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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

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  • $\begingroup$ 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] $\endgroup$
    – Carser
    Feb 5 '19 at 17:52
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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 $c$.

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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 $I_3$.

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  • $\begingroup$ This doesn’t seem to address the actual question being asked: “... how one can do row reduction with letters.” $\endgroup$
    – amd
    Feb 5 '19 at 21:11
  • $\begingroup$ 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. $\endgroup$
    – Carser
    Feb 5 '19 at 21:16
  • $\begingroup$ 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. $\endgroup$
    – amd
    Feb 5 '19 at 21:18
  • $\begingroup$ 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. $\endgroup$
    – Carser
    Feb 5 '19 at 21:25

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