"If A=21 and B=2−3 8 6-1-2 Find: a)..." solved and explained

High School
Algebra
Algebra I
Quadratic function and equation

ukelangf0

Answered question

2021-12-04

A = 21

and

B = 2 - 3

8 6-1-2
Find:
a) BA
(b) determinant B
(c) Adjoint B
(d) A-1

Answer & Explanation

Wasither1957

Beginner2021-12-05Added 17 answers

Step 1
Given:
$A = [\begin{matrix} 2 & 1 \\ 8 & 6 \end{matrix}] a n d B = [\begin{matrix} 2 & - 3 \\ - 1 & - 2 \end{matrix}]$
a) $B A = [\begin{matrix} 2 & - 3 \\ - 1 & - 2 \end{matrix}] \cdot [\begin{matrix} 2 & 1 \\ 8 & 6 \end{matrix}]$
$B A = [\begin{matrix} 4 + (- 24) & 2 + (- 18) \\ (- 2) + (- 16) & (- 1) + (- 12) \end{matrix}]$
$B A = [\begin{matrix} - 20 & - 16 \\ - 18 & - 13 \end{matrix}]$
b) Determinant of B is:
$d e t (B) = [\begin{matrix} 2 & - 3 \\ - 1 & - 2 \end{matrix}]$
$det (B) = [(2) \times (- 2)] - [(- 3) \times (- 1)]$
$det (B) = - 4 - 3$
$det (B) = - 7$
c) Now, the Adjoint B.
Since, $B = [\begin{matrix} 2 & - 3 \\ - 1 & - 2 \end{matrix}]$
Therefore, Adjoint $B = [\begin{matrix} - 2 & 3 \\ 1 & 2 \end{matrix}]$
d) The inverse of A i.e., $A^{- 1}$ .
$A^{- 1} = \frac{A d j \oint A}{[A]}$
So,
$d e t (A) = [\begin{matrix} 2 & 1 \\ 8 & 6 \end{matrix}]$
$det (A) = 12 - 8$
$det (A) = 4$
or, $[A] = 4$
Also,
Adjoint of A is:
Adjoint $A = [\begin{matrix} 6 & - 1 \\ - 8 & 2 \end{matrix}]$
Therefore,
$A^{- 1} = \frac{1}{4} [\begin{matrix} 6 & - 1 \\ - 8 & 2 \end{matrix}]$
$A^{- 1} = [\begin{matrix} \frac{6}{4} & \frac{- 1}{4} \\ \frac{- 8}{4} & \frac{2}{4} \end{matrix}]$
$A^{- 1} = [\begin{matrix} \frac{3}{2} & \frac{- 1}{4} \\ - 2 & \frac{1}{2} \end{matrix}]$

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