Exercise 138Linear algebra › Linear mapsL1Added 3 Nov 2026 Composition of linear maps #composition#matrix product#linear maps StatementLet f:R2→R3f:\mathbb{R}^2\to\mathbb{R}^3f:R2→R3 have matrixA=(120110),A=\begin{pmatrix} 1 & 2 \\ 0 & 1 \\ 1 & 0 \end{pmatrix},A=101210,and g:R3→R2g:\mathbb{R}^3\to\mathbb{R}^2g:R3→R2 have matrixB=(101210).B=\begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \end{pmatrix}.B=(120110).Compute the matrix of g∘fg\circ fg∘f. HintThe matrix of g∘fg\circ fg∘f is BABABA. Show solutionHide solutionSolutionProduct.BA=(101210)(120110)=(1+0+12+0+02+0+04+1+0)=(2225).BA = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1+0+1 & 2+0+0 \\ 2+0+0 & 4+1+0 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & 5 \end{pmatrix}.BA=(120110)101210=(1+0+12+0+02+0+04+1+0)=(2225).Check on the basis. f(1,0)=(1,0,1)f(1,0)=(1,0,1)f(1,0)=(1,0,1) and g(1,0,1)=(1+1, 2)=(2,2)g(1,0,1)=(1+1,\ 2)=(2,2)g(1,0,1)=(1+1, 2)=(2,2): first column. f(0,1)=(2,1,0)f(0,1)=(2,1,0)f(0,1)=(2,1,0) and g(2,1,0)=(2, 4+1)=(2,5)g(2,1,0)=(2,\ 4+1)=(2,5)g(2,1,0)=(2, 4+1)=(2,5): second column.Conclusion. The matrix of g∘fg\circ fg∘f is (2225)\begin{pmatrix} 2 & 2 \\ 2 & 5 \end{pmatrix}(2225).