Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 23

MATH152 April 2016

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Question A 23
Find the inverse of the matrix $A={\begin{bmatrix}2&1&3\\1&0&1\\0&1&2\end{bmatrix}}.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
1. For each entry of the matrix, ignore the values on the current row and column, and compute the determinant of the remaining minor matrix. 2. With values from step (1), make the matrix of cofactors (Don't forget the alternating signs). 3. Transpose the Matrix of cofactors. 4. Multiply that by ${\frac {1}{\det(A)}}$.

Hint 2

Instead of doing this for each column, we can row reduce all these systems simultaneously, by attaching all columns of $I$ (i.e. the whole matrix $I$) on the right of $A$ in the augmented matrix and obtaining all columns of $X$ (i.e. the whole inverse matrix) on the right of the identity matrix in the row-equivalent matrix: $[A|I]\rightarrow [I|X]$. If this procedure works out, i.e. if we are able to convert $A$ to identity using row operations, then $A$is invertible and $A^{-1}=X$.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 1st entry: $a_{11}=2$, the determinant of the minor matrix is $-1$, and the sign for this entry is $(-1)^{1+1}=1\longrightarrow -1.$ 2nd entry: $a_{12}=1$, the determinant of the minor matrix is $2$, and the sign for this entry is $(-1)^{1+2}=-1\longrightarrow -2.$ 2nd entry: $a_{13}=3$, the determinant of the minor matrix is $1$, and the sign for this entry is $(-1)^{1+3}=1\longrightarrow 1.$ Continue with this method, we have the cofactor matrix ${\begin{bmatrix}-1&-2&1\\1&4&-2\\1&1&-1\end{bmatrix}}.$ Transpose the cofactor matrix to find the adjugate/adjoint matrix: ${\begin{bmatrix}-1&1&1\\-2&4&1\\1&-2&-1\end{bmatrix}}.$ We now find $\det(A)$: $\det(A)=2\times (0-1)-1\times (2-3)+0=-1$. Therefore, $A^{-1}=(-1){\begin{bmatrix}-1&1&1\\-2&4&1\\1&-2&-1\end{bmatrix}}={\begin{bmatrix}1&-1&-1\\2&-4&-1\\-1&2&1\end{bmatrix}}.$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Alternatively, we can also use row reduction. Form the augmented matrix $[A|I]=$ $\left[{\begin{array}{ccc|ccc}2&1&3&1&0&0\\1&0&1&0&1&0\\0&1&2&0&0&1\end{array}}\right]$ Interchange $R_{2}$ and $R_{1}$ : $\left[{\begin{array}{ccc|ccc}1&0&1&0&1&0\\2&1&3&1&0&0\\0&1&2&0&0&1\end{array}}\right]$ $R_{2}-2R_{1}\rightarrow R_{2}$ : $\left[{\begin{array}{ccc|ccc}1&0&1&0&1&0\\0&1&1&1&-2&0\\0&1&2&0&0&1\end{array}}\right]$ $R_{3}-R_{2}\rightarrow R_{3}$ : $\left[{\begin{array}{ccc|ccc}1&0&1&0&1&0\\0&1&1&1&-2&0\\0&0&1&-1&2&1\end{array}}\right]$ $R_{1}-R_{3}\rightarrow R_{1}$ : $\left[{\begin{array}{ccc|ccc}1&0&0&1&-1&-1\\0&1&1&1&-2&0\\0&0&1&-1&2&1\end{array}}\right]$ $R_{2}-R_{3}\rightarrow R_{2}$ : $\left[{\begin{array}{ccc|ccc}1&0&0&1&-1&-1\\0&1&0&2&-4&-1\\0&0&1&-1&2&1\end{array}}\right]$ Therefore, $A^{-1}={\begin{bmatrix}1&-1&-1\\2&-4&-1\\-1&2&1\end{bmatrix}}.$