The Matrix Method and Cramer's Rule

Transcript

0:00Introduction

Hello everyone! Okay, something very simple first. If small a times x equals b, what is x? Easy, right? x equals b over small a, as long as small a is not zero. Now the puzzle. Three equations, three unknowns, x, y and z, all mixed together. Can we solve all three in one single line, just like that? Sounds impossible, right? Today you will see it is the same idea, one level up, with two methods: the matrix method, and Cramer's rule. Both come up often in JEE, so let's begin.

0:39One matrix equation for the whole system

So, here is a general system: three equations in three unknowns, x, y and z. The small numbers at the bottom right are subscripts, okay? Small a one, b one, c one are the coefficients of the first equation, and d one is its right side. Subscript two means the second equation, and three the third. Clear? Now look at these coefficients. They sit in neat rows and columns, right? So tell me, which tool from this unit can hold all of them together? Yes, exactly, a matrix! Put all the coefficients in matrix A, in the square brackets, one row for each equation. Capital X is the column of unknowns, and capital B is the column of right side numbers, d one, d two, d three. Now multiply: row one of matrix A times the column X gives small a one x, plus b one y, plus c one z, the left side of the first equation! So the whole system becomes one line, A X equals B, just like small a x equals b! Beautiful, right? Now, two words you must know. If the system has at least one solution, we call it consistent. If it has no solution at all, it is inconsistent. Got it? And if every right side number is zero, so that B is the zero matrix, written capital O, the system is called homogeneous. With me?

2:04The matrix method: X = A inverse times B

Alright, the matrix method. For small a times x equals b, with small a not zero, we divide: x equals b over small a. The slashed equals sign means not equal, and the double arrow means gives. Okay? So for A X equals B, can we just write capital X equals B over A? Can we divide by a matrix? No! There is no division for matrices. But we have the inverse. Our matrix A is square, three by three, and if its determinant is not zero, A inverse exists. The straight bars mean determinant, and the minus one at the top right means inverse. Clear? So, multiply both sides on the left by A inverse. The round brackets keep A X together. On the left, A inverse times A gives capital I, the identity matrix. So I X equals A inverse times B, and the identity matrix times X is just X itself. With me? Now, one student writes the answer as X equals B times A inverse. Same thing, right? No, no, careful! That is the trap. B is a column, three by one, and A inverse is three by three, so B times A inverse is not even defined! We multiplied on the left, so A inverse stays on the left. So, the result: when the determinant of A is not zero, there is exactly one solution, X equals A inverse times B, okay? Every solution must equal this one product. Got it?

3:38Cramer's rule

Now the second method, Cramer's rule: only determinants. Delta, the Greek capital letter that looks like a little triangle, is the determinant of A, right? Then delta one, delta two, delta three: replace the first, second or third column of A by the column B, okay? So for delta one, the first column, the x column, goes out, and d one, d two, d three come in. The other two columns stay as they were. Right? Now you tell me. For delta two, which column does B replace? And careful: a column, not a row. The second column, the y column. Correct! So here is the rule. When delta is not zero, x equals delta one over delta, y equals delta two over delta, and z equals delta three over delta. Delta always sits at the bottom. Flip it, and you fall into a trap! Clear? And the same rule works for two equations in two unknowns, with two by two determinants; the little cross is read as by. With me?

4:38Example 1: solve 3x + 4y = 11, 2x - y = 0 by Cramer's rule

Alright, Example one, an easy warm up. Solve three x plus four y equals eleven, and two x minus y equals zero, by Cramer's rule. Pause the video, and try it first. Done? Okay, let's check. First delta: three, four on top, two, minus one below. Three times minus one, minus four times two: minus three minus eight, that is minus eleven. Not zero, so Cramer's rule works, and there is exactly one solution. Okay? Now delta one. The right side numbers are eleven and zero. Where do they go? Into the first column, the x column. Yes, exactly! So delta one has eleven and four on top, zero and minus one below. Eleven times minus one, minus four times zero: minus eleven. Right? And delta two: now eleven and zero replace the second column, the y column. Three times zero, minus eleven times two: minus twenty two. Got it? So x equals minus eleven over minus eleven, which is one, and y equals minus twenty two over minus eleven, which is two. Check: three plus eight is eleven, and two minus two is zero. Perfect! One wrong sign in delta, and both answers flip. So guard your signs!

6:03Example 2: solve a 3x3 system with X = A inverse B

Now, remember our puzzle? Let's solve three equations in one line. Solve x plus two y plus three z equals one, y plus four z equals two, and five x plus six y equals three, using A inverse from chapter three. Pause the video, and try it. Done? Okay, first we need matrix A, from the coefficients. But look at the second equation. There is no x at all! So what do we write in that place? Zero, exactly! Never skip it; a missing term means a zero coefficient. So row one is one, two, three, and row two is zero, one, four. The third equation has no z, so row three is five, six, zero. And B is the column one, two, three. Okay? And from chapter three, here is A inverse, on screen. It exists, so the determinant of A is not zero, and the system has exactly one solution. Right? Now X equals A inverse times B, with A inverse on the left. Row one times the column: minus twenty four, plus thirty six, plus fifteen, that is twenty seven. Row two: twenty, minus thirty, minus twelve, that is minus twenty two. Row three: minus five, plus eight, plus three, that is six. Clear? Now a topper's habit: always check in one equation. First equation: twenty seven, plus two times minus twenty two, that is minus forty four, plus three times six, that is eighteen. Total: one. Correct! So x equals twenty seven, y equals minus twenty two, and z equals six. See? Three equations, solved in one line, X equals A inverse times B, just like x equals b over small a. Puzzle solved!

7:53Quick revision: AX = B, A inverse B and Cramer's rule

So, quick revision. One: write the system as A X equals B; a missing term goes in as zero. Two: when the determinant of A is not zero, there is exactly one solution, X equals A inverse times B, inverse on the left. Three: Cramer's rule, when delta is not zero, x equals delta one over delta, and so on, where B replaces one column. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!