different dimensions.

Step 2: Recover the whole solution family. The equation $Ax=b$ says $x_1+x_3=b_1$ and $x_2+x_3=b_2$. Hence $$\boxed{x=(b_1-t,b_2-t,t)^T,\qquad t\in\mathbb R.}$$ Every such vector works, and every solution has this form. The vector $Bb=(b_1,b_2,0)^T$ selects $t=0$; it is a choice, not the only inverse image.

Step 3: Rule out a left inverse. The nonzero vector $n=(-1,-1,1)^T$ satisfies $An=0$. If $CA=I_3$, then $n=I_3n=CAn=C0=0$, a contradiction. Thus no left inverse exists, even though $AB=I_2$.

Step 4: Interpret the repeated operation. Associativity gives $P^2=B(AB)A=BI_2A=P$. For $x=(x_1,x_2,x_3)^T$, $Px=(x_1+x_3,x_2+x_3,0)^T$. Consequently $Px=x$ exactly when $x_3=0$, and $$\boxed{P(2,-1,3)^T=(5,2,0)^T.}$$ The discarded difference is $x-Px=x_3(-1,-1,1)^T$, invisible to $A$. Indeed $APx=Ax$. This is an idempotent projection onto the plane $x_3=0$, but it is not the perpendicular projection: it changes the first two coordinates.