[amp_mcq option1=”(ATA)-1b” option2=”-(ATA)-1b” option3=”\\[ – \\left( {\\frac{{{{\\text{A}}^{ – 1}}{\\text{b}}}}{2}} \\right)\\]” option4=”\\[\\frac{{{{\\text{A}}^{ – 1}}{\\text{b}}}}{2}\\]” correct=”option4″]<\/p>\n
The correct answer is $\\boxed{\\frac{{{{\\text{A}}^{ – 1}}{\\text{b}}}}{2}}$.<\/p>\n
Let $X$ be a vector of dimension $n \\times 1$. The function $f(X)$ is defined as follows:<\/p>\n
$$f(X) = XTAX + bTX + c$$<\/p>\n
where $A$ is a symmetric positive definite matrix with dimension $n \\times n$; $b$ and $x$ are vectors of dimension $n \\times 1$.<\/p>\n
We can write the function $f(X)$ as follows:<\/p>\n
$$f(X) = X^T(A^T + A)X + b^TX + c$$<\/p>\n
Since $A$ is a symmetric positive definite matrix, it is invertible. Therefore, we can write the function $f(X)$ as follows:<\/p>\n
$$f(X) = (X^TA^TX + b^TX + c)(A^{-1})^T$$<\/p>\n
We can minimize the function $f(X)$ by setting its derivative to zero. The derivative of $f(X)$ with respect to $X$ is given by:<\/p>\n
$$\\frac{\\partial f(X)}{\\partial X} = 2(A^T + A)X + b^T = 2AX + b^T$$<\/p>\n
Setting the derivative to zero and solving for $X$, we get:<\/p>\n
$$X = -(A^T + A)^{-1}b^T = -A^{-1}b^T$$<\/p>\n
Therefore, the minimum value of $f(X)$ will occur when $X = -A^{-1}b^T$.<\/p>\n
The other options are incorrect because they do not minimize the function $f(X)$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”(ATA)-1b” option2=”-(ATA)-1b” option3=”\\[ – \\left( {\\frac{{{{\\text{A}}^{ – 1}}{\\text{b}}}}{2}} \\right)\\]” option4=”\\[\\frac{{{{\\text{A}}^{ – 1}}{\\text{b}}}}{2}\\]” correct=”option4″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[489],"tags":[],"class_list":["post-20030","post","type-post","status-publish","format-standard","hentry","category-linear-algebra","no-featured-image-padding"],"yoast_head":"\n