Question

assignment 7: problem 11
(1 point)
let ( a=left\begin{array}{cc}1 & 5 \\ 2 & -3end{array}
ight ).
(a) find the determinant of ( a ).
( operatorname{det}(a)= )
(b) find the matrix of cofactors of ( a ).
( c=left\begin{array}{ll}square & square \\ square & squareend{array}
ight )
(c) find the adjoint of ( a ).
( operatorname{adj}(a)=left\begin{array}{ll}square & square \\ square & squareend{array}
ight )
(d) find the inverse of ( a ).
( a^{-1}=left\begin{array}{ll}square & square \\ square & squareend{array}
ight )

Explanation:

Step1: Calculate the determinant of \(A\)

For a \(2\times2\) matrix \(A=

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

\), the determinant is \(det(A)=ad - bc\).
Here \(a = 1\), \(b = 5\), \(c = 2\), \(d=-3\).

$$
LATEXBLOCK1
$$

Step2: Find the matrix of co - factors \(C\)

For a \(2\times2\) matrix \(A=

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

\), the co - factor matrix \(C=

$$\begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}$$

\), where \(C_{ij}=(-1)^{i + j}M_{ij}\) ( \(M_{ij}\) is the minor).
\(M_{11}=d=-3\), \(C_{11}=(-1)^{1 + 1}M_{11}=-3\)
\(M_{12}=c = 2\), \(C_{12}=(-1)^{1+2}M_{12}=-2\)
\(M_{21}=b = 5\), \(C_{21}=(-1)^{2 + 1}M_{21}=-5\)
\(M_{22}=a = 1\), \(C_{22}=(-1)^{2+2}M_{22}=1\)
So \(C=

$$\begin{bmatrix}-3&-2\\-5&1\end{bmatrix}$$

\)

Step3: Find the adjoint of \(A\)

The adjoint of \(A\), \(adj(A)\) is the transpose of the co - factor matrix \(C\).
If \(C=

$$\begin{bmatrix}-3&-2\\-5&1\end{bmatrix}$$

\), then \(adj(A)=C^{T}=

$$\begin{bmatrix}-3&-5\\-2&1\end{bmatrix}$$

\)

Step4: Find the inverse of \(A\)

The inverse of \(A\) is \(A^{-1}=\frac{1}{det(A)}adj(A)\)
Since \(det(A)=-13\) and \(adj(A)=

$$\begin{bmatrix}-3&-5\\-2&1\end{bmatrix}$$

\)

$$ A^{-1}=\frac{1}{-13}
LATEXBLOCK8
=
LATEXBLOCK9
$$

Answer:

(a) \(-13\)
(b) \(

$$\begin{bmatrix}-3&-2\\-5&1\end{bmatrix}$$

\)
(c) \(

$$\begin{bmatrix}-3&-5\\-2&1\end{bmatrix}$$

\)
(d) \(

$$\begin{bmatrix}\frac{3}{13}&\frac{5}{13}\\\frac{2}{13}&-\frac{1}{13}\end{bmatrix}$$

\)