Inverse of matrix with unknown values

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Matrix B is invertible when its determinant is nonzero. The determinant calculation shows that B is singular at λ = -2 and λ = 3, meaning it has an inverse for all other values of λ. For part B of the problem, determining the number of solutions for the system Bx = 0 when λ = 2 involves understanding the nullspace, which may not have been covered yet. Multiplying both sides of the equation by B^(-1) can help in finding the solution. Overall, the key takeaway is that B is invertible for λ ≠ -2, 3.
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Homework Statement



Let B=<br /> \begin{pmatrix}<br /> λ-3 &amp; 12 &amp; -1 \\<br /> 0 &amp; λ+2 &amp; λ \\<br /> 0 &amp; 0 &amp; 5<br /> \end{pmatrix}<br />

For what values of λ does matrix B have an inverse?


Homework Equations





The Attempt at a Solution



I first tried to calculate det(B)=(λ-3)(5λ+10) +0 -0

to get x=-2 or 3.

I'm unsure if this calculation helps me, and what to do beyond this point.
 
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ryan1180 said:

Homework Statement



Let B=<br /> \begin{pmatrix}<br /> λ-3 &amp; 12 &amp; -1 \\<br /> 0 &amp; λ+2 &amp; λ \\<br /> 0 &amp; 0 &amp; 5<br /> \end{pmatrix}<br />

For what values of λ does matrix B have an inverse?


Homework Equations





The Attempt at a Solution



I first tried to calculate det(B)=(λ-3)(5λ+10) +0 -0

to get x=-2 or 3.

I'm unsure if this calculation helps me, and what to do beyond this point.

Well, if det(B)=0 then it doesn't have an inverse. If det(B) is nonzero, then it does. So?
 
ryan1180 said:

Homework Statement



Let B=<br /> \begin{pmatrix}<br /> λ-3 &amp; 12 &amp; -1 \\<br /> 0 &amp; λ+2 &amp; λ \\<br /> 0 &amp; 0 &amp; 5<br /> \end{pmatrix}<br />

For what values of λ does matrix B have an inverse?

Homework Equations


The Attempt at a Solution



I first tried to calculate det(B)=(λ-3)(5λ+10) +0 -0

to get x=-2 or 3.

I'm unsure if this calculation helps me, and what to do beyond this point.

Yes, this calculation helps you, because the matrix is invertible if and only if det(B) is nonzero. Your calculation tells you exactly what values of \lambda make det(B) zero.
 
Okay, so since the matrix is singular when λ=-2 or 3, my answer then be that the matrix has an inverse at λ≠-2,3, correct?
 
ryan1180 said:
Okay, so since the matrix is singular when λ=-2 or 3, my answer then be that the matrix has an inverse at λ≠-2,3, correct?

Correct.
 
Thanks! It looks like I just failed to make a simple connection between concepts.

There is also a part B that asks: How many solutions will the system Bx = 0 have if λ = 2?
I've tried to find an answer for this part, but haven't had much success. Looking ahead in my book seems to indicate that finding the "nullspace" will help to solve that. Since we haven't done that yet, is there any other way that I would be able to solve that part?
 
ryan1180 said:
Thanks! It looks like I just failed to make a simple connection between concepts.

There is also a part B that asks: How many solutions will the system Bx = 0 have if λ = 2?
I've tried to find an answer for this part, but haven't had much success. Looking ahead in my book seems to indicate that finding the "nullspace" will help to solve that. Since we haven't done that yet, is there any other way that I would be able to solve that part?

You've decided B is invertible if λ = 2, right? Multiply both sides of Bx=0 by B^(-1).
 
ryan1180 said:

Homework Statement



Let B=<br /> \begin{pmatrix}<br /> λ-3 &amp; 12 &amp; -1 \\<br /> 0 &amp; λ+2 &amp; λ \\<br /> 0 &amp; 0 &amp; 5<br /> \end{pmatrix}<br />

For what values of λ does matrix B have an inverse?


Homework Equations





The Attempt at a Solution



I first tried to calculate det(B)=(λ-3)(5λ+10) +0 -0

to get x=-2 or 3.

I'm unsure if this calculation helps me, and what to do beyond this point.

Question: if you were unsure whether the determinant calculation could help you, why did you do it?

RGV
 

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