# write B as a linear combination of the other matrices, if possible. B=[[2,-2,3],[0,0,-2],[0,0,2]] A_1=[[1,0,0],[0,1,0],[0,0,1]] A_2=[[0,1,1],[0,0,1],[0,0,0]] A_3=[[-1,0,-1],[0,1,0],[0,0,-1]] A_4=[[1,-1,1],[0,-1,-1],[0,0,1]]

Carol Gates 2021-01-13 Answered
write B as a linear combination of the other matrices, if possible.
$B=\left[\left[2,-2,3\right],\left[0,0,-2\right],\left[0,0,2\right]\right]$
${A}_{1}=\left[\left[1,0,0\right],\left[0,1,0\right],\left[0,0,1\right]\right]$
${A}_{2}=\left[\left[0,1,1\right],\left[0,0,1\right],\left[0,0,0\right]\right]$
${A}_{3}=\left[\left[-1,0,-1\right],\left[0,1,0\right],\left[0,0,-1\right]\right]$
${A}_{4}=\left[\left[1,-1,1\right],\left[0,-1,-1\right],\left[0,0,1\right]\right]$
You can still ask an expert for help

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it

## Expert Answer

hosentak
Answered 2021-01-14 Author has 100 answers
Matrix solution:

We have step-by-step solutions for your answer!

Jeffrey Jordon
Answered 2021-09-30 Author has 2495 answers

Consider the linear combination of matrices as:

$B=\alpha {A}_{1}+B{A}_{2}+\gamma {A}_{3}+\delta {A}_{4}$, where $\alpha ,\beta ,\gamma ,\delta$ are constants.

Substitute the matrices of $B,{A}_{1},{A}_{2},{A}_{3},{A}_{4}$ to find constants:

$⇒\left[\begin{array}{ccc}2& -2& 3\\ 0& 0& -2\\ 0& 0& 2\end{array}\right]=\alpha \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]+\beta \left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]+\gamma \left[\begin{array}{ccc}-1& 0& -0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right]+\delta \left[\begin{array}{ccc}1& -1& 1\\ 0& -1& -1\\ 0& 0& 1\end{array}\right]$

$⇒\left[\begin{array}{ccc}2& -2& 3\\ 0& 0& -2\\ 0& 0& 2\end{array}\right]=\left[\begin{array}{ccc}\alpha -\gamma +\delta & \beta -\delta & \beta -\gamma +\delta \\ 0& \alpha +\gamma -\delta & \beta -\delta \\ 0& 0& \alpha -\gamma +\delta \end{array}\right]$

By equation component equations:

$\alpha -\gamma +\delta =2$ (1)

$\beta -\delta =-2$ (2)

$\beta -\gamma +\delta =3$ (3)
$\alpha +\gamma -\delta =0$ (4)

Adding (1) and (4) $⇒2\alpha =2⇒\alpha =1$

From equation (2) $\beta =\delta -2$

Therefore, equation (3) becomes

$\delta -2-\gamma +\delta =3$

$⇒2\delta =5+\gamma$

$⇒\delta =\frac{5+\gamma }{2}$

Substitute above value in equation (4), if becomes,

$1+\gamma -\frac{5+\gamma }{2}=0$

$⇒2+2\gamma -5-\gamma =0$

$⇒\gamma =3$

Therefore, $\delta =\frac{5+\gamma }{2}=\frac{5+3}{2}=4$

Frome equation (2),

$\beta =4-2=2$

Substitute the values of constants

Therefore, the linear combination of matrix B is given by

$B=1{A}_{1}+2{A}_{2}+3{A}_{3}+4{A}_{4}$

We have step-by-step solutions for your answer!

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it