I don't know how you calculate the coefficients, using a formula or other method. I post here an answer initially as Example therein : Total spin of two spin-$1/2$ particles, but I deleted as so extensive. Ignore any reference to equations not appearing in the Example.
The coefficient you are trying to calculate appears at the end in equation (Ex-26.2) which is repeated here for convenience :
\begin{equation}
\mathbf{\left|\tfrac{1}{2}\;,-\tfrac{1}{2}\right\rangle_{[1]}}
=\sqrt{\tfrac{2}{3}}\cdot\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!-\!1\right\rangle_{\!b}-\sqrt{\tfrac{1}{3}}\cdot\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:0\:\right\rangle_{\!b}
\tag{Ex-26.2}
\end{equation}
This coefficient is $\:+\sqrt{\tfrac{2}{3}}$. But it could be $\:-\sqrt{\tfrac{2}{3}}$ also, since no-one could forbid us to use as basic states of the product system the opposite ones given in equations (Ex-26.2). It's a matter of convention.
You can find calculations of Clebsch-Gordan coefficients using a general formula in Examples therein :How to determine whether an eigenstate of total spin is symmetric or antisymmetric?
Example
$\qquad j_{\alpha}=\frac{1}{2}\:, \quad j_{\beta}=1$
Let the system $\;\alpha\;$ be a particle $\;p_{\alpha}\;$ with spin $\;j_{\alpha}=1/2\;$ and the system $\;\beta\;$ be a particle $\;p_{\beta}\;$ with orbital angular momentum or spin $\;j_{\beta}=1$. Alternatively, the system $\;\beta\;$ may be the same particle $\;p_{\alpha}\;$ with orbital angular momentum $\;j_{\beta}=1$.
So, in system $\;\alpha\;$
\begin{equation}
J^{\boldsymbol{\alpha}}_{1}=\tfrac{1}{2}
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}, \quad
J^{\boldsymbol{\alpha}}_{2}=\tfrac{1}{2}
\begin{bmatrix}
0 & \!\!\!-i \\
i & 0
\end{bmatrix}, \quad
J^{\boldsymbol{\alpha}}_{3}=\tfrac{1}{2}
\begin{bmatrix}
1 & 0 \\
0 & \!\!\!-1
\end{bmatrix}
\tag{Ex-01}
\end{equation}
and
\begin{equation} \left(\mathbf{J}^{\boldsymbol{\alpha}}\right)^{2}=\left(J^{\boldsymbol{\alpha}}_{1}\right)^{2}+\left(J^{\boldsymbol{\alpha}}_{2}\right)^{2}+\left(J^{\boldsymbol{\alpha}}_{3}\right)^{2}=
j_{\alpha}\left( j_{\alpha}+1\right)\cdot \mathrm{I}_{\mathbf{a}}=\tfrac{3}{4}
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\tag{Ex-02}
\end{equation}
The basic vectors $\mathbf{a}_{\imath}\: (\imath=1,2)$ are the common eigenvectors of $\left(\mathbf{J}^{\alpha}\right)^{2}$ and $J^{\alpha}_{3}$ :
\begin{align}
\mathbf{a}_{1} & = \left|j_{\alpha},m^{\alpha}_{1} \right\rangle_{\!a}=\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}
=
\begin{bmatrix}
1 \\
0
\end{bmatrix}_{\!a}
\tag{Ex-03.1}\\
\mathbf{a}_{2} & = \left|j_{\alpha},m^{\alpha}_{2} \right\rangle_{\!a}=\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}
=
\begin{bmatrix}
0 \\
1
\end{bmatrix}_{\!a}
\tag{Ex-03.2}
\end{align}
A state of system $\alpha$ is represented by a 2-dimensional complex vector $\boldsymbol{\xi}$
\begin{align}
\boldsymbol{\xi} & = \xi_{1}\mathbf{a}_{1}\!\!+\!\xi_{2}\mathbf{a}_{2}=\xi_{1}\left|j_{\alpha},m^{\alpha}_{1} \right\rangle_{\!a}\!\!+\!\xi_{2}\left|j_{\alpha},m^{\alpha}_{2} \right\rangle_{\!a}
\nonumber\\
& =
\xi_{1}\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\!\!+\!\xi_{2}\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}
=
\xi_{1}\!\!
\begin{bmatrix}
1 \\
0
\end{bmatrix}_{\!a}
\!\!+\!
