KCET · Maths · Matrices
The inverse of the matrix \( \left[\begin{array}{rrr}2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3\end{array}\right] \) is
- A \( \left[\begin{array}{ccc}3 & -5 & 5 \\ -1 & -6 & -2 \\ 1 & -5 & 2\end{array}\right] \)
- B \( \left[\begin{array}{ccc}3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & -2\end{array}\right] \)
- C \( \left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right] \)
- D \( \left[\begin{array}{ccc}3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2\end{array}\right] \)
Answer & Solution
Correct Answer
(D) \( \left[\begin{array}{ccc}3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2\end{array}\right] \)
Step-by-step Solution
Detailed explanation
(D)
\[
\begin{array}{l}
\text { Let } A=\left[\begin{array}{ccc}
2 & 5 & 0 \\
0 & 1 & 1 \\
-1 & 0 & 3
\end{array}\right] \\
|A|=2(3-0)-5(0+1) \\
=6-5=1 \\
\therefore|A|=1
\end{array}
\]
\[
\begin{array}{l}
\operatorname{adj} A=\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]^{T}=\left[\begin{array}{ccc}
3 & -15 & 5 \\
-1 & 6 & -2 \\
1 & -5 & 2
\end{array}\right] \\
A^{-1}=\frac{a d j A}{|A|}=\left[\begin{array}{ccc}
3 & -15 & 5 \\
-1 & 6 & -2 \\
1 & -5 & 2
\end{array}\right]
\end{array}
\]
\[
\begin{array}{l}
\text { Let } A=\left[\begin{array}{ccc}
2 & 5 & 0 \\
0 & 1 & 1 \\
-1 & 0 & 3
\end{array}\right] \\
|A|=2(3-0)-5(0+1) \\
=6-5=1 \\
\therefore|A|=1
\end{array}
\]
\[
\begin{array}{l}
\operatorname{adj} A=\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]^{T}=\left[\begin{array}{ccc}
3 & -15 & 5 \\
-1 & 6 & -2 \\
1 & -5 & 2
\end{array}\right] \\
A^{-1}=\frac{a d j A}{|A|}=\left[\begin{array}{ccc}
3 & -15 & 5 \\
-1 & 6 & -2 \\
1 & -5 & 2
\end{array}\right]
\end{array}
\]
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