(d) Check the linear dependency of the matrices B,C,D in the...

(d) Check the linear dependency of the matrices B,C,D in the next problem. (e) Let $v_1 = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$, $v_2 = \begin{bmatrix} 2 & 5 & 6 \end{bmatrix}$, $u = \begin{bmatrix} 2 & 1 & 0 \end{bmatrix}$ and $w = \begin{bmatrix} 1 & 1 & 1 \end{bmatrix}$. Is $u \in \text{span}(v_1, v_2)$? Is $w \in \text{span}(v_1, v_2)$? If yes, write u or w as a linear combination of $v_1$ and $v_2$.

Transcript

00:01 In this problem we are given the set of all matrices, matrix a, that is negative 1 -1, negative -3 -3, matrix b, which is 5011, matrix c, which is 0 -3 -9 -5 -1, and matrix d, which is negative -1, and matrix d, which is negative -10, negative -7 -17.

00:24 We are asked to determine whether this set is a linearly independent or dependent set.

00:29 If it is a linearly dependent set we are asked to express one matrix as a linear combination of the others.

00:35 So consider the linear combination c1 times negative 1 1 negative 3 3 plus c2 times 5 -11 plus c3 3 times matrix c which is 0 3 negative 5 1 plus c 4 times the matrix d which is negative 1 negative 717 is equal to the zero metric 0 -000 -0 and solve the corresponding system of equations if the system has only trivial solution then it is linearly independent otherwise it is dependent so from here we get the equations negative c1 plus 5 c2 minus c4 is equal to 0 c1 plus 3 c3 is equal to 0 negative 3 c1 plus c2 minus 5c3 minus 7 c4 is equal to 0 and 3c1 plus c2 plus c3 plus 17 c4 is equal to 0.

01:44 Now to solve this system we can consider the corresponding augmented metric that is negative 1 -50 -90 -10 -1 -3 -0 negative 3 -1 negative 5 negative 70 and 3 -1 -1 -17 -0 now we need to convert this into the corresponding row reduced form for that first perform r2 interchanges with r1 to get the matrix 1 -3 -0 negative 1 negative 1 5 0 0 negative 1 negative 3 1 negative 5 negative 7 0 and 31117 0 now make all the other elements below this pivot 1 0 in the first column to get the matrix 1 -3000 0 0 0 5 3 negative 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 14 negative 70, 01 negative 8, 17 .0.

03:09 Now interchange r2 and r3 to get the matrix, 1 -3 -3 -0 -3 -0 -3 -0 -0 -1 -3 -70 -0.

03:19 0 -1 -4 -negative -70, 0 -5 -3 -1 -8 -17 -0.

03:30 Now this is the pivot 1 in the second row.

03:34 Make all the entries below the pivot 1 in the pivot column 0 by row operations...

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