\xi_{2}\!\!
\begin{bmatrix}
0 \\
1
\end{bmatrix}_{\!a}
=
\begin{bmatrix}
\xi_{1} \\
\xi_{2}
\end{bmatrix}_{\!a}
\tag{Ex-04}
\end{align}
in Hilbert space
\begin{equation}
\mathsf{H}_{\alpha}\equiv\left\{\boldsymbol{\xi}\in \mathbb{C}^{\boldsymbol{2}}: \boldsymbol{\xi}= \xi_{1}\mathbf{a}_{1}+\xi_{2}\mathbf{a}_{2}
\right\}
\tag{Ex-05}
\end{equation}
In system $\;\beta\;$
\begin{equation}
J^{\boldsymbol{\beta}}_{1}=\sqrt{\tfrac{1}{2}}
\begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{bmatrix}, \quad
J^{\boldsymbol{\beta}}_{2}=\sqrt{\tfrac{1}{2}}
\begin{bmatrix}
0 & \!\!\!-i & 0 \\
i & 0 & \!\!\!-i \\
0 & i & 0
\end{bmatrix}, \quad
J^{\boldsymbol{\beta}}_{3}=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & \!\!\!-1
\end{bmatrix}
\tag{Ex-06}
\end{equation}
and
\begin{equation} \left(\mathbf{J}^{\boldsymbol{\beta}}\right)^{2}=\left(J^{\boldsymbol{\beta}}_{1}\right)^{2}+\left(J^{\boldsymbol{\beta}}_{2}\right)^{2}+\left(J^{\boldsymbol{\beta}}_{3}\right)^{2}=
j_{\beta}\left( j_{\beta}+1\right)\cdot \mathrm{I}_{\mathbf{b}}=2
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\tag{Ex-07}
\end{equation}
The basic vectors $\mathbf{b}_{\jmath}\: (\jmath=1,2,3)$ are the common eigenvectors of $\left(\mathbf{J}^{\beta}\right)^{2}$ and $J^{\beta}_{3}$ :
\begin{align}
\mathbf{b}_{1} & = \left|j_{\beta},m^{\beta}_{1} \right\rangle_{\!b}=\left|1,\:\!\!+\!1\right\rangle_{\!b}
=
\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}_{\!b}
\tag{Ex-08.1}\\
\mathbf{b}_{2} & = \left|j_{\beta},m^{\beta}_{2} \right\rangle_{\!b}=\left|1,\:0\:\right\rangle_{\!b}
=
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}_{\!b}
\tag{Ex-08.2}\\
\mathbf{b}_{3} & = \left|j_{\beta},m^{\beta}_{3} \right\rangle_{\!b}=\left|1,\:\!\!-\!1\right\rangle_{\!b}
=
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}_{\!b}
\tag{Ex-08.3}
\end{align}
A state of system $\beta$ is represented by a 3-dimensional complex vector $\boldsymbol{\eta}$
\begin{align}
\boldsymbol{\eta} & =\eta_{1}\mathbf{b}_{1}\!+\!\eta_{2}\mathbf{b}_{2}\!+\!\eta_{3}\mathbf{b}_{3}=
\eta_{1}\left|j_{\beta},m^{\beta}_{1}\right\rangle_{\!b}\!+\!\eta_{2} \left|j_{\beta},m^{\beta}_{2} \right\rangle_{\!b}\!+\!\eta_{3}\left|j_{\beta},m^{\beta}_{3} \right\rangle_{\!b}
\nonumber\\
& =
\eta_{1}\!\left|1,\:\!\!+\!1\right\rangle_{\!b}\!+\!\eta_{2}\!\left|1,\:0\:\right\rangle_{\!b}\!+\!\eta_{3}\!\left|1,\:\!\!-\!1\right\rangle_{\!b}
\!=\!
\eta_{1}\!\!
\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}_{\!b}
\!\!\!\!+\!\eta_{2}\!\!
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}_{\!b}
\!\!\!\!+\!\eta_{3}\!\!
