Question

1. consider the system of equations

2x + y - z = 5
3x + 2y + z = 8
x - 3y + 2z = -8
the system can be written in matrix form, ax = b, where a is the coefficient matrix, x is the variable matrix, and b is the constant matrix.
which of the following are the entries in the third row of the inverse of matrix a?
the first entry in the third row of the inverse matrix is
the second entry in the third row of the inverse matrix is
the third entry in the third row of the inverse matrix is
options: -1/10, -5/10, -13/10, -6/10, -1/10, -7/10, -1/10

Explanation:

Step1: Write the system in standard form

First, rewrite the system of equations in standard \(Ax = b\) form. The given equations are:
1. \(2x + y - z = 5\)
2. \(3x + 2y + z = 8\)
3. \(x - 3y + 2z = - 8\) (rewriting \(x=3y - 2z - 8\) as \(x - 3y + 2z=-8\))

So the coefficient matrix \(A=\begin{bmatrix}2&1&- 1\\3&2&1\\1&-3&2\end{bmatrix}\)

Step2: Find the determinant of \(A\)

The determinant of a \(3\times3\) matrix \(\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\) is given by:
\(\det(A)=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})\)

For \(A=\begin{bmatrix}2&1&-1\\3&2&1\\1&-3&2\end{bmatrix}\):
\[ \begin{align*} \det(A)&=2(2\times2 - 1\times(-3))-1(3\times2 - 1\times1)+(-1)(3\times(-3)-2\times1)\\ &=2(4 + 3)-1(6 - 1)-1(-9 - 2)\\ &=2\times7-1\times5 - 1\times(-11)\\ &=14-5 + 11\\ &=20 \end{align*} \]

Step3: Find the matrix of minors

The minor of an element \(a_{ij}\) is the determinant of the sub - matrix obtained by removing the \(i\) - th row and \(j\) - th column.

- Minor of \(a_{11}\) (element 2, row 1, column 1): Remove row 1 and column 1, sub - matrix \(\begin{bmatrix}2&1\\-3&2\end{bmatrix}\), \(\det\begin{bmatrix}2&1\\-3&2\end{bmatrix}=2\times2-1\times(-3) = 4 + 3=7\)
- Minor of \(a_{12}\) (element 1, row 1, column 2): Remove row 1 and column 2, sub - matrix \(\begin{bmatrix}3&1\\1&2\end{bmatrix}\), \(\det\begin{bmatrix}3&1\\1&2\end{bmatrix}=3\times2 - 1\times1=6 - 1 = 5\)
- Minor of \(a_{13}\) (element - 1, row 1, column 3): Remove row 1 and column 3, sub - matrix \(\begin{bmatrix}3&2\\1&-3\end{bmatrix}\), \(\det\begin{bmatrix}3&2\\1&-3\end{bmatrix}=3\times(-3)-2\times1=-9 - 2=-11\)
- Minor of \(a_{21}\) (element 3, row 2, column 1): Remove row 2 and column 1, sub - matrix \(\begin{bmatrix}1&-1\\-3&2\end{bmatrix}\), \(\det\begin{bmatrix}1&-1\\-3&2\end{bmatrix}=1\times2-(-1)\times(-3)=2 - 3=-1\)
- Minor of \(a_{22}\) (element 2, row 2, column 2): Remove row 2 and column 2, sub - matrix \(\begin{bmatrix}2&-1\\1&2\end{bmatrix}\), \(\det\begin{bmatrix}2&-1\\1&2\end{bmatrix}=2\times2-(-1)\times1 = 4 + 1=5\)
- Minor of \(a_{23}\) (element 1, row 2, column 3): Remove row 2 and column 3, sub - matrix \(\begin{bmatrix}2&1\\1&-3\end{bmatrix}\), \(\det\begin{bmatrix}2&1\\1&-3\end{bmatrix}=2\times(-3)-1\times1=-6 - 1=-7\)
- Minor of \(a_{31}\) (element 1, row 3, column 1): Remove row 3 and column 1, sub - matrix \(\begin{bmatrix}1&-1\\2&1\end{bmatrix}\), \(\det\begin{bmatrix}1&-1\\2&1\end{bmatrix}=1\times1-(-1)\times2=1 + 2=3\)
- Minor of \(a_{32}\) (element - 3, row 3, column 2): Remove row 3 and column 2, sub - matrix \(\begin{bmatrix}2&-1\\3&1\end{bmatrix}\), \(\det\begin{bmatrix}2&-1\\3&1\end{bmatrix}=2\times1-(-1)\times3=2 + 3=5\)
- Minor of \(a_{33}\) (element 2, row 3, column 3): Remove row 3 and column 3, sub - matrix \(\begin{bmatrix}2&1\\3&2\end{bmatrix}\), \(\det\begin{bmatrix}2&1\\3&2\end{bmatrix}=2\times2-1\times3=4 - 3=1\)

