Matrices

Do you mean that the determinant of that matrix is -51 and you need to find the value of a?
 
julianna5704 said:
How do I solve a Matrix such as -51=
-210
a24
-2-15
Click to expand...
Please ask the question using standard well-known mathematical notation.
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How does a 3x3 matrix equal a number? What does it mean? Do you want the trace to equal -51? If yes, then the answer is that it can't.
 
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To find the determinant ∣210a24−2−15∣\displaystyle \left|\begin{array}{ccc} {_2} & 1 & 0 \\ a & 2 & 4 \\ {-2} & -1 & 5 \end{array}\right|, seeing that "0" on the right of the first row, I would "expand" on either the first row or the third column.

Expanding on the first row,
∣−210a24−2−15∣=−2∣24−15∣−∣a4−25∣=−2(10+4)−(5a+8)=−28−5a−8=−36−5a=−51\displaystyle \left|\begin{array}{ccc} {-2} & 1 & 0 \\ a & 2 & 4 \\ {-2} & -1 & 5 \end{array}\right|= -2\left|\begin{array}{cc} 2 & 4 \\ -1 & 5\end{array}\right|- \left|\begin{array}{cc} a & 4 \\ -2 & 5\end{array}\right|= -2(10+ 4)- (5a+ 8)= -28- 5a- 8= -36- 5a= -51 so -5a= 36- 51= -15. a= 3.

Expanding on the third column,
∣−210a24−2−15∣=−4∣a2−2−1∣+5∣−21a2∣=−4(−a+4)+5(−4−a)=4a−16−20−5a=−36−5a=−51\displaystyle \left|\begin{array}{ccc} {-2} & 1 & 0 \\ a & 2 & 4 \\ {-2} & -1 & 5 \end{array}\right|= -4 \left|\begin{array}{cc}a & 2 \\ -2 & -1\end{array}\right|+ 5\left|\begin{array}{cc}-2 & 1 \\ a & 2 \end{array}\right|= -4(-a+ 4)+ 5(-4- a)= 4a- 16- 20- 5a= -36- 5a= -51 so, again, 5a= 51- 36= 15 and a= 3.

(Does anyone know why I am getting the "-" in "-2" moved like that?)
 
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