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Yes, SR matrices come in handy for that type of calculations. This:

# data
Qf = 1000000*matrix([[0,0],[0,1]]);
Xf = matrix([[1],[1675]]);

# matrix with symbolic coefficients
X = matrix([[var('x1')], [var('x2')]]);

# quadratic function
f = ((X-Xf).transpose()*Qf*(X-Xf));

# see result
Qf, Xf, X, f

produces

$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(\left(\begin{array}{rr} 0 & 0 \\ 0 & 1000000 \end{array}\right), \left(\begin{array}{r} 1 \\ 1675 \end{array}\right), \left(\begin{array}{r} x_{1} \\ x_{2} \end{array}\right), \left(\begin{array}{r} 1000000 \, {\left(x_{2} - 1675\right)}^{2} \end{array}\right)\right). $$

To evaluate $f$, do f(x1=1,x2=1).

More generally, to define your $X$ it can be useful to do something like:

# create a coefficients matrix of m rows and n columns
m = 4; n = 2; 
xij = [[var('x'+str(1+i)+str(1+j)) for j in range(n)] for i in range(m)]
X = matrix(SR, xij)
X

$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} x_{11} & x_{12} \\ x_{21} & x_{22} \\ x_{31} & x_{32} \\ x_{41} & x_{42} \end{array}\right). $$

The quadratic function $f$ defined above is a $1\times 1$ matrix (convince yourself, for instance by reading to output of type(f)). To take the gradient this is one possible way:

# passing from a 1x1 matrix to a scalar
f = f[0, 0]

grad_f = f.gradient([x1, x2])

# see result
grad_f

$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(0, 2000000 x_{2} - 3350000000\right). $$

Yes, SR matrices come in handy for that type of calculations. This:calculations.

# data
Qf = 1000000*matrix([[0,0],[0,1]]);
Xf = matrix([[1],[1675]]);

# matrix with symbolic coefficients
X = matrix([[var('x1')], [var('x2')]]);

# quadratic function
f = ((X-Xf).transpose()*Qf*(X-Xf));

# see result
Qf, Xf, X, f

produces

$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(\left(\begin{array}{rr} 0 & 0 \\ 0 & 1000000 \end{array}\right), \left(\begin{array}{r} 1 \\ 1675 \end{array}\right), \left(\begin{array}{r} x_{1} \\ x_{2} \end{array}\right), \left(\begin{array}{r} 1000000 \, {\left(x_{2} - 1675\right)}^{2} \end{array}\right)\right). $$

To evaluate $f$, do f(x1=1,x2=1).

More generally, to define your $X$ it can be useful to do something like:

# create a coefficients matrix of m rows and n columns
m = 4; n = 2; 
xij = [[var('x'+str(1+i)+str(1+j)) for j in range(n)] for i in range(m)]
X = matrix(SR, xij)
X

$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} x_{11} & x_{12} \\ x_{21} & x_{22} \\ x_{31} & x_{32} \\ x_{41} & x_{42} \end{array}\right). $$

The quadratic function $f$ defined above is a $1\times 1$ matrix (convince yourself, for instance by reading to output of type(f)). To take the gradient this is one possible way:

# passing from a 1x1 matrix to a scalar
f = f[0, 0]

grad_f = f.gradient([x1, x2])

# see result
grad_f

$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(0, 2000000 x_{2} - 3350000000\right). $$

Yes, SR matrices come in handy for that type of calculations.

# data
Qf = 1000000*matrix([[0,0],[0,1]]);
Xf = matrix([[1],[1675]]);

# matrix with symbolic coefficients
X = matrix([[var('x1')], [var('x2')]]);

# quadratic function
f = ((X-Xf).transpose()*Qf*(X-Xf));

# see result
Qf, Xf, X, f

produces

$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(\left(\begin{array}{rr} 0 & 0 \\ 0 & 1000000 \end{array}\right), \left(\begin{array}{r} 1 \\ 1675 \end{array}\right), \left(\begin{array}{r} x_{1} \\ x_{2} \end{array}\right), \left(\begin{array}{r} 1000000 \, {\left(x_{2} - 1675\right)}^{2} \end{array}\right)\right). $$

To evaluate $f$, do f(x1=1,x2=1).

More generally, to define your $X$ it can be useful to do something like:

# create a coefficients coefficient matrix of m rows and n columns
m = 4; n = 2; 
xij = [[var('x'+str(1+i)+str(1+j)) for j in range(n)] for i in range(m)]
X = matrix(SR, xij)
X

$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} x_{11} & x_{12} \\ x_{21} & x_{22} \\ x_{31} & x_{32} \\ x_{41} & x_{42} \end{array}\right). $$

The quadratic function $f$ defined above is a $1\times 1$ matrix (convince yourself, for instance by reading to output of type(f)). To take the gradient this is one possible way:

# passing from a 1x1 matrix to a scalar
f = f[0, 0]

grad_f = f.gradient([x1, x2])

# see result
grad_f

$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(0, 2000000 x_{2} - 3350000000\right). $$

Yes, SR matrices come in handy for that type of calculations.

# data
Qf = 1000000*matrix([[0,0],[0,1]]);
Xf = matrix([[1],[1675]]);

# matrix with symbolic coefficients
X = matrix([[var('x1')], [var('x2')]]);

# quadratic function
f = ((X-Xf).transpose()*Qf*(X-Xf));

# see result
Qf, Xf, X, f

produces

$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(\left(\begin{array}{rr} 0 & 0 \\ 0 & 1000000 \end{array}\right), \left(\begin{array}{r} 1 \\ 1675 \end{array}\right), \left(\begin{array}{r} x_{1} \\ x_{2} \end{array}\right), \left(\begin{array}{r} 1000000 \, {\left(x_{2} - 1675\right)}^{2} \end{array}\right)\right). $$

To evaluate $f$, do f(x1=1,x2=1).

More generally, to define your $X$ it can be useful to do something like:

# create a coefficient matrix of m rows and n columns
m = 4; n = 2; 
xij = [[var('x'+str(1+i)+str(1+j)) for j in range(n)] for i in range(m)]
X = matrix(SR, xij)
X

$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} x_{11} & x_{12} \\ x_{21} & x_{22} \\ x_{31} & x_{32} \\ x_{41} & x_{42} \end{array}\right). $$

The quadratic function $f$ defined above is a $1\times 1$ matrix (convince yourself, for instance by reading to the output of type(f)). To take the gradient this is one possible way:

# passing from a 1x1 matrix to a scalar
f = f[0, 0]

grad_f = f.gradient([x1, x2])

# see result
grad_f

$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(0, 2000000 x_{2} - 3350000000\right). $$