write B as a linear combination of the other matrices, if possible. B=[23−42],A1=[1001],A2=[0−110],A3=[1101]

Given The given matrix is B=[23−42],A1=[1001],A2=[0−110],A3=[1101] Calculation: Let us represent B as a linear combination of A1,A2,A3 B=c1A1+c2A2+c3A3 [23−42]=c1[1001]+c2[0−110]+c3[1101] =[c100c1]+[0−c2c20]+[c3c30c3] =[c1+0+c30−c2+c30+c2+0c1+0+c3] Step 3 Now , by comparing both sides. c1+0+c3=2…(1) 0−c2+c3=3…(2) 0+c2+0=−4…(3) c1+0+c3=2…(4) from equation (3) and (2) c2=−4 and c3=−1 substitute c3=−1 in (1) we get c1=3 Thus ,we can say B=3A1−4A2−A3

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waigaK

Answered question

2020-11-07

write B as a linear combination of the other matrices, if possible.

B = [\begin{matrix} 2 & 3 \\ - 4 & 2 \end{matrix}], A_{1} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}], A_{2} = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}], A_{3} = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]

Answer & Explanation

Arnold Odonnell

Skilled2020-11-08Added 109 answers

Given
The given matrix is

B = [\begin{matrix} 2 & 3 \\ - 4 & 2 \end{matrix}], A_{1} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}], A_{2} = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}], A_{3} = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]

Calculation:
Let us represent B as a linear combination of

A_{1}, A_{2}, A_{3}

B = c_{1} A_{1} + c_{2} A_{2} + c_{3} A_{3}

[\begin{matrix} 2 & 3 \\ - 4 & 2 \end{matrix}] = c_{1} [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + c_{2} [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] + c_{3} [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]

= [\begin{matrix} c_{1} & 0 \\ 0 & c_{1} \end{matrix}] + [\begin{matrix} 0 & - c_{2} \\ c_{2} & 0 \end{matrix}] + [\begin{matrix} c_{3} & c_{3} \\ 0 & c_{3} \end{matrix}]

= [\begin{matrix} c_{1} + 0 + c_{3} & 0 - c_{2} + c_{3} \\ 0 + c_{2} + 0 & c_{1} + 0 + c_{3} \end{matrix}]

Step 3
Now , by comparing both sides.

c_{1} + 0 + c_{3} = 2 \dots (1)

0 - c_{2} + c_{3} = 3 \dots (2)

0 + c_{2} + 0 = - 4 \dots (3)

c_{1} + 0 + c_{3} = 2 \dots (4)

from equation (3) and (2)

c_{2} = - 4 and c_{3} = - 1

substitute

c_{3} = - 1 in (1) we get c_{1} = 3

Thus ,we can say

B = 3 A_{1} - 4 A_{2} - A_{3}

Jeffrey Jordon

Expert2022-01-27Added 2605 answers

Answer is given below (on video)

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