Science:Math Exam Resources/Courses/MATH221/April 2009/Question 10

MATH221 April 2009

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Question 10
Find the inverses of $A$ and $AA^{T}$, where $A={\begin{bmatrix}1&3&-1\\0&1&2\\0&0&1\end{bmatrix}}$

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Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. (i) find inverse of $A$ To see if inverse matrix exists, we need to verify if determinant is not zero. Matrix is upper triangular, therefore we can simply take the product of the diagonal entries as the determinant of the matrix, so $det(A)=1*1*1=1\neq 0$ Therefore, inverse matrix exists. We need $A^{-1}A=I$, so with $\left.{\begin{matrix}1&3&-1\\0&1&2\\0&0&1\end{matrix}}\right|{\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}}$ $R1+R3$ $R2-2R3$ $R1-3R2$ then we get, $\left.{\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}}\right|{\begin{matrix}1&-3&7\\0&1&-2\\0&0&1\end{matrix}}$ Therefore, inverse of A is ${\begin{bmatrix}1&-3&7\\0&1&-2\\0&0&1\end{bmatrix}}$ (ii) Find inverse of $AA^{T}$ we get $AA^{T}={\begin{bmatrix}1&3&-1\\0&1&2\\0&0&1\end{bmatrix}}{\begin{bmatrix}1&0&0\\3&1&0\\-1&2&1\end{bmatrix}}={\begin{bmatrix}11&1&-1\\1&5&2\\-1&2&1\\\end{bmatrix}}$ Again, We need $A^{-1}A=I$ , so with $\left.{\begin{matrix}11&1&-1\\1&5&2\\-1&2&1\\\end{matrix}}\right|{\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}}$ Swap 2nd and the 1st rows, eliminate 1st column R2-11R1 R3+R1 R2 divided by -54, eliminate 2nd column R3-7R2 R1-5R2 R3 divided by (1/54), eliminate 3rd column R1-(-7/54)R3 R2-(23/54)R3 yields $\left.{\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}}\right|{\begin{matrix}1&-3&7\\-3&10&-23\\7&-23&54\end{matrix}}$ Thus the inverse of $AA^{T}$ is ${\begin{bmatrix}1&-3&7\\-3&10&-23\\7&-23&54\end{bmatrix}}$