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}_{\!b}
\!=\!
\begin{bmatrix}
\eta_{1} \\
\eta_{2} \\
\eta_{3}
\end{bmatrix}_{\!b}
\tag{Ex-09}
\end{align}
in Hilbert space
\begin{equation}
\mathsf{H}_{\beta}\equiv\left\{\boldsymbol{\eta}\in \mathbb{C}^{\boldsymbol{3}}: \boldsymbol{\eta} =\eta_{1}\mathbf{b}_{1}+\eta_{2}\mathbf{b}_{2}+\eta_{3}\mathbf{b}_{3}
\right\}
\tag{Ex-10}
\end{equation}
According to the general equations (15) a product state of the composite system is
\begin{equation}
\boldsymbol{\chi} = \boldsymbol{\xi} \boldsymbol{\otimes} \boldsymbol{\eta}=\left( \sum_{\imath=1}^{\imath=2}\xi_{\imath}\mathbf{a}_{\imath}\right) \boldsymbol{\otimes}\left( \sum_{\jmath=1}^{\jmath=3}\eta_{\jmath}\mathbf{b}_{\jmath}\right)= \sum_{\imath,\jmath=1,1}^{\imath,\jmath=2,3}\xi_{\imath}\eta_{\jmath}\left( \mathbf{a}_{\imath} \boldsymbol{\otimes }\mathbf{b}_{\jmath}\right)
\tag{Ex-11}
\end{equation}
with matrix representation, in agreement with equation (18)
\begin{equation}
\boldsymbol{\chi}=
\begin{bmatrix}
\begin{array}{c}
\chi_{1} \\
\chi_{2} \\
\chi_{3} \\
\chi_{4}\\
\chi_{5}\\
\chi_{6}
\end{array}
\end{bmatrix}_{\!e}=
\begin{bmatrix}
\begin{array}{c}
\xi_{1}\eta_{1} \\
\xi_{1}\eta_{2} \\
\xi_{1}\eta_{3} \\
\xi_{2}\eta_{1} \\
\xi_{2}\eta_{2} \\
\xi_{2}\eta_{3}
\end{array}
\end{bmatrix}_{\!e}= \boldsymbol{\xi} \boldsymbol{\otimes} \boldsymbol{\eta}
\tag{Ex-12}
\end{equation}
This representation is relatively to the basis $\:\left\lbrace \mathbf{e}_{k}\right\rbrace \:$ defined according to the general equations (16):
\begin{align}
\mathbf{e}_{1} & \equiv \mathbf{a}_{1}\boldsymbol{\otimes} \mathbf{b}_{1}=\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\boldsymbol{\otimes}\left|1,\:\!\!+\!1\right\rangle_{\!b}=
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}^{\!\mathsf{T}}
\tag{Ex-13.1}\\
\mathbf{e}_{2} & \equiv \mathbf{a}_{1}\boldsymbol{\otimes} \mathbf{b}_{2}=\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\boldsymbol{\otimes}\left|1,\:0\:\right\rangle_{\!b}=
\begin{bmatrix}
0 & 1 & 0 & 0 & 0 & 0
\end{bmatrix}^{\!\mathsf{T}}
\tag{Ex-13.2}\\
\mathbf{e}_{3} & \equiv \mathbf{a}_{1}\boldsymbol{\otimes} \mathbf{b}_{3}=\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\boldsymbol{\otimes}\left|1,\:\!\!-\!1\right\rangle_{\!b}=
\begin{bmatrix}
0 & 0 & 1 & 0 & 0 & 0
\end{bmatrix}^{\!\mathsf{T}}
\tag{Ex-13.3}\\
\mathbf{e}_{4} & \equiv \mathbf{a}_{2}\boldsymbol{\otimes} \mathbf{b}_{1}=\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\boldsymbol{\otimes}\left|1,\:\!\!+\!1\right\rangle_{\!b}=
\begin{bmatrix}
0 & 0 & 0 & 1 & 0 & 0
\end{bmatrix}^{\!\mathsf{T}}
\tag{Ex-13.4}\\
\mathbf{e}_{5} & \equiv \mathbf{a}_{2}\boldsymbol{\otimes} \mathbf{b}_{2}=\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\boldsymbol{\otimes}\left|1,\:0\:\right\rangle_{\!b}=
\begin{bmatrix}
0 & 0 & 0 & 0 & 1 & 0
\end{bmatrix}^{\!\mathsf{T}}
\tag{Ex-13.5}\\
\mathbf{e}_{6} & \equiv \mathbf{a}_{2}\boldsymbol{\otimes} \mathbf{b}_{3}=\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\boldsymbol{\otimes}\left|1,\:\!\!-\!1\right\rangle_{\!b}=
\begin{bmatrix}
0 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}^{\!\mathsf{T}}
\tag{Ex-13.6}
\end{align}
where the symbol $\:\mathsf{T}\:$ means the Transpose.
