solve this ---**Question Stem:**
What is the determinant of the matrix $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$?
**Matrix:**
$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$
**Options:**
a) 1
b) -1
c) 10
d) -10
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We need to find the determinant of a 3 by 3 matrix. The matrix A has elements 1, 2, 3 in the first row, 0, 1, 4 in the second row, and 5, 6, 0 in the third row. We have four options: 1, negative 1, 10, or negative 10.
To find the determinant of a 3 by 3 matrix, we use the standard formula. For a general matrix with elements a through i, the determinant equals a times the quantity e i minus f h, minus b times the quantity d i minus f g, plus c times the quantity d h minus e g. For our specific matrix, we identify a equals 1, b equals 2, c equals 3, d equals 0, e equals 1, f equals 4, g equals 5, h equals 6, and i equals 0.
Now let's substitute our values into the formula step by step. First, we calculate 1 times the quantity 1 times 0 minus 4 times 6, minus 2 times the quantity 0 times 0 minus 4 times 5, plus 3 times the quantity 0 times 6 minus 1 times 5. This simplifies to 1 times 0 minus 24, minus 2 times 0 minus 20, plus 3 times 0 minus 5. Which equals 1 times negative 24, minus 2 times negative 20, plus 3 times negative 5. This gives us negative 24 plus 40 minus 15, which equals 16 minus 15, and finally equals 1.
Let's verify our answer using an alternative method called cofactor expansion along the first column. We expand the determinant as 1 times the 2 by 2 determinant of elements 1, 4, 6, 0, minus 0 times another 2 by 2 determinant, plus 5 times the 2 by 2 determinant of elements 2, 3, 1, 4. This gives us 1 times the quantity 0 minus 24, plus 5 times the quantity 8 minus 3. Which equals negative 24 plus 25, confirming our answer of 1.
Both calculation methods confirm that the determinant of matrix A equals 1. Looking at our multiple choice options, we can see that option a, which is 1, matches our calculated result. Therefore, the correct answer is option a, 1. This demonstrates how we can use either the standard determinant formula or cofactor expansion to solve 3 by 3 matrix determinant problems.