What would the current $I_1$ be in Fig. 19–61 if the $12$-$\Omega$ resistor is shorted out (resistance = 0)? Let $r = 1.0 \; \Omega$.

Hi jlara4606, thanks for pointing out the textbook mistake, which is indeed a mistake. To check the work in this question, or any "systems of equations" problem, substitute the answers back into the original. Continuing on to find $I_2$ and $I_3$ we can substitute into the original three equations.
Equation 1: $24-23I_1-29I_2 = 0$
Equation 2: $18-23I_1-28I_3 = 0$
Equation 3 (rearranged slightly for convenience): $I_1 - I_2 - I_3 = 0$

Here are the results if $I_1 = 0.5642$:
I1 0.5624
I2 0.38156056
I3 0.18094464
Equation 1: -0.00045624
Equation 2: -0.00164992
Equation 3: -0.0001052

All three equations are expected to be zero, and it's close enough since there was some intermediate rounding in my work.

Compare that if $I_1 = 0.71$, then:
I1 0.71
I2 0.264499
I3 0.059706
Equation 1: -0.000471
Equation 2: -0.001768
Equation 3: 0.385795

And it's the result for Equation 3, $0.385795$ which is the red flag that $0.71$ is not correct, since it's supposed to be zero.

The take-away message is the strategy for checking your work with systems of equations by substituting the answers back into the original equations and check that the equations equal what they are supposed to (zero in each case here).

All the best with your studies,
Mr. Dychko