How to find left and right inverse of a 3x2 matrix.

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Let A be a 3×2 matrix, L its left inverse, and R its right inverse. L and R are then matrices such that LA = I₂ (the 2×2 identity matrix) and AR = I₃ (the 3×3 identity matrix). Clearly L must be 2×3 and R must be 3×2 in order for the matrix products to be defined.

To find L and R, we start by introducing a square matrix on the the left sides of either equation above. In particular, we uniformly multiply both sides by the transpose of A, then solve for the inverse.

For the left inverse, we have

$LA=I$

$(LA)A^\top = IA^\top$

$L\left(AA^\top\right) = A^\top$

$\left(L\left(AA^\top\right)\right)\left(AA^\top\right)^{-1} = A^\top \left(AA^\top\right)^{-1}$

$L\left(\left(AA^\top\right)\left(AA^\top\right)^{-1}\right) = A^\top \left(AA^\top\right)^{-1}$

$LI = A^\top \left(AA^\top\right)^{-1}$

$L = A^\top \left(AA^\top\right)^{-1}$

We do the same thing for the right inverse, but take care with how we multiply both sides of AR = I₃.

$AR=I$

$A^\top(AR)=A^\top I$

$\left(A^\top A\right)R = A^\top$

$\left(A^\top A\right)^{-1} \left(\left(A^\top A\right)R\right) = \left(A^\top A\right)^{-1} A^\top$

$\left(\left(A^\top A\right)^{-1} \left(A^\top A\right)\right) R = \left(A^\top A\right)^{-1} A^\top$

$IR = \left(A^\top A\right)^{-1} A^\top$

$R = \left(A^\top A\right)^{-1} A^\top$

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AfilCa [17]

Check out the attached image.

Figure 1 moves to figure 2 after the translation rule (x,y) ---> (x+1, y+2)

Figure 2 moves to figure 3 after the rotation 90 degrees clockwise around the origin

Figure 3 moves to figure 4 after the translation rule (x,y) ---> (x+2, y-3)

Figure 4 is in quadrant IV. The size does not change

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Answer: Choice B) Quadrant IV; no