According to the general equation (23) the product space is the 6-dimensional complex Hilbert space
\begin{equation}
\mathsf{H}_{f}=\mathsf{H}_{\alpha}\boldsymbol{\otimes}\mathsf{H}_{\beta}\equiv \lbrace \; \boldsymbol{\chi} \; : \;\boldsymbol{\chi}=\sum_{k=1}^{k=6}\chi_{k}\mathbf{e}_{k},\;\chi_{k} \in \mathbb{C} \rbrace
\tag{Ex-14}
\end{equation}
identical to $\mathbb{C}^{6}$.
From equation (69c) with the help of (47), both repeated here for convenience
\begin{equation}
J_{3} = \Bigl(J^{\boldsymbol{\alpha}}_{3}\boldsymbol{\otimes}\mathrm{I}_{\boldsymbol{\beta}}\Bigr)+\Bigl(\mathrm{I}_{\boldsymbol{\alpha}}\boldsymbol{\otimes} J^{\boldsymbol{\beta}}_{3}\Bigr)
\tag{69c}
\end{equation}
\begin{equation}
\mathrm{C}=\mathrm{A}\boldsymbol{\otimes} \mathrm{B} =
\begin{bmatrix}
a_{11}\mathrm{B} & a_{12}\mathrm{B} & \cdots & a_{1 \rho}\mathrm{B} & \cdots & a_{1r}\mathrm{B} \\
a_{21}\mathrm{B} & a_{22}\mathrm{B} & \cdots & a_{2 \rho}\mathrm{B} & \cdots & a_{2r}\mathrm{B} \\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{\imath 1}\mathrm{B} & a_{\imath 2}\mathrm{B} & \cdots & a_{\imath \rho}\mathrm{B} & \cdots & a_{\imath r}\mathrm{B}\\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{r1}\mathrm{B} & a_{r2}\mathrm{B} & \cdots & a_{r \rho}\mathrm{B} & \cdots & a_{rr}\mathrm{B}
\end{bmatrix}
\tag{47}
\end{equation}
and the matrix expressions of $\:J^{\boldsymbol{\alpha}}_{3}\:$ and $\:J^{\boldsymbol{\beta}}_{3}\:$ in equations (Ex-01) and (Ex-06) respectively, we have
\begin{align}
\Bigl(J^{\boldsymbol{\alpha}}_{3}\boldsymbol{\otimes}\mathrm{I}_{\boldsymbol{\beta}}\Bigr) & =
\tfrac{1}{2}
\begin{bmatrix}
1 & 0 \\
& \\
0 & \!\!\!-1
\end{bmatrix}
\boldsymbol{\otimes}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
=
\begin{bmatrix}
\tfrac{1}{2} & 0 & 0 & 0 & 0 & 0\\
0 & \tfrac{1}{2} & 0 & 0 & 0 & 0\\
0 & 0 & \tfrac{1}{2} & 0 & 0 & 0\\
0 & 0 & 0 & \!\!\!-\tfrac{1}{2} & 0 & 0\\
0 & 0 & 0 & 0 & \!\!\!-\tfrac{1}{2} & 0\\
0 & 0 & 0 & 0 & 0 & \!\!\!-\tfrac{1}{2}
\end{bmatrix}
\tag{Ex-15.1}\\
\Bigl(\mathrm{I}_{\boldsymbol{\alpha}}\boldsymbol{\otimes} J^{\boldsymbol{\beta}}_{3}\Bigr)& =\quad \!\!
