Solveeit Logo
Login

Question

Question: Inverse of the matrix \(\begin{bmatrix} 3 & –2 & –1 \\ –4 & 1 & –1 \\ 2 & 0 & 1 \end{bmatrix}\) is...

Inverse of the matrix [3–2–1–41–1201]\begin{bmatrix} 3 & –2 & –1 \\ –4 & 1 & –1 \\ 2 & 0 & 1 \end{bmatrix} is

A

[123337–2–4–5]\begin{bmatrix} 1 & 2 & 3 \\ 3 & 3 & 7 \\ –2 & –4 & –5 \end{bmatrix}

B

[123257–2–4–5]\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ –2 & –4 & –5 \end{bmatrix}

C

[1–35746427]\begin{bmatrix} 1 & –3 & 5 \\ 7 & 4 & 6 \\ 4 & 2 & 7 \end{bmatrix}

D

[12–48–4–5352]\begin{bmatrix} 1 & 2 & –4 \\ 8 & –4 & –5 \\ 3 & 5 & 2 \end{bmatrix}

Answer

[123257–2–4–5]\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ –2 & –4 & –5 \end{bmatrix}

Explanation

Solution

|A| = ∣3–2–1–41–1201∣\left| \begin{matrix} 3 & –2 & –1 \\ –4 & 1 & –1 \\ 2 & 0 & 1 \end{matrix} \right|

= 3(1) + 2 (– 4 + 2) – 1 (0 – 2)

= 3 – 4 + 2 = 1

adj A = [12–225–437–5]T\begin{bmatrix} 1 & 2 & –2 \\ 2 & 5 & –4 \\ 3 & 7 & –5 \end{bmatrix}^{T} = [123257–2–4–5]\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ –2 & –4 & –5 \end{bmatrix}

A–1 =AdjA∣A∣\frac{AdjA}{|A|} = [123257–2–4–5]\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ –2 & –4 & –5 \end{bmatrix}