Recent questions tagged matrix-algebra / GO Instrumentation

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GATE2020: 3
A set of linear equations is given in the form $Ax=b$, where A is a $2\times 4$ matrix with real number entries and $b\neq 0.$ will it be possible to solve for $x$ and obtain a unique solution by multiplying both left and right sides of the equation by $A^T$ (the super script $T$ denotes the transpose) and ... $A^T A$ is well conditioned Yes, can obtain a unique solution provided the matrix $A$ is well conditioned

A set of linear equations is given in the form $Ax=b$, where A is a $2\times 4$ matrix with real number entries and $b\neq 0.$ will it be possible to solve for $x$ and ob...

soujanyareddy13

2.7k points

soujanyareddy13 asked Nov 3, 2020

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2

GATE2017: 2
The figure shows a shape $\text ABC$ and its mirror image $\text A_1B_1C_1$ across the horizontal axis $\text (X-axis)$. The coordinate transformation matrix that maps $\text ABC$ to $\text A_1B_1C_1$ is $\begin{bmatrix}0&1\\1 &0\end{bmatrix}$ $\begin{bmatrix}0 &1\\-1 &0\end{bmatrix}$ $\begin{bmatrix}-1 &0\\0 &1\end{bmatrix}$ $\begin{bmatrix}1 &0\\0 &-1\end{bmatrix}$

The figure shows a shape $\text ABC$ and its mirror image $\text A_1B_1C_1$ across the horizontal axis $\text (X-axis)$. The coordinate transformation matrix that maps $\...

soujanyareddy13

2.7k points

soujanyareddy13 asked Nov 1, 2020

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3

GATE2014-26
A scalar valued function is defined as $f(X)=X^TAX+b^TX+c,$ where A is a symmetric positive definite matrix with dimension $n\times n$; b and X are vectors of dimension $n\times 1$. The minimum value of f(x) will occur when X equals $(A^TA)^{-1}b$ $-(A^TA)^{-1}b$ $-(\frac{A^{-1}b}{2})$ $\frac{A^{-1}b}{2}$

A scalar valued function is defined as $f(X)=X^TAX+b^TX+c,$ where A is a symmetric positive definite matrix with dimension $n\times n$; b and X are vectors of dimension $...

Milicevic3306

7.9k points

Milicevic3306 asked Mar 25, 2018

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4

GATE2012-27
Given that $A=\begin{bmatrix}-5 &-3\\2 &0\end{bmatrix}$ and $I=\begin{bmatrix}1 & 0\\0 &1\end{bmatrix}$, the value of $A^3$ is $15A+12I$ $19A+30I$ $17A+15I$ $17A+21I$

Given that$A=\begin{bmatrix}-5 &-3\\2 &0\end{bmatrix}$ and $I=\begin{bmatrix}1 & 0\\0 &1\end{bmatrix}$, the value of $A^3$ is$15A+12I$$19A+30I$$17A+15I$$17A+21I$

Milicevic3306

7.9k points

Milicevic3306 asked Mar 25, 2018