Review : Matrices & Vectors — Question 9

Question 9 – Solution

Strategy. Images of a basis determine a linear map, while repeated information along one input direction leaves freedom.

Step 1: Recover the columns from independent inputs. The first and second measurements are respectively the sum and difference of the columns of $A$. Half their sum gives column one; half their difference gives column two. Therefore $$\boxed{A=\begin{pmatrix}1&1\\2&-2\end{pmatrix},\qquad\det A=-4\ne 0.}$$ The column sum is $(2,0)^T$ and the difference is $(0,4)^T$, checking both measurements. Both columns are forced, so recovery is unique.

Step 2: Predict by linearity and test a new report. Because $(3,1)^T=2(1,1)^T+(1,-1)^T$, linearity predicts $\boxed{A(3,1)^T=2(2,0)^T+(0,4)^T=(4,4)^T}$. Thus a report of $(4,5)^T$ is inconsistent with the two original measurements for any linear map, not just with a particular computation of its entries.

Step 3: Describe the dependent-input experiment. Now the second input and output are exactly twice the first, so the second measurement adds no restriction. Write the first column as $(\alpha,\beta)^T$. The required column sum $(2,0)^T$ forces the second column, giving $$\boxed{A_{\alpha,\beta}=\begin{pmatrix}\alpha&2-\alpha\\ \beta&-\beta\end{pmatrix},\qquad \alpha,\beta\in\mathbb R.}$$ Every member gives both required outputs; every compatible matrix has this form.

Step 4: Classify inversion and the missing information. The determinant is $-\alpha\beta-(2-\alpha)\beta=-2\beta$. Thus exactly the members with $\boxed{\beta\ne 0}$ are invertible. If another input $w$ is outside the line through $(1,1)^T$, those two inputs form a basis. Each standard coordinate vector has a unique expression in that basis, so the two measured images uniquely determine both columns. Any assigned image for this independent input defines one linear map; invertibility would additionally require independent output images.