\begin{bmatrix}
1 & 0 \\
& \\
0 & 1
\end{bmatrix}
\boldsymbol{\otimes}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & \!\!\!-1
\end{bmatrix}
=
\begin{bmatrix}
\;1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & -1 & 0 & 0 & 0\\
0 & 0 & 0 & \;1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & -1
\end{bmatrix}
\tag{Ex-15.2}
\end{align}
Adding equations (15.1), (15.2) we find the following diagonal form of $\:J_{\boldsymbol{3}}\:$
\begin{equation}
J_{\boldsymbol{3}} =
\begin{bmatrix}
\begin{array}{cccccc}
\!\!+\frac{3}{2}&0&0&0&0&0\\
0&\!\!+\frac{1}{2}&0&0&0&0\\
0&0&\!\!-\frac{1}{2}&0&0&0\\
0&0&0&\!\!+\frac{1}{2}&0&0\\
0&0&0&0&\!\!-\frac{1}{2}&0\\
0&0&0&0&0&\!\!-\frac{3}{2}
\end{array}
\end{bmatrix}
\tag{Ex-16}
\end{equation}
As expected its diagonal elements, that is its eigenvalues, are all possible sums $\;\left(m^{\boldsymbol{\alpha}}_{\boldsymbol{\imath}}+m^{\boldsymbol{\beta}}_{\boldsymbol{\jmath}}\right) \;$ of the corresponding eigenvalues of its summands. These are the $\;2\cdot3\;$ combinations of
\begin{align}
m^{\boldsymbol{\alpha}}_{\boldsymbol{\imath}} & = +\tfrac{1}{2},-\tfrac{1}{2}
\tag{16.1a}\\
m^{\boldsymbol{\beta}}_{\boldsymbol{\jmath}} & = +1,0,-1
\tag{16.1b}
\end{align}
From general expression for $\:\mathbf{J}^{\boldsymbol{2}}\:$, equation (80), which we repeat here for convenience
\begin{equation}
\mathbf{J}^{\boldsymbol{2}} =\bigl[ j_{\alpha}(j_{\alpha}+1)+ j_{\beta}(j_{\beta}+1) \bigr] \mathrm{I}_{f} +2\sum_{q=1}^{q=3}\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{q}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{q}}\Bigr)
\tag{80}
\end{equation}
we have for the first term of the right hand side the following scalar multiple of the $\:6 \times 6\:$ identity matrix
\begin{equation}
\bigl[ j_{\alpha}(j_{\alpha}+1)+ j_{\beta}(j_{\beta}+1) \bigr] \mathrm{I}_{f}= \bigl(\tfrac{3}{4}+2 \bigr) \mathrm{I}_{f}=
\tfrac{11}{4}
\begin{bmatrix}
\begin{array}{cccccc}
1&0&0&0&0&0\\
0&1&0&0&0&0\\
0&0&1&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&1&0\\
0&0&0&0&0&1
\end{array}
\end{bmatrix}
\tag{Ex-17}
\end{equation}
while using the matrix representations of $\:J^{\boldsymbol{\alpha}}_{\boldsymbol{q}}\:$ and $\:J^{\boldsymbol{\beta}}_{\boldsymbol{q}}\:$ for the three terms in series, equations (Ex-01) and (Ex-06) respectively, we have successively
\begin{align}
J^{\boldsymbol{\alpha}}_{\boldsymbol{1}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{1}} & =
\tfrac{1}{2}
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
\boldsymbol{\otimes}
\sqrt{\tfrac{1}{2}}
\begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{bmatrix}
= \tfrac{1}{2\sqrt{2}}
\begin{bmatrix}
0&0&0&0&1&0\\
0&0&0&1&0&1\\
0&0&0&0&1&0\\
0&1&0&0&0&0\\
1&0&1&0&0&0\\
0&1&0&0&0&0
\end{bmatrix}
\tag{Ex-18.1}\\
J^{\boldsymbol{\alpha}}_{\boldsymbol{2}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{2}} & =
\tfrac{1}{2}
\begin{bmatrix}
0 & \!\!\!-i \\
i & 0
\end{bmatrix}
\boldsymbol{\otimes}
\sqrt{\tfrac{1}{2}}
\begin{bmatrix}
0 & \!\!\!-i & 0 \\
i & 0 & \!\!\!-i \\
0 & i & 0
\end{bmatrix}
=\tfrac{1}{2\sqrt{2}}\!\!
