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Consider the matrices

\(\mathbf{A} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \mathbf{B} = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 &3 \\ 1 & 3 & 5 \end{pmatrix}, \mathbf{C} = \begin{pmatrix} 1 & -2 & 1 \\ 2 & -4 &2 \\ 3 & -6 & 3 \end{pmatrix}, \mathbf{D} = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1\\ 1 & 1 & 0 \end{pmatrix}.\)

For each matrix in the list above, figure out whether the matrix is invertible.

 

For each matrix in the list above, enter in "point" if the output grid covers a point, "line" if the output grid covers a line, "plane" if the output grid covers a plane, and "3-space" if it covers all of three-dimensional space.

 Aug 15, 2019
 #1
avatar+6179 
+1

\(\text{The best way to proceed here is to row reduce each matrix to reveal it's rank}\\ \text{You'll need that info for the second part of the question anyway}\\~\\ \text{$\textbf{A}$ is the identify matrix. You should know immediately that it's rank 3 and thus invertible}\\~\\ \text{$\textbf{B}$ has to be row reduced}\\ \text{$\textbf{B} \sim \left( \begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ \end{array} \right)$}\\ \text{We see that $\textbf{B}$ is rank 2 and thus not invertible}\)

 

\(\text{$\textbf{C} \sim \left( \begin{array}{ccc} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)$}\\ \text{$\textbf{C}$ is seen to be rank 1 and thus not invertible}\)

 

\(\text{$\textbf{D}$ row reduces to the identity matrix and is thus rank 3 and invertible}\)

 

\(\text{As far as the output goes rank 3 outputs to all 3D space, rank 2 to a plane, rank 1 to a line}\\ \text{$\textbf{A}$ and $\textbf{D}$ output to all 3D space}\\ \text{$\textbf{B}$ outputs to a plane}\\ \text{$\textbf{C}$ outputs to a line}\)

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 Aug 15, 2019
edited by Rom  Aug 16, 2019
 #2
avatar+24388 
+1

Consider the matrices

\(\mathbf{A} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \mathbf{B} = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 &3 \\ 1 & 3 & 5 \end{pmatrix}, \mathbf{C} = \begin{pmatrix} 1 & -2 & 1 \\ 2 & -4 &2 \\ 3 & -6 & 3 \end{pmatrix}, \mathbf{D} = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1\\ 1 & 1 & 0 \end{pmatrix}.\)

 

\(\boxed{\text{Any matrix }\mathbf{A}\text{ can be inverted if the following applies: } \mathbf{det(A)\ne 0} } \)

 

 

\(\begin{array}{|rcll|} \hline \det(A) &=& \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ &=& 1\cdot 1 \cdot 1 +0\cdot 0 \cdot 0 +0\cdot 0 \cdot 0 -0\cdot 1 \cdot 0 -1\cdot 0 \cdot 0 -0\cdot 0 \cdot 1 \\ &=& 1+0+0-0-0-0 \\ &=& \mathbf{1} \quad | \quad \mathbf{A} \text{ is invertible } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \det(B) &=& \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 &3 \\ 1 & 3 & 5 \end{vmatrix} \\ &=& 1\cdot 2 \cdot 5 +1\cdot 1 \cdot 3 +1\cdot 3 \cdot 1 -1\cdot 2 \cdot 1 -1\cdot 3 \cdot 3 -1\cdot 1 \cdot 5 \\ &=& 10+3+3-2-9-5 \\ &=& \mathbf{0} \quad | \quad \mathbf{B} \text{ is not invertible } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \det(C) &=& \begin{vmatrix} 1 & -2 & 1 \\ 2 & -4 &2 \\ 3 & -6 & 3 \end{vmatrix} \\ &=& 1\cdot (-4) \cdot 3 +3\cdot (-2) \cdot 2 +2\cdot (-6) \cdot 1 -3\cdot (-4) \cdot 1 -1\cdot (-6) \cdot 2 -2\cdot (-2) \cdot 3 \\ &=& -12-12-12+12+12+12 \\ &=& \mathbf{0} \quad | \quad \mathbf{C} \text{ is not invertible } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \det(D) &=& \begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 1\\ 1 & 1 & 0 \end{vmatrix} \\ &=& 1\cdot 1 \cdot 0 +1\cdot 0 \cdot 1 +0\cdot 1 \cdot 1 -1\cdot 1 \cdot 1 -1\cdot 1 \cdot 1 -0\cdot 0 \cdot 0 \\ &=& 0+0+0-1-1-0 \\ &=& \mathbf{-2} \quad | \quad \mathbf{D} \text{ is invertible } \\ \hline \end{array}\)

 

laugh

 Aug 16, 2019
edited by heureka  Aug 16, 2019

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