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Let $$A = \begin{pmatrix} 1 & a+3 & 8 & 2 \\ 0 & 0 & 2 & 0 \\ 0 & a-2 & 5 & 0 \\ 0 & 10 & a & a+3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 7 \\ -2 \\ 0 \\ \end{pmatrix}$$

I'm a bit stuck on this question... I need to find values of a for which this matrix has unique solutions.

So far I found that:

$z=\dfrac{7}{2}$ $y= \dfrac{-39}{2(a-2)}$ $w= \dfrac{(390-(7a^2)+14a)}{ (2(a-2)(a+3))}$

Can someone help to solve it completely?

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A matrix equation $Ax = b$ has a unique solution iff the determinant of $A$ is non-zero. – Arthur Mar 8 '13 at 18:21
but how to find determinant when you don't know some of the values... – user65835 Mar 8 '13 at 18:25
Expand in minors along the first column, then along the last. – Robert Israel Mar 8 '13 at 18:25
@user65835 The determinant will be an expression containing $a$, and you will have to find the values of $a$ which makes the determinant non-zero. You calculate the determinant the same way you would for any matrix where you know all the entries. – Arthur Mar 8 '13 at 18:26
Solve the equation $2a^2 - 2a + 12 = 0$. Any $a$ that is not a solution is an $a$ such that the matrix has nonzero determinant. – Jim Mar 8 '13 at 18:28

$$\begin{align}\det A &= 1\cdot\det\left(\begin{array}{ccc}0 & 2 & 0\\a-2 & 3 & 0\\10 & a & a+3\end{array}\right)\quad (\text{expanded down first column})\\ &= 1\cdot(a+3)\cdot\det\left(\begin{array}{cc}0 & 2\\a-2 & 5\end{array}\right)\quad(\text{expanded down third column})\\ &= 1\cdot(a+3)\cdot\bigl(0-2(a-2)\bigr)\\ &= -2(a+3)(a-2).\end{align}$$

You'll have unique solutions if and only if $\det A\neq 0.$

In the comments below, you've come to see that as long as $a\neq 2$ and $a\neq -3$, then there will be a unique solution. Another way to see that is simply to look at the equations you've already generated: $$z=\frac72\tag{1}$$ $$y=\frac{-39}{2(a-2)}\tag{2}$$ $$w=\frac1{a+3}\left(\frac{195}{a-2}-\frac{7a}2\right)\tag{3}$$ Now $(2)$ doesn't even make sense when $a=2$, so in that case, we have no solutions at all. Likewise, $(3)$ makes no sense when $a=-3$, so we've no solutions in that case, either.

Otherwise, $(1)$ through $(3)$ make sense, and plugging the values thus determined into $$x=-(a+3)y-8z-2z\tag{4}$$ gives us the value of $x$ needed for the solution. Since $x,y,z,w$ are determined once we've plugged in our appropriate $a$, then the solution thus determined is unique.

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so for question like "for which values of a this matrix has unique solution? " Do I say a= -3 and a=2 ? – user65835 Mar 8 '13 at 18:34
@user65835: No, those are the values that make the determinant = 0, you want all other values. Is that clear? – Amzoti Mar 8 '13 at 18:41
so everything other than a=-3 and a=2 is a solution? or am I missing a point? – user65835 Mar 8 '13 at 18:42
@user65835: Now you have the point and you even calculated some of the values. Think of $a$ as a free variable that you can make anything you want, except for those "bad" values that make your determinant $0$. If you look at my comment, you have the general solution for your variables based on $a$, but make sure to avoid those values of bad $a's$ cause look what they do to the results. – Amzoti Mar 8 '13 at 18:46
Amzoti you lost me, so if a is NOT equal to -3 or 2 the matrix has a general solution positive? – user65835 Mar 8 '13 at 18:50

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