An algebra problem by Diana Ir

Algebra Level 2

Given that $a \begin{bmatrix}{2} && {6} \\ {1} && {4}\end{bmatrix} = \begin{bmatrix}{x} && {27} \\ {y} && {z}\end{bmatrix}$ for some real number $a$ , find $x+z$ .

A. $\frac43$
B. $\frac{27}2$
C. 26
D. 27
E. 48

E C D A B

1 solution

Diana Ir
Jan 22, 2016

the first thing you want to do is solve for A. to solve for A, you use the only known value which 27 how do you get 27 from your matrices? well A times 6 = 27. 6a=27 a=4.5 A is found, now solve for x and z.

A times 2 = x a = 4.5 4.5 * 2 = 9 A times 4 = z 4.5 * 4 = 18 18+9 = 27.

Another way to solve it is: So we know A = 27/6. Substitute A when solving for x and z. So we know that 2a = x. 2(27/6)=x. x =9. 4a=z. 4(27/6)=z z = 18. x+z = 27.

Sorry to say, but I don't think I'll agree with your solution. The question states that 'a' is a real number so it can't be a decimal. Hence the first step of your solution is itself, invalid. I also tried to solve it the same way but the restriction for 'a' being a real number held me back.

aalekh patel - 5 years, 4 months ago

Real numbers include all the rational numbers, such as the integer 5 and the fraction 4/3, all the irrational numbers, such as √2 (1.41421356…, the square root of 2, an irrational algebraic number) and all transcendental numbers, such as π (3.14159265…, a transcendental number).

Diana Ir - 5 years, 4 months ago

Oh I am so sorry I mistakenly confused the property of an integer with that of a real number. Apologies for claiming your answer as incorrect!

aalekh patel - 5 years, 4 months ago

@Aalekh Patel – No worries! It is a math community, so a great place to discuss answers, right :)

Diana Ir - 5 years, 4 months ago

An algebra problem by Diana Ir

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