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Problem 3

Find the adjoint of the matrix \(A .\) Then use the adjoint to find the inverse of \(A\) (if possible). $$A=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 2 & 6 \\ 0 & -4 & -12 \end{array}\right]$$

Problem 3

Determine which property of determinants the equation illustrates. $$\left|\begin{array}{rrr}1 & 4 & 2 \\\0 & 0 & 0 \\\5 & 6 & -7\end{array}\right|=0$$

Problem 4

Find the adjoint of the matrix \(A .\) Then use the adjoint to find the inverse of \(A\) (if possible). $$A=\left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & -1 \\ 2 & 2 & 2 \end{array}\right]$$

Problem 4

Determine which property of determinants the equation illustrates. $$\left|\begin{array}{rrr}-3 & 2 & 1 \\\6 & 0 & 0 \\\\-3 & 2 & 1\end{array}\right|=0$$

Problem 4

The Determinant of a Matrix Product In Exercises \(1-6,\) find \((a)|A|,\) (b) \(|B|,\) (c) \(A B,\) and \((d)|A B| .\) Then verify that \(|\boldsymbol{A}||\boldsymbol{B}|=|\boldsymbol{A B}|\). $$A=\left[\begin{array}{rrr}2 & 0 & 1 \\\1 & -1 & 2 \\\3 & 1 & 0\end{array}\right], \quad B=\left[\begin{array}{rrr}2 & -1 & 4 \\\0 & 1 & 3 \\\3 & -2 & 1\end{array}\right]$$

Problem 4

Find the determinant of the matrix. $$\left[\begin{array}{rr}-3 & 1 \\ 5 & 2\end{array}\right]$$

Problem 5

Determine which property of determinants the equation illustrates. $$\left|\begin{array}{rrr}1 & 3 & 4 \\\\-7 & 2 & -5 \\\6 & 1 & 2\end{array}\right|=-\left|\begin{array}{rrr}1 & 4 & 3 \\\\-7 & -5 & 2 \\\6 & 2 & 1\end{array}\right|$$

Problem 5

Find the determinant of the matrix. $$\left[\begin{array}{rr}5 & 2 \\ -6 & 3\end{array}\right]$$

Problem 5

Find the adjoint of the matrix \(A .\) Then use the adjoint to find the inverse of \(A\) (if possible). $$A=\left[\begin{array}{rrr} -3 & -5 & -7 \\ 2 & 4 & 3 \\ 0 & 1 & -1 \end{array}\right]$$

Problem 5

The Determinant of a Matrix Product In Exercises \(1-6,\) find \((a)|A|,\) (b) \(|B|,\) (c) \(A B,\) and \((d)|A B| .\) Then verify that \(|\boldsymbol{A}||\boldsymbol{B}|=|\boldsymbol{A B}|\). $$A=\left[\begin{array}{rrrr}2 & 0 & 1 & 1 \\\1 & -1 & 0 & 1 \\\2 & 3 & 1 & 0 \\\1 & 2 & 3 & 0\end{array}\right], \quad B=\left[\begin{array}{rrrr}1 & 0 & -1 & 1 \\\2 & 1 & 0 & 2 \\\1 & 1 & -1 & 0 \\\3 & 2 & 1 & 0\end{array}\right]$$

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