SOLVED: HALLENGE ACTIVITY 7.5.3: The Fundamental Theorem of Linear Algebra Jump to level Express the general solution of Ax = b where A = and 16 b = asx = Xr 4 Cxn where Xr is the row space solution a

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HALLENGE ACTIVITY
7.5.3: The Fundamental Theorem of Linear Algebra
Jump to level
Express the general solution of Ax = b where A =
and 16
b = asx = Xr 4 Cxn where Xr is the row space solution and Xn is a basis for null(A)
Ex:1.2 X =
Ex: 5 + c
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01:51

LENGE Ity9,4.3: The Fundamental Theorem of Linear Algebra_Jump to level 1Consider the system Ax = b where A = 58 ~12 24 and ‘b=[4] basis for the null space of A is given by {[E] andx = [27 Is a particular solution: Use this information to find the row space solution X to Ax = bEx42

05:05

Basis for solution space of Ax-00 0 0Find a basis for the solution space ofy 2 Wwhere A is the (4×4) matrix given below:2 3 0 2 5 2 2 4 7 -1 -2 0 3A =A=[1 0 ] [1 52 ] [2 1 ] [-1 22 3]

03:50

Exercise 5.45 If A is m X n and B is n X m, show that AB = 0 if and only if col B C null A.Exercise 5.4.13 Let A be an m X n matrix with columns C1, C2, Cn: If rank A = n, show that {AT c1, AT c2, AT cn} is a basis of Rn

06:11

Q4. Bases; Row, Column, and Null Space

For the following linear system:2×1 – x2 – 3×3 + 5×4 – txs = 0x1 – 2×2 + 3×4 – 2xs = 0xz – 3×3 – x4 – 3xs = 0xi + 8×2 – 9×3 + xs = 0

Represent this linear system in the form Ax = b.

(ii) Explain what the null space of the coefficient matrix A is in terms of the linear system.

(iii) Find a basis for the null space of A.

(iv) Find the rank and nullity of matrix A.

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You are watching: SOLVED: HALLENGE ACTIVITY 7.5.3: The Fundamental Theorem of Linear Algebra Jump to level Express the general solution of Ax = b where A = and 16 b = asx = Xr 4 Cxn where Xr is the row space solution a. Info created by THVinhTuy selection and synthesis along with other related topics.

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