\begin{bmatrix}
0&0&0&0&\!\!\!\!-\!1&0\\
0&0&0&1&0&\!\!\!\!-\!1\\
0&0&0&0&1&0\\
0&1&0&0&0&0\\
\!\!-\!1&0&1&0&0&0\\
0&\!\!\!\!-\!1&0&0&0&0
\end{bmatrix}
\tag{Ex-18.2}\\
J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{3}} & =
\tfrac{1}{2}
\begin{bmatrix}
1 & 0 \\
0 & \!\!\!-1
\end{bmatrix}
\boldsymbol{\otimes}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & \!\!\!-1
\end{bmatrix}
=\tfrac{1}{2}
\begin{bmatrix}
1&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&\!\!\!-\!1&0&0&0\\
0&0&0&\!\!\!-\!1&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&1
\end{bmatrix}
\tag{Ex-18.3}
\end{align}
so adding equations (18)
\begin{equation}
2\sum_{q=1}^{q=3}\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{q}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{q}}\Bigr)=
\begin{bmatrix}
1&0&0&0&0&0\\
0&0&0&\sqrt{2}&0&0\\
0&0&-1&0&\sqrt{2}&0\\
0&\sqrt{2}&0&-1&0&0\\
0&0&\sqrt{2}&0&0&0\\
0&0&0&0&0&1
\end{bmatrix}
\tag{Ex-19}
\end{equation}
while adding equations (17) and (19) we have finally for $\:\mathbf{J}^{\boldsymbol{2}}\:$
\begin{equation}
\mathbf{J}^{\boldsymbol{2}} =
\begin{bmatrix}
\frac{15}{4}&0&0&0&0&0\\
0&\frac{11}{4}&0&\sqrt{2}&0&0\\
0&0&\frac{7}{4}&0&\sqrt{2}&0\\
0&\sqrt{2}&0&\frac{7}{4}&0&0\\
0&0&\sqrt{2}&0&\frac{11}{4}&0\\
0&0&0&0&0&\frac{15}{4}
\end{bmatrix}
\tag{Ex-20}
\end{equation}
Now, to find the eigenvalues and eigenvectors of this $\;6\times 6 \;$ symmetric matrix $\:\mathbf{J}^{\boldsymbol{2}}\:$ is not so difficult as it seems from a first glance because :
1. The state $\:\mathbf{e}_{1}\:$ is a common eigenstate of $\:J_{\boldsymbol{3}}\:$ and $\:\mathbf{J}^{\boldsymbol{2}} \:$ of eigenvalue $\widetilde{m}_{1}=+\tfrac{3}{2}$ and $\:\widetilde{\lambda}_{1}=\tfrac{15}{4}=\tfrac{3}{2}\left(\tfrac{3}{2}+1\right)\:$ respectively :
\begin{align}
J_{\boldsymbol{3}}\mathbf{e}_{1} & = \widetilde{m}_{1}\cdot\mathbf{e}_{1} =\left( +\tfrac{3}{2}\right) \cdot\mathbf{e}_{1}
\tag{Ex-21a}\\
\mathbf{J}^{\boldsymbol{2}}\mathbf{e}_{1} & = \widetilde{\lambda}_{1}\cdot\mathbf{e}_{1}=\tfrac{15}{4}\cdot\mathbf{e}_{1}=\tfrac{3}{2}\left(\tfrac{3}{2}+1\right)\cdot\mathbf{e}_{1}
\tag{Ex-21b}
\end{align}
The operator $\:\mathbf{J}^{\boldsymbol{2}} \:$ leaves invariant the 1-dimensional eigenspace of $\:J_{\boldsymbol{3}}\:$ with eigenvalue $\:\widetilde{m}_{1}=+\tfrac{3}{2}$, that is the subspace spanned by the state $\:\left\lbrace \mathbf{e}_{1}\right\rbrace$.
2. The state $\:\mathbf{e}_{6}\:$ is a common eigenstate of $\:J_{\boldsymbol{3}}\:$ and $\:\mathbf{J}^{\boldsymbol{2}} \:$ of eigenvalue $\widetilde{m}_{6}=-\tfrac{3}{2}$ and $\:\widetilde{\lambda}_{6}=\tfrac{15}{4}=\tfrac{3}{2}\left(\tfrac{3}{2}+1\right)\:$ respectively :
\begin{align}
J_{\boldsymbol{3}}\mathbf{e}_{6} & = \widetilde{m}_{6}\cdot\mathbf{e}_{6}=\left( -\tfrac{3}{2}\right) \cdot\mathbf{e}_{6}
\tag{Ex-22a}\\
\mathbf{J}^{\boldsymbol{2}}\mathbf{e}_{6} & = \widetilde{\lambda}_{6}\cdot\mathbf{e}_{6}=\tfrac{15}{4}\cdot\mathbf{e}_{6}=\tfrac{3}{2}\left(\tfrac{3}{2}+1\right)\cdot\mathbf{e}_{6}
\tag{Ex-22b}
\end{align}
The operator $\:\mathbf{J}^{\boldsymbol{2}} \:$ leaves invariant the 1-dimensional eigenspace of $\:J_{\boldsymbol{3}}\:$ with eigenvalue $\:\widetilde{m}_{6}=-\tfrac{3}{2}$, that is the subspace spanned by the state $\:\left\lbrace \mathbf{e}_{6}\right\rbrace$.
