Let k, h be unknown constants and consider the linear...

Let $k$, $h$ be unknown constants and consider the linear system: $x - 3y - 5z = 3$ $y + z = 3$ $-x + 5y + hz = k$ 1. This system has infinitely many solutions whenever h select and k select 2. This system has no solution whenever h select and k select

Transcript

00:01 Let k and h be a non -constant and consider the linear system negative 4x plus 4y minus 6z equals 5 negative 7x minus 3y minus 9z equals negative 6 and negative 2x minus 18y plus h times z equals k so this system has a unique solution whenever h is different from and we want to find the value of a for which a is different from that value give us a unique solution and we know the unique solution depends on the coefficient matrix not on the right hand side so the value of k is not relevant here so what is going to decide the unique solution of this linear system is the matrix a defined as coefficients of the variable for each equation that is negative 4, 4 and negative 6 the order in which we consider the coefficients are x y and z for each equation so we get negative 4 for negative 6 negative 7 negative 3 negative 9 and negative 2 negative 18 and h here if we find that matrix and vector b to be equal to 5 negative 6 and k that is those values right here and the unknown vector x bar equal x y and z this then this linear system given above is the same as a times x bar equal b if you do this matrix vector operation and then put the vector v you get just equality of components of vectors means this linear system here this is the matrix vector form of writing the linear system this is the coefficient matrix the coefficient matrix this is the unknown vector each of its components are unknown variables and this is the right hand side vector or what is the same the vector of of independent terms of the linear equations and so we know that this linear system has a unique solution that's equivalent that it's equivalent to the matrix a is invariable and that's equivalent also to the determinant of a being different from zero so that's the condition we are going to find that is when this determinant of the matrix a is different from zero that will be the case when the linear system has a unique solution which is the one that we want to characterize good so all we got to do is to find the determinant of a and as you can see the h is going to be constrained by that condition but big value of k can be any value doesn't matter the value of k for the existence of a unique solution for the linear system so let's see what's the determinant of matrix a put that that matrix again here negative 4 4 6 negative 7 negative 3 negative 9 and negative 2 negative 18 h and we can develop that determinant by any of the cross columns let's choose the first row here so we get negative 4 times the determinant of the sub matrix obtained by removing the first column and row which is where this coefficient here is located and we get negative 3 negative 9 negative 18 h then we have this coefficient 4 but with a negative sign because the position or location of that coefficient 4 here is first row second column or 1 2 and those indices 1 2 add up to 3 which is an odd number in that case we're going to change the sign of the coefficient for the factor here…

Question

Let $k$, $h$ be unknown constants and consider the linear system: $x - 3y - 5z = 3$ $y + z = 3$ $-x + 5y + hz = k$ 1. This system has infinitely many solutions whenever h select and k select 2. This system has no solution whenever h select and k select

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