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# Consider the matrices

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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

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

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

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

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

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Aug 15, 2019
edited by Rom  Aug 16, 2019
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Consider the matrices

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

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

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

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

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

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

Aug 16, 2019
edited by heureka  Aug 16, 2019