Wierzycaz

2020-10-27

If possible, find scalars ${c}_{1},{c}_{2},\text{and}{c}_{3}$ so that

${c}_{1}\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right]+{c}_{2}\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]+{c}_{3}\left[\begin{array}{c}-1\\ 4\\ -1\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]$

Delorenzoz

Skilled2020-10-28Added 91 answers

Step 1

Consider the equation:

${c}_{1}\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right]+{c}_{2}\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]+{c}_{3}\left[\begin{array}{c}-1\\ 4\\ -1\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]\dots (1)$

Step 2

Now, when a matrix is multiplied by a scalar quantity, then all elements of matrix is multiplied by that scalar quantity.

Using this principle, rewrite equation (1) as:

$\left[\begin{array}{c}{c}_{1}\\ 2{c}_{1}\\ -3{c}_{1}\end{array}\right]+\left[\begin{array}{c}-{c}_{2}\\ {c}_{2}\\ {c}_{2}\end{array}\right]+\left[\begin{array}{c}-{c}_{3}\\ 4{c}_{3}\\ -{c}_{3}\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]\dots (2)$

step 3

Now again,

Addition of matrices can be done by adding same position of elements together, thus creating a new single matrix.

From (2):

$\left[\begin{array}{c}{c}_{1}-{c}_{2}-{c}_{3}\\ 2{c}_{1}+{c}_{2}+4{c}_{3}\\ -3{c}_{1}+{c}_{2}-{c}_{3}\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]$

Step 4

Equating rows from both sides of matrices:

${c}_{1}-{c}_{2}-{c}_{3}=2$

${c}_{1}=2+{c}_{2}+{c}_{3}\dots (3)$

and

$2{c}_{1}+{c}_{2}+4{c}_{3}=-2$

$2(2+{c}_{2}+{c}_{3})+{c}_{2}+4{c}_{3}=-2[\because \text{From}(3)]$

$3{c}_{2}+6{c}_{3}=-6$

${c}_{2}+2{c}_{3}=-3\dots (4)$

Step 5

and

$-3{c}_{1}+{c}_{2}-{c}_{3}=3$

$-3(2+{c}_{2}+{c}_{3})+{c}_{2}-{c}_{3}=3[\because \text{From}(3)]$

$-2{c}_{2}-4{c}_{3}=9$

${c}_{2}+2{c}_{3}=-4.5\dots (5)$

From (4) and (5): Since L.H.S. of both equations are same but R.H.S. are not same, It is impossible to find the scalar quantity${c}_{1},{c}_{2}\text{and}{c}_{3}$ .

Consider the equation:

Step 2

Now, when a matrix is multiplied by a scalar quantity, then all elements of matrix is multiplied by that scalar quantity.

Using this principle, rewrite equation (1) as:

step 3

Now again,

Addition of matrices can be done by adding same position of elements together, thus creating a new single matrix.

From (2):

Step 4

Equating rows from both sides of matrices:

and

Step 5

and

From (4) and (5): Since L.H.S. of both equations are same but R.H.S. are not same, It is impossible to find the scalar quantity

Jeffrey Jordon

Expert2022-01-29Added 2605 answers

Answer is given below (on video)

Jeffrey Jordon

Expert2022-01-29Added 2605 answers

Answer is given below (on video)

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