# Assuming a; c; f are all non-zero, use row reduction to work out the general form for $A^{-1}$

Let $$A = \left[ \begin{array} { c c c } { a } & { 0 } & { 0 } \\ { b } & { c } & { 0 } \\ { d } & { e } & { f } \end{array} \right]$$ where $$a , b , c , d , e , f$$ are real numbers.is called a lower triangular matrix.)

Assuming $$a , c , f$$ are all non-zero, use row reduction to work out the general form for $$A ^ { - 1 }$$ . (The answer will be a matrix each of whose entries is a formula involving a, $$b , c , d , e$$ and $$f$$ or some subset of these variables.

Could I get a hint on how one can do row reduction with letters

• Your row operations will result in expressions with variables in them. For example, if you something like row_3 - row_1, you'll get a row with [(d-a), e, f] Feb 5 '19 at 17:52

The general technique is to use row reduction to convert $$A$$ to the identity matrix while tracking the results of the same operations on $$I$$. So, for example, start by multiplying the first row (of both $$A$$ and $$I$$) by $$1/a$$.

The next step will be to subtract $$b$$ times the (new) first row from the second row, and then you'll divide the (new) second row by $$c$$.

The inverse $$B=A^{-1}$$ is the matrix such that $$\begin{bmatrix} a & 0 & 0 \\ b & c & 0 \\ d & e & f \end{bmatrix} B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ or putting in some variables $$\begin{bmatrix} a & 0 & 0 \\ b & c & 0 \\ d & e & f \end{bmatrix} \begin{bmatrix} B_{1} & B_{2} & B_{3} \\ B_{4} & B_{5} & B_{6} \\ B_{7} & B_{8} & B_{9} \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ You could write out the 9 equations that result from this matrix multiplication and you'll start getting $$B$$ in terms of $$a,b,c,d,e,f$$. For example, $$aB_1+0B_4+0B_7=1$$ This shows that $$B_1=\frac{1}{a}$$. Then you can keep solving this system to uncover $$B$$.

As @amd pointed to in the comments and as Robert also showed, the above could be written as an augmented matrix $$\left[\begin{array}{rrr|rrr} a & 0 & 0 & 1 & 0 & 0 \\ b & c & 0 & 0 & 1 & 0 \\ d & e & f & 0 & 0 & 1 \end{array}\right]$$ on which you can perform row operations. For example, start by multiplying the first row by $$\frac{1}{a}$$ to get $$\left[\begin{array}{rrr|rrr} 1 & 0 & 0 & \frac{1}{a} & 0 & 0 \\ b & c & 0 & 0 & 1 & 0 \\ d & e & f & 0 & 0 & 1 \end{array}\right]$$ From here you might multiply row two by $$\frac{1}{c}$$ to get $$\left[\begin{array}{rrr|rrr} 1 & 0 & 0 & \frac{1}{a} & 0 & 0 \\ \frac{b}{c} & 1 & 0 & 0 & \frac{1}{c} & 0 \\ d & e & f & 0 & 0 & 1 \end{array}\right]$$ Subtract $$\frac{b}{c}$$ times the first row from the second to get $$\left[\begin{array}{rrr|rrr} 1 & 0 & 0 & \frac{1}{a} & 0 & 0 \\ 0 & 1 & 0 & -\frac{b}{ac} & \frac{1}{c} & 0 \\ d & e & f & 0 & 0 & 1 \end{array}\right]$$ and keep working toward reducing the left hand side to $$I_3$$.

• This doesn’t seem to address the actual question being asked: “... how one can do row reduction with letters.”
– amd
Feb 5 '19 at 21:11
• I think it does hint at it, since the solution of this system I suggested for B is implicitly row operations. Do you think that the question intended for the row operations to apply to $A$? Possibly it did. Feb 5 '19 at 21:16
• I’d guess that the row reduction is meant to be applied to $[A\mid I]$, which is a standard method for computing $A^{-1}$. The OP’s issue, however, appears to be that some of the entries of $A$ are variables instead of specific numbers.
– amd
Feb 5 '19 at 21:18
• Right, what I have above is equivalent to what you've written as an augmented matrix, and the algebra is equivalent to the row-reductions on that augmented matrix. I can add some more to it to make this more clear. Feb 5 '19 at 21:25