56b254…: “Next to the famous set”

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For any $n\times n$ matrix $A$ and any vector $\vec {b} \in \RR ^n$ , $A_i(\vec {b})$ is the matrix obtained from $A$ be replacing column $i$ by $\vec {b}$ .

Given that $A = \begin {bmatrix} 1& 7\\ -2 &-3\end {bmatrix}$ and $\vec {b} = \begin {bmatrix} 2\\ 9\end {bmatrix}$ compute the following.

$A_1(\vec {b}) =\begin {bmatrix} \answer {2} &\answer {7}\\ \answer {9} & \answer {-3} \end {bmatrix}$

$A_2(\vec {b}) =\begin {bmatrix} \answer {1} &\answer {2}\\ \answer {-2} & \answer {9} \end {bmatrix}$

Use Cramer’s rule to determine the unique solution $\vec {x}$ to the equation: $\begin {bmatrix} -4 & 1 \\ 2& 12 \end {bmatrix}\vec {x} = \begin {bmatrix} 3\\-1\end {bmatrix}.$

$x_1=-\dfrac {\answer {37}}{\answer {50}}$
$x_2=\dfrac {\answer {1}}{\answer {25}}$ (reduce the fraction)

Use Cramer’s rule to determine the unique solution $\vec {x}$ to the equation: $\begin {bmatrix} 1 & -5 \\ 7& -8 \end {bmatrix}\vec {x} = \begin {bmatrix} 1\\1\end {bmatrix}.$

$x_1=-\dfrac {\answer {1}}{\answer {9}}$ (reduce the fraction)
$x_2=-\dfrac {\answer {2}}{\answer {9}}$ (reduce the fraction)

For which of the following matrices can you use cramer’s rule to find the solution to $A\vec {x} = \vec {b}$ for some vector $\vec {b}$ of the appropriate dimension?

$A = \begin {bmatrix} 1 &7\\ 3& 4\\ 0&7\end {bmatrix}$
$A = \begin {bmatrix} 1 &-9\\ 0& 1\end {bmatrix}$
$A = \begin {bmatrix} 1 &7&2&5\\ 0&1&1& 4\\ 0&0&0&-7\\0&0&0&5\end {bmatrix}$ }vspace5pt

For which of the following matrices can you use cramer’s rule to find the solution to $A\vec {x} = \vec {b}$ for some vector $\vec {b}$ of the appropriate dimension?

$A = \begin {bmatrix} 1 &0 &2\\ 0&7& 1 \\0&3&1\end {bmatrix}$
$A = \begin {bmatrix} 2 &-6\\ -3& 9\end {bmatrix}$

Theorem: Let $A$ be an invertible $n\times n$ matrix. Then $A^{-1} = \dfrac {1}{\det A} \adj A.$

Let $A = \begin {bmatrix} 2& 4& 7\\ 1& 3&1\\ -1&-1&2\end {bmatrix}$ . Compute the following.

$\det A = \answer {16}$

$\adj A = \begin {bmatrix} \answer {7} & \answer {15} &\answer {-17}\\ \answer {3} & \answer {11} &\answer {-5}\\ \answer {2} & \answer {2} &\answer {2} \end {bmatrix}$

$A^{-1} = \begin {bmatrix} \frac {\answer {7}}{\answer {16}} & -\frac {\answer {15}}{\answer {16}} &-\frac {\answer {17}}{\answer {16}}\\ \frac {\answer {3}}{\answer {16}}& \frac {\answer {11}}{\answer {16}} &-\frac {\answer {5}}{\answer {16}}\\ \frac {\answer {1}}{\answer {8} }& \frac {\answer {1}}{\answer {8}} &\frac {\answer {1}}{\answer {8}} \end {bmatrix}$