Write the matrix A = [ 3 , − 1 , 1 , − 2 ] A = [3, -1, 1, -2] A = [ 3 , − 1 , 1 , − 2 ] as a linear combination of A 1 = [ 1 , 1 , 0 , − 1 ] A1 = [1, 1, 0, -1] A 1 = [ 1 , 1 , 0 , − 1 ] , A 2 = [ 1 , 1 , − 1 , 0 ] A2 = [1, 1, -1, 0] A 2 = [ 1 , 1 , − 1 , 0 ] and A 3 = [ 1 , − 1 , 0 , 0 ] A3 = [1, -1, 0, 0] A 3 = [ 1 , − 1 , 0 , 0 ]
**Solution:**
We need to present the matrix A A A in the form:
A = x ∗ A 1 + y ∗ A 2 + z ∗ A 3 A = x * A_1 + y * A_2 + z * A_3 A = x ∗ A 1 + y ∗ A 2 + z ∗ A 3
where x , y , z x, y, z x , y , z some constants.
x ∗ A 1 + y ∗ A 2 + z ∗ A 3 = x ∗ [ 1 , 1 , 0 , − 1 ] + y ∗ [ 1 , 1 , − 1 , 0 ] + z ∗ [ 1 , − 1 , 0 , 0 ] = [ x , x , 0 , − x ] + [ y , y , − y , 0 ] + [ z , − z , 0 , 0 ] = [ x + y + z , x + y − z , 0 − y + 0 , − x + 0 + 0 ] = [ x + y + z , x + y − z , − y , − x ] \begin{array}{l}
x * A_1 + y * A_2 + z * A_3 = x * [1, 1, 0, -1] + y * [1, 1, -1, 0] + z * [1, -1, 0, 0] \\
= [x, x, 0, -x] + [y, y, -y, 0] + [z, -z, 0, 0] \\
= [x + y + z, x + y - z, 0 - y + 0, -x + 0 + 0] \\
= [x + y + z, x + y - z, -y, -x]
\end{array} x ∗ A 1 + y ∗ A 2 + z ∗ A 3 = x ∗ [ 1 , 1 , 0 , − 1 ] + y ∗ [ 1 , 1 , − 1 , 0 ] + z ∗ [ 1 , − 1 , 0 , 0 ] = [ x , x , 0 , − x ] + [ y , y , − y , 0 ] + [ z , − z , 0 , 0 ] = [ x + y + z , x + y − z , 0 − y + 0 , − x + 0 + 0 ] = [ x + y + z , x + y − z , − y , − x ]
So we have that
[ x + y + z , x + y − z , − y , − x ] = A [x + y + z, x + y - z, -y, -x] = A [ x + y + z , x + y − z , − y , − x ] = A [ x + y + z , x + y − z , − y , − x ] = [ 3 , − 1 , 1 , − 2 ] [x + y + z, x + y - z, -y, -x] = [3, -1, 1, -2] [ x + y + z , x + y − z , − y , − x ] = [ 3 , − 1 , 1 , − 2 ]
We have next system of linear equation:
{ x + y + z = 3 x + y − z = − 1 − y = 1 − x = − 2 \left\{
\begin{array}{l}
x + y + z = 3 \\
x + y - z = -1 \\
-y = 1 \\
-x = -2
\end{array}
\right. ⎩ ⎨ ⎧ x + y + z = 3 x + y − z = − 1 − y = 1 − x = − 2 { 2 z = 4 y = − 1 x = 2 \left\{
\begin{array}{l}
2z = 4 \\
y = -1 \\
x = 2
\end{array}
\right. ⎩ ⎨ ⎧ 2 z = 4 y = − 1 x = 2 { x = 2 y = − 1 z = 2 \left\{
\begin{array}{l}
x = 2 \\
y = -1 \\
z = 2
\end{array}
\right. ⎩ ⎨ ⎧ x = 2 y = − 1 z = 2
So matrix A A A can be present as
A = 2 A 1 − A 2 + 2 A 3 A = 2A_1 - A_2 + 2A_3 A = 2 A 1 − A 2 + 2 A 3
**Answer:** A = 2 A 1 − A 2 + 2 A 3 A = 2A_1 - A_2 + 2A_3 A = 2 A 1 − A 2 + 2 A 3