If [5132] and B=[3−61−3] Find: a) A2 b) determinant A c) At d) B−1

Step 1 Given that A[5132] B=[3−61−3] a) A2=[5132] [5132] =[5132] [5132] =[25−35−215−63−4] =[2239−1] b) |A|=[5132] =5[2]−1[3] =10−3 =7 c) ∵(A)T=[abba] AT=[abba] At=[5132] =[5312] d) B−1=Ady(B)|A| |B|=3[−3]+6[1] =−9+6=−3 Step 2 (Ady B)=b11=−3 b12=1 b21=−6 b22=3 Ady(B)=[−31−63] B−1=Ady(B)|B| =[−31−63] Ady(B)=[−3−613] A−1=Ady(B)|B| =[−3−613]−3 =[−3−3−6−3−13−33] =[−12−131]

"If [5132] and B=[3−61−3] Find: a) A2 b)..." was answered by our community

High School
Algebra
Algebra I
Quadratic function and equation

urca3m403

Answered question

2021-12-03

[\begin{matrix} 5 & 1 \\ 3 & 2 \end{matrix}] a n d B = [\begin{matrix} 3 & - 6 \\ 1 & - 3 \end{matrix}]

Find:
a)

A^{2}

b) determinant A
c)

A^{t}

B^{- 1}

Answer & Explanation

Margaret Plemons

Beginner2021-12-04Added 13 answers

Step 1
Given that
$A [\begin{matrix} 5 & 1 \\ 3 & 2 \end{matrix}]$
$B = [\begin{matrix} 3 & - 6 \\ 1 & - 3 \end{matrix}]$
a) $A^{2} = [\begin{matrix} 5 & 1 \\ 3 & 2 \end{matrix}] [\begin{matrix} 5 & 1 \\ 3 & 2 \end{matrix}]$
$= [\begin{matrix} 5 & 1 \\ 3 & 2 \end{matrix}] [\begin{matrix} 5 & 1 \\ 3 & 2 \end{matrix}]$
$= [\begin{matrix} 25 - 3 & 5 - 2 \\ 15 - 6 & 3 - 4 \end{matrix}]$
$= [\begin{matrix} 22 & 3 \\ 9 & - 1 \end{matrix}]$
b) $| A | = [\begin{matrix} 5 & 1 \\ 3 & 2 \end{matrix}]$
$= 5 [2] - 1 [3]$
$= 10 - 3$
$= 7$
c) $∵ (A)^{T} = [\begin{matrix} a & b \\ b & a \end{matrix}]$
$A^{T} = [\begin{matrix} a & b \\ b & a \end{matrix}]$
$A^{t} = [\begin{matrix} 5 & 1 \\ 3 & 2 \end{matrix}]$
$= [\begin{matrix} 5 & 3 \\ 1 & 2 \end{matrix}]$
d) $B^{- 1}$ = $\frac{A d y (B)}{| A |}$
$| B | = 3 [- 3] + 6 [1]$
$= - 9 + 6 = - 3$
Step 2
$(A d y B) = b_{11} = - 3$
$b_{12} = 1$
$b_{21} = - 6$
$b_{22} = 3$
$A d y (B) = [\begin{matrix} - 3 & 1 \\ - 6 & 3 \end{matrix}]$
$B^{- 1} = \frac{A d y (B)}{| B |}$
$= [\begin{matrix} - 3 & 1 \\ - 6 & 3 \end{matrix}]$
$A d y (B) = [\begin{matrix} - 3 & - 6 \\ 1 & 3 \end{matrix}]$
$A^{- 1} = \frac{A d y (B)}{| B |}$
$= \frac{[\begin{matrix} - 3 & - 6 \\ 1 & 3 \end{matrix}]}{- 3}$
$= [\begin{matrix} \frac{- 3}{- 3} & \frac{- 6}{- 3} \\ \frac{- 1}{3} & \frac{- 3}{3} \end{matrix}]$
$= [\begin{matrix} - 1 & 2 \\ - \frac{1}{3} & 1 \end{matrix}]$

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