3.The eigenstates $\:\mathbf{e}_{2}\:$ and $\:\mathbf{e}_{4}\:$ of $\:J_{\boldsymbol{3}}\:$ with eigenvalue $\:+\tfrac{1}{2}\:$ are transformed by $\:\mathbf{J}^{\boldsymbol{2}} \:$ to linear combinations of these same eigenstates
\begin{align}
\mathbf{J}^{\boldsymbol{2}}\mathbf{e}_{2} & = \tfrac{11}{4}\cdot\mathbf{e}_{2}+\sqrt{2}\cdot\mathbf{e}_{4}
\tag{Ex-23a}\\
\mathbf{J}^{\boldsymbol{2}}\mathbf{e}_{4} & = \sqrt{2}\cdot\mathbf{e}_{2}+\tfrac{7}{4}\cdot\mathbf{e}_{4}
\tag{Ex-23b}
\end{align}
The operator $\:\mathbf{J}^{\boldsymbol{2}} \:$ leaves invariant
the 2-dimensional eigenspace of $\:J_{\boldsymbol{3}}\:$ with eigenvalue $\:+\tfrac{1}{2}=\widetilde{m}_{2}=\widetilde{m}_{4}$, that is the subspace spanned by states $\:\left\lbrace \mathbf{e}_{2},\mathbf{e}_{4} \right\rbrace \:$. Its restriction on this subspace is represented by a real symmetric $\:2 \times 2\:$ matrix, so it has in this subspace two real eigenvalues which moreover are positive and different, see in the following.
4.The eigenstates $\:\mathbf{e}_{3}\:$ and $\:\mathbf{e}_{5}\:$ of $\:J_{\boldsymbol{3}}\:$ with eigenvalue $\:-\tfrac{1}{2}\:$ are transformed by $\:\mathbf{J}^{\boldsymbol{2}} \:$ to linear combinations of these same eigenstates
\begin{align}
\mathbf{J}^{\boldsymbol{2}}\mathbf{e}_{3} & = \tfrac{7}{4}\cdot\mathbf{e}_{3}+\sqrt{2}\cdot\mathbf{e}_{5}
\tag{Ex-24a}\\
\mathbf{J}^{\boldsymbol{2}}\mathbf{e}_{5} & = \sqrt{2}\cdot\mathbf{e}_{3}+\tfrac{11}{4}\cdot\mathbf{e}_{5}
\tag{Ex-24b}\\
\end{align}
The operator $\:\mathbf{J}^{\boldsymbol{2}} \:$ leaves invariant the 2-dimensional eigenspace of $\:J_{\boldsymbol{3}}\:$ with eigenvalue $\:-\tfrac{1}{2}=\widetilde{m}_{3}=\widetilde{m}_{5}$, that is the subspace spanned by states $\:\left\lbrace \mathbf{e}_{3},\mathbf{e}_{5}\right\rbrace$. Its restriction on this subspace is represented by a real symmetric $\:2 \times 2\:$ matrix, so it has in this subspace two real eigenvalues which moreover are positive and different, see in the following.
