Find the values of p, q, r and s if \(\begin{bmatrix} p^2 -1 & \ 0 &-3

1.	Find the values of p, q, r and s if \(\begin{bmatrix} p^2 -1 & \ 0 &-31 - q^3 \\[0.3em] 7 &r + 1& 9\\[0.3em] -2 &8 & s - 1 \end{bmatrix}\)= \(\begin{bmatrix} 1& \ 0 &-4 \\[0.3em] 7 &3/2& 9\\[0.3em] -2 &8 & π \end{bmatrix}\)
Answer» When two matrices (of same order) are equal then their corresponding entries are equal. Here \(\begin{bmatrix} p^2 -1 & \ 0 &-31 - q^3 \\[0.3em] 7 &r + 1& 9\\[0.3em] -2 &8 & s - 1 \end{bmatrix}\)= \(\begin{bmatrix} 1& \ 0 &-4 \\[0.3em] 7 &3/2& 9\\[0.3em] -2 &8 & π \end{bmatrix}\) ⇒ p² – 1 = 1 ⇒ p² = 1 + 1 = 2 p = ± √2 -31 – q³ = -4 -q³ = -4 + 31 = 27 q³ = -27 = (-3)³ ⇒ q = -3 r + 1 = 3/2 ⇒ r = 3/2 – 1 = 3 - 2/2 = 1/2 s – 1 = π ⇒ s = – π + 1 (i.e.,) s = 1 – π So, p = ± √2, q = -3, r = 1/2 and s = 1 – π

Find the values of p, q, r and s if \(\begin{bmatrix} p^2 -1 & \ 0 &-31 - q^3 \\[0.3em] 7 &r + 1& 9\\[0.3em] -2 &8 & s - 1 \end{bmatrix}\)= \(\begin{bmatrix} 1& \ 0 &-4 \\[0.3em] 7 &3/2& 9\\[0.3em] -2 &8 & π \end{bmatrix}\)

Answer»

When two matrices (of same order) are equal then their corresponding entries are equal.

Here \(\begin{bmatrix} p^2 -1 & \ 0 &-31 - q^3 \\[0.3em] 7 &r + 1& 9\\[0.3em] -2 &8 & s - 1 \end{bmatrix}\)= \(\begin{bmatrix} 1& \ 0 &-4 \\[0.3em] 7 &3/2& 9\\[0.3em] -2 &8 & π \end{bmatrix}\)

⇒ p² – 1 = 1

⇒ p² = 1 + 1 = 2

p = ± √2

-31 – q³ = -4

-q³ = -4 + 31 = 27

q³ = -27 = (-3)³

⇒ q = -3

r + 1 = 3/2

⇒ r = 3/2 – 1 = 3 - 2/2 = 1/2

s – 1 = π

⇒ s = – π + 1 (i.e.,) s = 1 – π

So, p = ± √2, q = -3, r = 1/2 and s = 1 – π

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Find the values of p, q, r and s if \(\begin{bmatrix} p^2 -1 &amp; \ 0 &amp;-31 - q^3 \\[0.3em] 7 &amp;r + 1&amp; 9\\[0.3em] -2 &amp;8 &amp; s - 1 \end{bmatrix}\)= \(\begin{bmatrix} 1&amp; \ 0 &amp;-4 \\[0.3em] 7 &amp;3/2&amp; 9\\[0.3em] -2 &amp;8 &amp; π \end{bmatrix}\)

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Find the values of p, q, r and s if \(\begin{bmatrix} p^2 -1 & \ 0 &-31 - q^3 \\[0.3em] 7 &r + 1& 9\\[0.3em] -2 &8 & s - 1 \end{bmatrix}\)= \(\begin{bmatrix} 1& \ 0 &-4 \\[0.3em] 7 &3/2& 9\\[0.3em] -2 &8 & π \end{bmatrix}\)