intersection. Hence, write the equation of a line passing through the point of intersection and perpendicular to both the lines. 35. (a) If P = $\begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}$ and Q = $\begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}$, find (QP) and hence solve the following system of equations using matrices: $x - y = 3$, $2x + 3y + 4z = 17$, $y + 2z = 7$ OR (b) Obtain the value of $\Delta = \begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix}$ in terms of x, y and z. Further, if $\Delta = 0$ and x, y, z are non-zero real numbers, prove that $x^{-1} + y^{-1} + z^{-1} = -1$. (matrix, determinant, system of equations, inverse matrix, linear equations)

Step 1: Define the Determinant

We are given a $3 \times 3$ determinant and asked to find its value in terms of $x$, $y$, and $z$. This is the first step in solving the problem.

Step 2: Apply Column Operations to Simplify

To simplify the calculation of the determinant, we apply column operations. Subtracting the third column from the first and second columns will introduce zeros, making the expansion easier.

Step 3: Evaluate the Determinant after Column Operations

After applying the column operations, the determinant transforms into this new form. Notice that the first two columns now contain some zeros, which will simplify the expansion process.

Step 4: Expand the Determinant along the First Row

We expand the determinant along the first row. This involves multiplying each element in the first row by its corresponding cofactor. The terms with zero elements simplify the expansion by eliminating those cofactors immediately.

Step 5: Calculate the 2x2 Determinants

Now we calculate the $2 \times 2$ determinants resulting from the expansion. This involves the cross-multiplication of elements in each minor matrix and then combining them with the original row elements.

Step 6: Simplify and Express in Terms of x, y, z

We simplify the expression by performing the multiplications and combining like terms. This gives us the value of the determinant in terms of $x$, $y$, and $z$.

Step 7: Consider the Condition $\Delta = 0$

The problem states that if $\Delta = 0$, we need to prove a certain relationship. We set our derived expression for $\Delta$ equal to zero and proceed to the next step.

Step 8: Divide by $xyz$

Since $x, y, z$ are non-zero real numbers, we can divide the entire equation by $xyz$. This operation is crucial for transforming the expression into the desired form.

Step 9: Simplify the Expression

After dividing each term by $xyz$, we simplify the fractions. This results in an expression containing the reciprocals of $x, y,$ and $z$, along with the constant 1.

Step 10: Rearrange to Prove the Identity

Finally, we rearrange the terms to isolate the sum of reciprocals on one side of the equation. This proves the required identity, demonstrating that $x^{-1} + y^{-1} + z^{-1} = -1$.

More from this chapter

• Evaluate the determinant 1 & 0 & 4 3 & 5 & -1 0 & 1 & 2 (i) with the help of second row. (ii) with the help of third column.
• If A = [[1, 2], [3, 4]] find det(A) and adj(A).
• The value of 1 & 2 & 3 4 & 5 & 6 7 & 8 & 9 is:
• Using Cramer's rule, show that the following system of linear equations is consistent and hence solve it: x + y + z = 1, 2x + 3y + 2z = 2, x + y + 2z = 4.
• Let a R and A be a matrix of order 3 3 such that det(A)=-4 and A+I= 1 & a & 1 2 & 1 & 0 a & 1 & 2, where I is the 3 3 identity. If det ((a+1)adj((a-1)A))=2^m3^n
• Let A be a 3 × 3 matrix such that |adj (adj(adj A))| = 81. If S = n Z: |adj(adj A)|^(((n-1)²)/(2)) = |A|³n² - 5n - 4, then Σ_n S | A^(n² + n) | is equal to:
• Find the equation of the line joining A(1,3) and B(0,0) using determinants and find k if D(k, 0) is a point such that area of triangle ABD is 3 sq units.
• If the system of linear equations 3x + y + β z = 3, 2x + α y - z = -3, x + 2y + z = 4 has infinitely many solutions, then the value of 22β - 9α is:

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