The matrix of minors is \(\begin{bmatrix}7&5&-11\\-1&5&-7\\3&5&1\end{bmatrix}\)

Step4: Find the matrix of cofactors

The cofactor matrix \(C\) is obtained by multiplying the minors by \((-1)^{i + j}\), where \(i\) is the row number and \(j\) is the column number.

- \(C_{11}=(-1)^{1 + 1}\times7 = 7\), \(C_{12}=(-1)^{1+2}\times5=-5\), \(C_{13}=(-1)^{1 + 3}\times(-11)=-11\)
- \(C_{21}=(-1)^{2 + 1}\times(-1)=1\), \(C_{22}=(-1)^{2+2}\times5 = 5\), \(C_{23}=(-1)^{2 + 3}\times(-7)=7\)
- \(C_{31}=(-1)^{3 + 1}\times3=3\), \(C_{32}=(-1)^{3+2}\times5=-5\), \(C_{33}=(-1)^{3 + 3}\times1 = 1\)

The cofactor matrix is \(\begin{bmatrix}7&-5&-11\\1&5&7\\3&-5&1\end{bmatrix}\)

Step5: Find the adjugate matrix (transpose of cofactor matrix)

The adjugate matrix \(adj(A)\) is the transpose of the cofactor matrix. So we swap rows and columns:

\(adj(A)=\begin{bmatrix}7&1&3\\-5&5&-5\\-11&7&1\end{bmatrix}\)

Step6: Find the inverse matrix

The inverse of a matrix \(A\) is given by \(A^{-1}=\frac{1}{\det(A)}adj(A)\). Since \(\det(A) = 20\), we have:

\(A^{-1}=\frac{1}{20}\begin{bmatrix}7&1&3\\-5&5&-5\\-11&7&1\end{bmatrix}=\begin{bmatrix}\frac{7}{20}&\frac{1}{20}&\frac{3}{20}\\-\frac{5}{20}&\frac{5}{20}&-\frac{5}{20}\\-\frac{11}{20}&\frac{7}{20}&\frac{1}{20}\end{bmatrix}=\begin{bmatrix}\frac{7}{20}&\frac{1}{20}&\frac{3}{20}\\-\frac{1}{4}&\frac{1}{4}&-\frac{1}{4}\\-\frac{11}{20}&\frac{7}{20}&\frac{1}{20}\end{bmatrix}\)

Now, we look at the third row of \(A^{-1}\):

- The first entry in the third row: \(-\frac{11}{20}\)
- The second entry in the third row: \(\frac{7}{20}\)
- The third entry in the third row: \(\frac{1}{20}\)

Answer:

The first entry in the third row of the inverse matrix is \(-\frac{11}{20}\)
The second entry in the third row of the inverse matrix is \(\frac{7}{20}\)
The third entry in the third row of the inverse matrix is \(\frac{1}{20}\)