\begin{equation}
\widehat{\mathbf{J}}^{\boldsymbol{2}} =
\begin{bmatrix}
\begin{array}{cc|cccc}
\frac{3}{4}& & & & & \\
&\frac{3}{4} & & & &\\
\hline
& &\frac{15}{4}& & & \\
& & &\frac{15}{4}& & \\
& & & &\frac{15}{4}& \\
& & & & &\frac{15}{4}
\end{array}
\end{bmatrix}
\tag{Ex-25}
\end{equation}
\begin{equation}
\widehat{J}_{\boldsymbol{1}} =
\begin{bmatrix}
\begin{array}{cc|cccc}
0 & \tfrac{1}{2}& & & & \\
\tfrac{1}{2}&0& & & & \\
\hline
& &0&\tfrac{\sqrt{3}}{2}&0&0 \\
& &\tfrac{\sqrt{3}}{2}&0&1&0 \\
& &0&1&0&\tfrac{\sqrt{3}}{2} \\
& &0&0&\tfrac{\sqrt{3}}{2}&0
\end{array}
\end{bmatrix}
\tag{Ex-25.1}
\end{equation}
\begin{equation}
\widehat{J}_{\boldsymbol{2}} =
\begin{bmatrix}
\begin{array}{cc|cccc}
0 &\!\!\!-i\tfrac{1}{2}& & & & \\
i\tfrac{1}{2} & 0& & & & \\
\hline
& &0&\!\!\!\!-i\tfrac{\sqrt{3}}{2}&0&0\\
& &\!\!\!i\tfrac{\sqrt{3}}{2}&0&-i&0 \\
& &0&i&0&\!\!\!\!-i\tfrac{\sqrt{3}}{2} \\
& &0&0&\!\!\!i\tfrac{\sqrt{3}}{2}&0
\end{array}
\end{bmatrix}
\tag{Ex-25.2}
\end{equation}
\begin{equation}
\widehat{J}_{\boldsymbol{3}} =
\begin{bmatrix}
\begin{array}{cc|cccc}
\frac{1}{2}& & & & & \\
&-\frac{1}{2} & & & & \\
\hline
& &\frac{3}{2}& & & \\
& & &\frac{1}{2}& & \\
& & & &-\frac{1}{2}& \\
& & & & &-\frac{3}{2}
\end{array}
\end{bmatrix}
\tag{Ex-25.3}
\end{equation}
Above matrix representations are with respect to the basis $\:\lbrace\mathbf{f}_{k}, k=1,2,3,4,5,6 \rbrace$ (see Table 2).
\begin{align}
\mathbf{f}_{1} & = \mathbf{\left|\tfrac{1}{2}\;,+\tfrac{1}{2}\right\rangle_{[1]}}
=\sqrt{\tfrac{1}{3}}\cdot\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:0\:\right\rangle_{\!b}-\sqrt{\tfrac{2}{3}}\cdot\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!+\!1\right\rangle_{\!b}
\tag{Ex-26.1}\\
\mathbf{f}_{2} & = \mathbf{\left|\tfrac{1}{2}\;,-\tfrac{1}{2}\right\rangle_{[1]}}
=\sqrt{\tfrac{2}{3}}\cdot\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!-\!1\right\rangle_{\!b}-\sqrt{\tfrac{1}{3}}\cdot\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:0\:\right\rangle_{\!b}
\tag{Ex-26.2}\\
\mathbf{f}_{3} & = \mathbf{\left|\tfrac{3}{2}\;,+\tfrac{3}{2}\right\rangle_{[2]}}
= \left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!+\!1\right\rangle_{\!b}
\tag{Ex-26.3}\\
\mathbf{f}_{4} & = \mathbf{\left|\tfrac{3}{2}\;,+\tfrac{1}{2}\right\rangle_{[2]}}
=\sqrt{\tfrac{2}{3}}\cdot\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:0\:\right\rangle_{\!b}+\sqrt{\tfrac{1}{3}}\cdot\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!+\!1\right\rangle_{\!b}
\tag{Ex-26.4}\\
\mathbf{f}_{5} & = \mathbf{\left|\tfrac{3}{2}\;,-\tfrac{1}{2}\right\rangle_{[2]}}
=\sqrt{\tfrac{1}{3}}\cdot\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!-\!1\right\rangle_{\!b}+\sqrt{\tfrac{2}{3}}\cdot\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:0\:\right\rangle_{\!b}
\tag{Ex-26.5}\\
\mathbf{f}_{6} & = \mathbf{\left|\tfrac{3}{2}\;,-\tfrac{3}{2}\right\rangle_{[2]}}
= \left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!-\!1\right\rangle_{\!b}
\tag{Ex-26.6}
\end{align}
Expressions are simplified by omitting the symbol $\;'\boldsymbol{\otimes}'\;$ in the product of states.
