Question

select the correct answer.
what is the inverse of this matrix?
\

$$\begin{bmatrix}1 & 5 & 2\\\\1 & 1 & 7\\\\0 & -3 & 4\\end{bmatrix}$$

\bigcirc a. \

$$\begin{bmatrix}-2 & 6 & 3\\\\4 & -4 & 5\\\\3 & -3 & 4\\end{bmatrix}$$

\bigcirc b. this matrix doesnt have an inverse.
\bigcirc c. \

$$\begin{bmatrix}-25 & 26 & -33\\\\4 & -4 & 5\\\\3 & -3 & 4\\end{bmatrix}$$

\bigcirc d. \

$$\begin{bmatrix}3 & -3 & 4\\\\4 & 4 & -5\\\\-1 & 3 & 2\\end{bmatrix}$$

Explanation:

Step1: Calculate matrix determinant

For matrix $A=

$$\begin{bmatrix}1&5&2\\1&1&7\\0&-3&4\end{bmatrix}$$

$, expand along row 3:

$$\det(A)=0\cdot
LATEXBLOCK2
- (-3)\cdot
LATEXBLOCK3
+ 4\cdot
LATEXBLOCK4
$$

$$=3\cdot(1\cdot7 - 2\cdot1) + 4\cdot(1\cdot1 - 5\cdot1)$$
$$=3\cdot(7-2)+4\cdot(1-5)$$
$$=3\cdot5 + 4\cdot(-4)=15-16=-1$$

Step2: Find adjugate matrix

First compute cofactor matrix:

• $C_{11}=(-1)^{1+1}

$$\begin{vmatrix}1&7\\-3&4\end{vmatrix}$$

=4+21=25$

• $C_{12}=(-1)^{1+2}

$$\begin{vmatrix}1&7\\0&4\end{vmatrix}$$

=-(4-0)=-4$

• $C_{13}=(-1)^{1+3}

$$\begin{vmatrix}1&1\\0&-3\end{vmatrix}$$

=-3-0=-3$

• $C_{21}=(-1)^{2+1}

$$\begin{vmatrix}5&2\\-3&4\end{vmatrix}$$

=-(20+6)=-26$

• $C_{22}=(-1)^{2+2}

$$\begin{vmatrix}1&2\\0&4\end{vmatrix}$$

=4-0=4$

• $C_{23}=(-1)^{2+3}

$$\begin{vmatrix}1&5\\0&-3\end{vmatrix}$$

=-(-3-0)=3$

• $C_{31}=(-1)^{3+1}

$$\begin{vmatrix}5&2\\1&7\end{vmatrix}$$

=35-2=33$

• $C_{32}=(-1)^{3+2}

$$\begin{vmatrix}1&2\\1&7\end{vmatrix}$$

=-(7-2)=-5$

• $C_{33}=(-1)^{3+3}

$$\begin{vmatrix}1&5\\1&1\end{vmatrix}$$

=1-5=-4$

Cofactor matrix:

$$\begin{bmatrix}25&-4&-3\\-26&4&3\\33&-5&-4\end{bmatrix}$$

Adjugate is transpose of cofactor matrix:

$$\text{adj}(A)=
LATEXBLOCK14
$$

Step3: Compute inverse matrix

$A^{-1}=\frac{1}{\det(A)}\cdot\text{adj}(A)=\frac{1}{-1}

$$\begin{bmatrix}25&-26&33\\-4&4&-5\\-3&3&-4\end{bmatrix}$$

$

$$=
LATEXBLOCK16
$$

Answer:

C.

$$\begin{bmatrix}-25&26&-33\\4&-4&5\\3&-3&4\end{bmatrix}$$