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by **littlebu** on Tue Dec 07, 2010 10:03 pm

I got the four equations as

a - b = -5
c -d = 7
4b = -4
4d = 12

solving these I get A as

A = [-6 -1]
A = [10 3]

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by **stapel_eliz** on Wed Dec 08, 2010 11:52 am

littlebu wrote:I got the four equations as

a - b = -5
c -d = 7
4b = -4
4d = 12

Excellent! That's exactly right. Then:

. . . . .4b = -4, so b = -1
. . . . .a - (-1) = -5, so a = -6
. . . . .4d = 12, so d = 3
. . . . .c - (3) = 7, so c = 10

And then you can quickly check:

$AX\, =\, \left[\begin{array}{cc}-6&-1\\10&3\end{array}\right]\left[\begin{array}{cc}1&0\\-1&4\end{array}\right]\, =\, \left[\begin{array}{cc}-6(1)\, +\, (-1)(-1)&-6(0)\, +\, (-1)(4)\\10(1)\, +\, 3(-1)&10(0)\, +\, 3(4)\end{array}\right]$

. . . . . $=\, \left[\begin{array}{cc}-6\, +\, 1&0\, -\, 4\\10\, -\, 3&0\, +\, 12\end{array}\right]\, =\, \left[\begin{array}{cc}-5&-4\\7&12\end{array}\right]$

Since the matrices multiply to the desired result, this confirms that your answer is correct. Good work! :wink:

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